Revision notes, worked problems, a question bank, assertion and reason questions and answers to the NCERT exercises. Tap a group to open it; opening one closes the others.
If d/dx F(x) = f(x), then F is an antiderivative of f and we write
| Function | Integral |
|---|---|
| xⁿ (n ≠ −1) | xⁿ⁺¹/(n + 1) + C |
| 1/x | log|x| + C |
| eˣ | eˣ + C |
| aˣ | aˣ/log a + C |
| sin x, cos x | −cos x + C, sin x + C |
| sec²x, cosec²x | tan x + C, −cot x + C |
| sec x tan x, cosec x cot x | sec x + C, −cosec x + C |
| 1/√(1 − x²) | sin⁻¹x + C |
| 1/(1 + x²) | tan⁻¹x + C |
| tan x, cot x | log|sec x| + C, log|sin x| + C |
| sec x, cosec x | log|sec x + tan x| + C, log|cosec x − cot x| + C |
Always keep +C for an indefinite integral and drop it for a definite one.
Put t = g(x) so that dt = g′(x) dx. Then
| Integrand | Rewrite using |
|---|---|
| sin²x, cos²x | (1 − cos 2x)/2, (1 + cos 2x)/2 |
| sin³x, cos³x | split off one factor: sin³x = sin x (1 − cos²x) |
| sin mx cos nx | product to sum: ½[sin(m + n)x + sin(m − n)x] |
| tan²x, cot²x | sec²x − 1, cosec²x − 1 |
| Integral | Result |
|---|---|
| ∫ dx/(x² − a²) | (1/2a) log|(x − a)/(x + a)| + C |
| ∫ dx/(a² − x²) | (1/2a) log|(a + x)/(a − x)| + C |
| ∫ dx/(x² + a²) | (1/a) tan⁻¹(x/a) + C |
| ∫ dx/√(x² − a²) | log|x + √(x² − a²)| + C |
| ∫ dx/√(a² − x²) | sin⁻¹(x/a) + C |
| ∫ dx/√(x² + a²) | log|x + √(x² + a²)| + C |
For px + q over ax² + bx + c, complete the square, or write px + q = A·(derivative of the quadratic) + B and split.
For a proper rational function P(x)/Q(x) split the denominator into factors, then match:
| Factor in the denominator | Partial fraction |
|---|---|
| (x − a) | A/(x − a) |
| (x − a)² | A/(x − a) + B/(x − a)² |
| (x − a)(x − b) | A/(x − a) + B/(x − b) |
| x² + bx + c (irreducible) | (Ax + B)/(x² + bx + c) |
From the product rule:
| Integral | Result |
|---|---|
| ∫ √(a² − x²) dx | (x/2)√(a² − x²) + (a²/2) sin⁻¹(x/a) + C |
| ∫ √(x² − a²) dx | (x/2)√(x² − a²) − (a²/2) log|x + √(x² − a²)| + C |
| ∫ √(x² + a²) dx | (x/2)√(x² + a²) + (a²/2) log|x + √(x² + a²)| + C |
If the quantity under the root is ax² + bx + c, complete the square first.
The definite integral is the limit of sums of thin strips, and it equals the area under the curve when f ≥ 0:
Example: ∫₀¹ x/(x² + 1) dx: t = x² + 1, limits 1 to 2, giving ½ ∫₁² dt/t = ½ log 2.
| Property | |
|---|---|
| P₀ | ∫ₐᵇ f(x) dx = ∫ₐᵇ f(t) dt |
| P₁ | ∫ₐᵇ f dx = −∫ᵇₐ f dx, and ∫ₐᵃ f dx = 0 |
| P₂ | ∫ₐᵇ f dx = ∫ₐᶜ f dx + ∫ᶜᵇ f dx (split at c) |
| P₃ | ∫ₐᵇ f(x) dx = ∫ₐᵇ f(a + b − x) dx |
| P₄ | ∫₀ᵃ f(x) dx = ∫₀ᵃ f(a − x) dx |
| P₅ | ∫₀²ᵃ f dx = ∫₀ᵃ f dx + ∫₀ᵃ f(2a − x) dx |
| P₆ | ∫₀²ᵃ f dx = 2∫₀ᵃ f dx if f(2a − x) = f(x), and 0 if f(2a − x) = −f(x) |
| P₇ | ∫₋ₐᵃ f dx = 2∫₀ᵃ f dx if f is even, and 0 if f is odd |
| Situation | Try |
|---|---|
| f′(x)/f(x) | log|f(x)| + C |
| Function and its derivative present | substitution t = g(x) |
| Rational function | divide, then partial fractions |
| Product of unlike functions | by parts, I L A T E |
| eˣ times (f + f′) | eˣ f(x) + C |
| √(quadratic) | complete the square, then a √ standard form |
| ∫ₐᵇ with a sum of two similar parts | P₃ or P₄, then add |
Try each problem first, then open the solution to check your working.
Find ∫ (4x³ − 3x² + 2x − 5) dx.
Use: ∫xⁿ dx = xⁿ⁺¹/(n + 1)
Integrate term by term: 4x⁴/4 − 3x³/3 + 2x²/2 − 5x.
Answer: x⁴ − x³ + x² − 5x + C
Find ∫ ((2x + 3)⁴) dx.
Use: ∫ f(ax + b) dx = F(ax + b)/a
Put t = 2x + 3, dt = 2 dx. Then ½∫t⁴ dt = t⁵/10.
Answer: (2x + 3)⁵/10 + C
Find ∫ x²/(x³ + 5) dx.
Use: ∫ f′/f dx = log|f|
The derivative of x³ + 5 is 3x², so the numerator is ⅓ of it.
∫ = ⅓ log|x³ + 5|.
Answer: log(|x³ + 5|)/3 + C
Find ∫ sin⁴x cos x dx.
Put t = sin x, dt = cos x dx. ∫t⁴ dt = t⁵/5.
Answer: sin⁵x/5 + C
Find ∫ 1/(9 + 4x²) dx.
Use: ∫dx/(a² + u²) = (1/a) tan⁻¹(u/a)
9 + 4x² = 4(x² + 9/4) = 4(x² + (3/2)²). So ∫ = (1/4)·(2/3) tan⁻¹(2x/3).
Equivalently put t = 2x, giving ½∫dt/(9 + t²) = (1/6) tan⁻¹(t/3).
Answer: tan⁻¹(2x/3)/6 + C
Find ∫ cos²x dx.
Use: cos²x = (1 + cos 2x)/2
∫ = ½∫(1 + cos 2x) dx = x/2 + sin 2x/4.
Answer: x/2 + sin(2x)/4 + C
Find ∫ (2x + 1)/((x − 1)(x + 3)) dx.
Use: A/(x − 1) + B/(x + 3)
2x + 1 = A(x + 3) + B(x − 1). x = 1 gives 3 = 4A, so A = 3/4. x = −3 gives −5 = −4B, so B = 5/4.
∫ = (3/4) log|x − 1| + (5/4) log|x + 3|.
Answer: 3 log(|x − 1|)/4 + 5 log(|x + 3|)/4 + C
Find ∫ 1/((x + 1)(x² + 1)) dx.
Use: A/(x + 1) + (Bx + C)/(x² + 1)
1 = A(x² + 1) + (Bx + C)(x + 1). x = −1 gives A = 1/2. Comparing x²: A + B = 0, so B = −1/2. Constants: A + C = 1, so C = 1/2.
∫ = ½ log|x + 1| − ¼ log(x² + 1) + ½ tan⁻¹x.
Answer: log(|x + 1|)/2 − log(x² + 1)/4 + tan⁻¹ x/2 + C
Find ∫ x cos x dx.
Use: u = x, dv = cos x dx
∫ = x sin x − ∫sin x dx = x sin x + cos x.
Answer: x sin x + cos x + C
Find ∫ x e^(3x) dx.
Use: u = x, dv = e³ˣ dx
∫ = x e³ˣ/3 − ∫e³ˣ/3 dx = x e³ˣ/3 − e³ˣ/9 = e³ˣ(3x − 1)/9.
Answer: e^(3x)(3x − 1)/9 + C
Find ∫ (eˣ(cos x − sin x)) dx.
Use: ∫eˣ[f + f′] dx = eˣ f
With f = cos x, f′ = −sin x, so the integrand is eˣ[f + f′].
∫ = eˣ cos x.
Answer: eˣ cos x + C
Find ∫ (x² e^(−x)) dx.
Use: by parts twice
∫x²e⁻ˣ dx = −x²e⁻ˣ + 2∫x e⁻ˣ dx. And ∫x e⁻ˣ dx = −x e⁻ˣ − e⁻ˣ.
So ∫ = −e⁻ˣ(x² + 2x + 2).
Answer: −e^(−x)(x² + 2x + 2) + C
Find ∫ (√(16 − x²)) dx.
Use: ∫√(a² − x²) dx with a = 4
∫ = (x/2)√(16 − x²) + (16/2) sin⁻¹(x/4).
Answer: x √(16 − x²)/2 + 8 sin⁻¹(x/4) + C
Evaluate ∫ from 1 to 4 of (2x + 1) dx.
Use: F(b) − F(a)
F(x) = x² + x. F(4) = 20, F(1) = 2.
Answer: 18
Evaluate ∫ from 0 to π/3 of sec x tan x dx.
F(x) = sec x. F(π/3) − F(0) = 2 − 1.
Answer: 1
Evaluate ∫ from 0 to 1 of x√(1 + x²) dx.
Use: t = 1 + x²
dt = 2x dx. Limits: x = 0 gives t = 1; x = 1 gives t = 2.
½∫₁² √t dt = ½ · (2/3)[t^(3/2)]₁² = (1/3)(2√2 − 1).
Answer: (2 √2 − 1)/3
Evaluate ∫ from 1 to e of (log x)/x dx.
Use: t = log x
dt = dx/x. Limits: 1 → 0 and e → 1.
∫₀¹ t dt = 1/2.
Answer: 1/2
Evaluate ∫ from −2 to 2 of (x⁵ + x²) dx.
Use: P₇
x⁵ is odd, so its integral vanishes. x² is even, so ∫₋₂² x² dx = 2∫₀² x² dx = 2 · 8/3 = 16/3.
Answer: 16/3
Evaluate ∫ from 0 to 3 of √x/(√x + √(3 − x)) dx.
Use: P₃ (or P₄)
By P₄, I = ∫₀³ √(3 − x)/(√(3 − x) + √x) dx.
Add: 2I = ∫₀³ dx = 3, so I = 3/2.
Answer: 3/2
Evaluate ∫ from 0 to 2 of |x² − 1| dx.
Use: split at x = 1 (P₂)
x² − 1 ≤ 0 on [0, 1] and ≥ 0 on [1, 2].
∫₀¹ (1 − x²) dx = 1 − 1/3 = 2/3. ∫₁² (x² − 1) dx = (8/3 − 2) − (1/3 − 1) = 2/3 + 2/3 = 4/3.
Sum = 2.
Answer: 2
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d/dx(cos 2x) = −2 sin 2x, so sin 2x = d/dx(−½ cos 2x).
Answer: −cos(2x)/2 + C
d/dx(sin 3x) = 3 cos 3x, so cos 3x = d/dx(⅓ sin 3x).
Answer: sin(3x)/3 + C
d/dx(e²ˣ) = 2e²ˣ, so e²ˣ = d/dx(½ e²ˣ).
Answer: e^(2x)/2 + C
d/dx(ax + b)³ = 3a(ax + b)², so divide by 3a.
Answer: (ax + b)³/(3a) + C
Use the antiderivative of each term: −½ cos 2x for sin 2x and (1/3)e³ˣ for e³ˣ, so −4e³ˣ gives −(4/3)e³ˣ.
Answer: −cos(2x)/2 − 4 e^(3x)/3 + C
∫e³ˣ dx = e³ˣ/3 and ∫1 dx = x.
Answer: 4 e^(3x)/3 + x + C
Simplify first: x²(1 − 1/x²) = x² − 1.
Answer: x³/3 − x + C
Integrate term by term using ∫xⁿ dx = xⁿ⁺¹/(n + 1).
Answer: ax³/3 + bx²/2 + cx + C
∫2x² dx = 2x³/3 and ∫eˣ dx = eˣ.
Answer: 2x³/3 + eˣ + C
Expand: (√x − 1/√x)² = x − 2 + 1/x.
Then ∫x dx = x²/2, ∫2 dx = 2x, ∫(1/x) dx = log|x|.
Answer: x²/2 − 2x + log |x| + C
Divide term by term: x + 5 − 4x⁻².
∫x dx = x²/2, ∫5 dx = 5x, ∫−4x⁻² dx = 4/x.
Answer: x²/2 + 5x + 4/x + C
Divide: x^(5/2) + 3x^(1/2) + 4x^(−1/2).
Raise each power by 1 and divide by the new power: (2/7)x^(7/2) + 2x^(3/2) + 8x^(1/2).
Answer: 2x^(7/2)/7 + 2x^(3/2) + 8 √x + C
Factor the numerator: x³ − x² + x − 1 = x²(x − 1) + (x − 1) = (x − 1)(x² + 1).
So the integrand is x² + 1 for x ≠ 1.
Answer: x³/3 + x + C
Expand: x^(1/2) − x^(3/2).
Integrate each power.
Answer: 2x^(3/2)/3 − 2x^(5/2)/5 + C
Expand: 3x^(5/2) + 2x^(3/2) + 3x^(1/2).
Integrate each power.
Answer: 6x^(7/2)/7 + 4x^(5/2)/5 + 2x^(3/2) + C
∫2x dx = x², ∫cos x dx = sin x, ∫eˣ dx = eˣ.
Answer: x² − 3 sin x + eˣ + C
∫sin x dx = −cos x, so −3 sin x gives +3 cos x.
∫5x^(1/2) dx = 5 · (2/3)x^(3/2).
Answer: 2x³/3 + 3 cos x + 10x^(3/2)/3 + C
Expand: sec²x + sec x tan x.
∫sec²x dx = tan x and ∫sec x tan x dx = sec x.
Answer: tan x + sec x + C
sec²x/cosec²x = tan²x = sec²x − 1.
∫sec²x dx = tan x, ∫1 dx = x.
Answer: tan x − x + C
Split: 2 sec²x − 3 (sin x/cos²x) = 2 sec²x − 3 sec x tan x.
∫sec²x dx = tan x, ∫sec x tan x dx = sec x.
Answer: 2 tan x − 3 sec x + C
√x + 1/√x = x^(1/2) + x^(−1/2). Integrating gives x^(3/2)/(3/2) + x^(1/2)/(1/2) = (2/3)x^(3/2) + 2x^(1/2).
Answer: (C) (2/3)x^(3/2) + 2x^(1/2) + C
f(x) = ∫(4x³ − 3x⁻⁴) dx = x⁴ + x⁻³ + C.
f(2) = 16 + 1/8 + C = 0, so C = −129/8.
Answer: (A) x⁴ + 1/x³ − 129/8
Put t = 1 + x², so dt = 2x dx.
The integral becomes ∫dt/t = log|t| = log(1 + x²) since 1 + x² > 0.
Answer: log(1 + x²) + C
Put t = log x, so dt = dx/x.
∫t² dt = t³/3.
Answer: (log x)³/3 + C
x + x log x = x(1 + log x). Put t = 1 + log x, dt = dx/x.
∫dt/t = log|t|.
Answer: log(|1 + log x|) + C
Put t = cos x, so dt = −sin x dx.
−∫sin t dt = cos t = cos(cos x).
Answer: cos(cos x) + C
sin θ cos θ = ½ sin 2θ, with θ = ax + b.
∫½ sin(2ax + 2b) dx = −cos(2ax + 2b)/(4a).
Answer: −cos(2ax + 2b)/(4a) + C
Put t = ax + b, dt = a dx.
(1/a)∫t^(1/2) dt = (2/(3a)) t^(3/2).
Answer: 2(ax + b)^(3/2)/(3a) + C
Put t = x + 2, so x = t − 2 and dx = dt.
∫(t − 2)t^(1/2) dt = ∫(t^(3/2) − 2t^(1/2)) dt = (2/5)t^(5/2) − (4/3)t^(3/2).
Answer: 2(x + 2)^(5/2)/5 − 4(x + 2)^(3/2)/3 + C
Put t = 1 + 2x², dt = 4x dx.
(1/4)∫t^(1/2) dt = (1/6)t^(3/2).
Answer: (1 + 2x²)^(3/2)/6 + C
4x + 2 = 2(2x + 1), the derivative of x² + x + 1 is 2x + 1.
Put t = x² + x + 1, dt = (2x + 1) dx: 2∫t^(1/2) dt = (4/3)t^(3/2).
Answer: 4(x² + x + 1)^(3/2)/3 + C
Put x = t², dx = 2t dt (t > 0).
∫2t dt/(t² − t) = 2∫dt/(t − 1) = 2 log|t − 1|.
Answer: 2 log(|√x − 1|) + C
Put t = x + 4, x = t − 4. ∫(t − 4)t^(−1/2) dt = (2/3)t^(3/2) − 8t^(1/2).
Answer: 2(x + 4)^(3/2)/3 − 8 √(x + 4) + C
Write x⁵ = x³ · x². Put t = x³ − 1, dt = 3x² dx, and x³ = t + 1.
(1/3)∫t^(1/3)(t + 1) dt = (1/3)[(3/7)t^(7/3) + (3/4)t^(4/3)].
Answer: (x³ − 1)^(7/3)/7 + (x³ − 1)^(4/3)/4 + C
Put t = 2 + 3x³, dt = 9x² dx.
(1/9)∫t⁻³ dt = −1/(18t²).
Answer: −1/(18(2 + 3x³)²) + C
Put t = log x, dt = dx/x; ∫t⁻ᵐ dt = t^(1−m)/(1 − m) for m ≠ 1.
Answer: (log x)^(1 − m)/(1 − m) + C
Put t = 9 − 4x², dt = −8x dx.
−(1/8)∫dt/t = −(1/8) log|t|.
Answer: −log(|9 − 4x²|)/8 + C
Put t = 2x + 3, dt = 2 dx.
Answer: e^(2x + 3)/2 + C
x/eˣ² = x e^(−x²). Put t = −x², dt = −2x dx.
−½∫eᵗ dt = −½ e^(−x²).
Answer: −e^(−x²)/2 + C
Put t = tan⁻¹ x, dt = dx/(1 + x²). ∫eᵗ dt = eᵗ.
Answer: e^(tan⁻¹ x) + C
Divide top and bottom by eˣ: (eˣ − e⁻ˣ)/(eˣ + e⁻ˣ). Put t = eˣ + e⁻ˣ, dt = (eˣ − e⁻ˣ) dx.
∫dt/t = log(eˣ + e⁻ˣ).
Since log(eˣ + e⁻ˣ) = log(e²ˣ + 1) − x, the same answer can be written log(e²ˣ + 1) − x.
Answer: log(e^(2x) + e^(0)) − x + C
Put t = e²ˣ + e⁻²ˣ, dt = 2(e²ˣ − e⁻²ˣ) dx.
½∫dt/t = ½ log t.
Answer: log(e^(2x) + e^(−2x))/2 + C
tan²θ = sec²θ − 1, with θ = 2x − 3.
∫sec²(2x − 3) dx = ½ tan(2x − 3) and ∫1 dx = x.
Answer: tan(2x − 3)/2 − x + C
∫sec²(7 − 4x) dx: put t = 7 − 4x, dt = −4 dx, giving −¼ tan t.
Answer: −tan(7 − 4x)/4 + C
Put t = sin⁻¹x, dt = dx/√(1 − x²). ∫t dt = t²/2.
Answer: (sin⁻¹ x)²/2 + C
The denominator is 2(3 cos x + 2 sin x), and the derivative of 3 cos x + 2 sin x is −3 sin x + 2 cos x, which is the numerator.
So the integral is ½∫dt/t with t = 3 cos x + 2 sin x: ½ log|3 cos x + 2 sin x|.
Answer: log(|3 cos x + 2 sin x|)/2 + C
Put t = 1 − tan x, dt = −sec²x dx.
−∫dt/t² = 1/t.
Answer: 1/(1 − tan x) + C
Put t = √x, dt = dx/(2√x), so dx/√x = 2 dt.
2∫cos t dt = 2 sin t.
Answer: 2 sin(√x) + C
Put t = sin 2x, dt = 2 cos 2x dx.
½∫t^(1/2) dt = (1/3)t^(3/2).
Answer: (sin(2x))^(3/2)/3 + C
Put t = 1 + sin x, dt = cos x dx. ∫t^(−1/2) dt = 2t^(1/2).
Answer: 2 √(1 + sin x) + C
Put t = log sin x, dt = cot x dx. ∫t dt = t²/2.
Answer: (log(sin x))²/2 + C
Put t = 1 + cos x, dt = −sin x dx. −∫dt/t = −log t (t > 0).
Answer: −log(1 + cos x) + C
Put t = 1 + cos x, dt = −sin x dx. −∫t⁻² dt = 1/t.
Answer: 1/(1 + cos x) + C
1/(1 + cot x) = sin x/(sin x + cos x).
Write sin x = ½[(sin x + cos x) − (cos x − sin x)], so the integrand is ½ − ½·(cos x − sin x)/(sin x + cos x).
Since (sin x + cos x)′ = cos x − sin x, the second part gives −½ log|sin x + cos x|.
Answer: x/2 − log(|sin x + cos x|)/2 + C
1/(1 − tan x) = cos x/(cos x − sin x).
Write cos x = ½[(cos x − sin x) + (cos x + sin x)], so the integrand is ½ + ½·(cos x + sin x)/(cos x − sin x).
Since (cos x − sin x)′ = −(sin x + cos x), the second part gives −½ log|cos x − sin x|.
Answer: x/2 − log(|cos x − sin x|)/2 + C
sin x cos x = tan x cos²x, so the integrand is sec²x/(√(tan x) · tan x)·tan x... simplify: 1/(sin x cos x) = sec²x/tan x.
Put t = tan x, dt = sec²x dx: ∫t^(−1/2) dt = 2√t.
Answer: 2 √(tan x) + C
Put t = 1 + log x, dt = dx/x. ∫t² dt = t³/3.
Answer: (1 + log x)³/3 + C
(x + 1)/x = 1 + 1/x, which is the derivative of x + log x.
Put t = x + log x: ∫t² dt = t³/3.
Answer: (x + log x)³/3 + C
Put t = tan⁻¹(x⁴), dt = 4x³ dx/(1 + x⁸).
¼∫sin t dt = −¼ cos t.
Answer: −cos(tan⁻¹(x⁴))/4 + C
The numerator is the derivative of x¹⁰ + 10ˣ.
So the integral is log(x¹⁰ + 10ˣ) + C.
Answer: (D) log(10ˣ + x¹⁰) + C
1/(sin²x cos²x) = (sin²x + cos²x)/(sin²x cos²x) = sec²x + cosec²x.
So the integral is tan x − cot x + C.
Answer: (B) tan x − cot x + C
sin²θ = (1 − cos 2θ)/2, with θ = 2x + 5.
∫½ dx = x/2 and ∫½ cos(4x + 10) dx = sin(4x + 10)/8, so subtract.
Answer: x/2 − sin(4x + 10)/8 + C
2 sin A cos B = sin(A + B) + sin(A − B): sin 3x cos 4x = ½[sin 7x + sin(−x)] = ½ sin 7x − ½ sin x.
∫ = −cos 7x/14 + cos x/2.
Answer: cos x/2 − cos(7x)/14 + C
cos 2x cos 4x = ½(cos 6x + cos 2x). Multiply by cos 6x: ½(cos²6x + cos 2x cos 6x) = ¼(1 + cos 12x) + ¼(cos 8x + cos 4x).
Integrate each term.
Answer: sin(12x)/48 + sin(8x)/32 + sin(4x)/16 + x/4 + C
sin³θ = (3 sin θ − sin 3θ)/4, with θ = 2x + 1.
∫ = ¼[−(3/2) cos θ + (1/6) cos 3θ].
Answer: −3 cos(2x + 1)/8 + cos(6x + 3)/24 + C
sin³x cos³x = sin³x(1 − sin²x) cos x. Put t = sin x, dt = cos x dx.
∫(t³ − t⁵) dt = t⁴/4 − t⁶/6.
Answer: sin⁴x/4 − sin⁶x/6 + C
sin x sin 3x = ½(cos 2x − cos 4x). Multiply by sin 2x: ½ sin 2x cos 2x − ½ sin 2x cos 4x.
½ sin 2x cos 2x = ¼ sin 4x and ½ sin 2x cos 4x = ¼(sin 6x − sin 2x).
So the product is ¼(sin 2x + sin 4x − sin 6x).
Integrate: ¼[−cos 2x/2 − cos 4x/4 + cos 6x/6].
Answer: −cos(2x)/8 − cos(4x)/16 + cos(6x)/24 + C
sin A sin B = ½[cos(A − B) − cos(A + B)]: sin 4x sin 8x = ½(cos 4x − cos 12x).
∫ = ½(sin 4x/4 − sin 12x/12).
Answer: sin(4x)/8 − sin(12x)/24 + C
(1 − cos x)/(1 + cos x) = tan²(x/2) = sec²(x/2) − 1.
∫sec²(x/2) dx = 2 tan(x/2).
Answer: 2 tan(x/2) − x + C
cos x/(1 + cos x) = 1 − 1/(1 + cos x) = 1 − ½ sec²(x/2).
∫ = x − tan(x/2).
Answer: x − tan(x/2) + C
sin⁴x = ((1 − cos 2x)/2)² = ¼(1 − 2 cos 2x + cos²2x) = ⅜ − ½ cos 2x + ⅛ cos 4x.
Integrate term by term.
Answer: 3x/8 − sin(2x)/4 + sin(4x)/32 + C
cos⁴θ = ⅜ + ½ cos 2θ + ⅛ cos 4θ, with θ = 2x.
∫ = 3x/8 + ½·sin 4x/4 + ⅛·sin 8x/8.
Answer: 3x/8 + sin(4x)/8 + sin(8x)/64 + C
sin²x = (1 − cos x)(1 + cos x), so the integrand is 1 − cos x.
∫ = x − sin x.
Answer: x − sin x + C
cos 2x − cos 2α = 2(cos²x − cos²α) = 2(cos x − cos α)(cos x + cos α).
The integrand is 2(cos x + cos α), so ∫ = 2 sin x + 2x cos α.
Answer: 2 sin x + 2x cos α + C
1 + sin 2x = (sin x + cos x)². Put t = sin x + cos x, dt = (cos x − sin x) dx.
∫dt/t² = −1/t.
Answer: −1/(cos x + sin x) + C
tan³θ sec θ = (sec²θ − 1) sec θ tan θ. Put t = sec 2x, dt = 2 sec 2x tan 2x dx.
½∫(t² − 1) dt = t³/6 − t/2.
Answer: sec³(2x)/6 − sec(2x)/2 + C
tan⁴x = tan²x(sec²x − 1) = tan²x sec²x − (sec²x − 1).
∫tan²x sec²x dx = tan³x/3, ∫sec²x dx = tan x, ∫1 dx = x.
Answer: tan³x/3 − tan x + x + C
Split: sin³x/(sin²x cos²x) + cos³x/(sin²x cos²x) = sin x/cos²x + cos x/sin²x = sec x tan x + cosec x cot x.
∫sec x tan x dx = sec x, ∫cosec x cot x dx = −cosec x.
Answer: sec x − cosec x + C
cos 2x + 2 sin²x = (1 − 2 sin²x) + 2 sin²x = 1.
So the integrand is sec²x and ∫ = tan x.
Answer: tan x + C
1/(sin x cos³x) = sec⁴x/tan x = (1 + tan²x) sec²x/tan x. Put t = tan x, dt = sec²x dx.
∫(1 + t²)/t dt = log|t| + t²/2.
Answer: log(|tan x|) + tan²x/2 + C
cos 2x = (cos x − sin x)(cos x + sin x), so the integrand is (cos x − sin x)/(cos x + sin x).
Put t = cos x + sin x, dt = (cos x − sin x) dx: ∫dt/t = log|t|.
Answer: log(|cos x + sin x|) + C
For 0 ≤ x ≤ π, sin⁻¹(cos x) = π/2 − x.
∫(π/2 − x) dx = πx/2 − x²/2.
Answer: πx/2 − x²/2 + C
Write 1 = sin[(x − a) − (x − b)]/sin(b − a). Then 1/(cos(x − a) cos(x − b)) = [tan(x − a) − tan(x − b)]/sin(b − a).
∫tan(x − a) dx = −log|cos(x − a)|, and similarly for b.
Answer: (log(|cos(x − b)|) − log(|cos(x − a)|))/sin(b − a) + C
The integrand is 1/cos²x − 1/sin²x = sec²x − cosec²x.
∫ = tan x + cot x + C.
Answer: (A) tan x + cot x + C
Put t = x eˣ, dt = eˣ(1 + x) dx.
∫sec²t dt = tan t = tan(x eˣ).
Answer: (B) tan(x eˣ) + C
Put t = x³, dt = 3x² dx. ∫dt/(t² + 1) = tan⁻¹ t.
Answer: tan⁻¹(x³) + C
Put t = 2x, dt = 2 dx. ½∫dt/√(t² + 1) = ½ log|t + √(t² + 1)|.
Answer: log(|2x + √(1 + 4x²)|)/2 + C
Put t = x − 2, dt = dx. Since (2 − x)² = (x − 2)², ∫dt/√(t² + 1) = log|t + √(t² + 1)|.
Answer: log(|x − 2 + √((2 − x)² + 1)|) + C
Put t = 5x, dt = 5 dx. (1/5)∫dt/√(9 − t²) = (1/5) sin⁻¹(t/3).
Answer: sin⁻¹(5x/3)/5 + C
Put t = x², dt = 2x dx. (3/2)∫dt/(1 + 2t²) = (3/2)·(1/√2) tan⁻¹(√2 t).
Answer: 3 tan⁻¹(√2x²)/(2 √2) + C
Put t = x³, dt = 3x² dx. (1/3)∫dt/(1 − t²) = (1/3)·½ log|(1 + t)/(1 − t)|.
Answer: log(|(1 + x³)/(1 − x³)|)/6 + C
Split: x/√(x² − 1) − 1/√(x² − 1).
∫x/√(x² − 1) dx = √(x² − 1) and ∫dx/√(x² − 1) = log|x + √(x² − 1)|.
Answer: √(x² − 1) − log(|x + √(x² − 1)|) + C
Put t = x³, dt = 3x² dx. (1/3)∫dt/√(t² + (a³)²) = (1/3) log|t + √(t² + a⁶)|.
Answer: log(|x³ + √(x⁶ + a⁶)|)/3 + C
Put t = tan x, dt = sec²x dx. ∫dt/√(t² + 4) = log|t + √(t² + 4)|.
Answer: log(|tan x + √(tan²x + 4)|) + C
x² + 2x + 2 = (x + 1)² + 1. Put t = x + 1: ∫dt/√(t² + 1).
Answer: log(|x + 1 + √(x² + 2x + 2)|) + C
9x² + 6x + 5 = (3x + 1)² + 4. Put t = 3x + 1, dt = 3 dx.
(1/3)∫dt/(t² + 4) = (1/3)(1/2) tan⁻¹(t/2).
Answer: tan⁻¹((3x + 1)/2)/6 + C
7 − 6x − x² = 16 − (x + 3)². Put t = x + 3: ∫dt/√(16 − t²) = sin⁻¹(t/4).
Answer: sin⁻¹((x + 3)/4) + C
(x − 1)(x − 2) = x² − 3x + 2 = (x − 3/2)² − 1/4.
Put t = x − 3/2: ∫dt/√(t² − (1/2)²) = log|t + √(t² − 1/4)|.
Answer: log(|x − 3/2 + √(x² − 3x + 2)|) + C
8 + 3x − x² = 41/4 − (x − 3/2)². Put t = x − 3/2: ∫dt/√((√41/2)² − t²) = sin⁻¹(2t/√41) = sin⁻¹((2x − 3)/√41).
Answer: sin⁻¹((2x − 3)/√41) + C
(x − a)(x − b) = (x − (a + b)/2)² − ((a − b)/2)².
Put t = x − (a + b)/2: ∫dt/√(t² − k²) = log|t + √(t² − k²)|, with k = (a − b)/2.
Answer: log(|x − (a + b)/2 + √((x − a)(x − b))|) + C
The numerator 4x + 1 is the derivative of 2x² + x − 3.
∫dt/√t = 2√t.
Answer: 2 √(2x² + x − 3) + C
Split: x/√(x² − 1) + 2/√(x² − 1).
∫x/√(x² − 1) dx = √(x² − 1) and ∫dx/√(x² − 1) = log|x + √(x² − 1)|.
Answer: √(x² − 1) + 2 log(|x + √(x² − 1)|) + C
5x − 2 = (5/6)(6x + 2) − 11/3, where 6x + 2 is the derivative of 3x² + 2x + 1.
∫(6x + 2)/(3x² + 2x + 1) dx = log(3x² + 2x + 1).
3x² + 2x + 1 = 3[(x + 1/3)² + 2/9], so ∫dx/(3x² + 2x + 1) = (1/3)·(3/√2) tan⁻¹((3x + 1)/√2) = (1/√2) tan⁻¹((3x + 1)/√2).
So the second part is −(11/3)(1/√2) tan⁻¹((3x + 1)/√2).
Answer: 5 log(3x² + 2x + 1)/6 − 11 tan⁻¹((3x + 1)/√2)/(3 √2) + C
(x − 5)(x − 4) = x² − 9x + 20. Write 6x + 7 = 3(2x − 9) + 34.
∫3(2x − 9)/√(x² − 9x + 20) dx = 6√(x² − 9x + 20).
x² − 9x + 20 = (x − 9/2)² − 1/4, so ∫34 dx/√(…) = 34 log|x − 9/2 + √(x² − 9x + 20)|.
Answer: 6 √(x² − 9x + 20) + 34 log(|x − 9/2 + √(x² − 9x + 20)|) + C
Write x + 2 = −½(4 − 2x) + 4, where 4 − 2x is the derivative of 4x − x².
−½∫(4 − 2x)/√(4x − x²) dx = −√(4x − x²).
4x − x² = 4 − (x − 2)², so ∫4 dx/√(4 − (x − 2)²) = 4 sin⁻¹((x − 2)/2).
Answer: −√(4x − x²) + 4 sin⁻¹((x − 2)/2) + C
x + 2 = ½(2x + 2) + 1. The first part gives √(x² + 2x + 3).
x² + 2x + 3 = (x + 1)² + 2, so ∫dx/√(…) = log|x + 1 + √(x² + 2x + 3)|.
Answer: √(x² + 2x + 3) + log(|x + 1 + √(x² + 2x + 3)|) + C
x + 3 = ½(2x − 2) + 4. The first part gives ½ log|x² − 2x − 5|.
x² − 2x − 5 = (x − 1)² − 6, so ∫4 dx/((x − 1)² − 6) = 4 · (1/(2√6)) log|(x − 1 − √6)/(x − 1 + √6)|.
Answer: log(|x² − 2x − 5|)/2 + (2/√6) log(|(x − 1 − √6)/(x − 1 + √6)|) + C
5x + 3 = (5/2)(2x + 4) − 7.
(5/2)∫(2x + 4)/√(x² + 4x + 10) dx = 5√(x² + 4x + 10).
x² + 4x + 10 = (x + 2)² + 6, so ∫−7 dx/√(…) = −7 log|x + 2 + √(x² + 4x + 10)|.
Answer: 5 √(x² + 4x + 10) − 7 log(|x + 2 + √(x² + 4x + 10)|) + C
x² + 2x + 2 = (x + 1)² + 1. Put t = x + 1: ∫dt/(t² + 1) = tan⁻¹ t.
Answer: (B) tan⁻¹(x + 1) + C
9x − 4x² = 4[81/64 − (x − 9/8)²] = 4[(9/8)² − (x − 9/8)²].
∫dx/(2√((9/8)² − (x − 9/8)²)) = ½ sin⁻¹((x − 9/8)/(9/8)) = ½ sin⁻¹((8x − 9)/9).
Answer: (B) (1/2) sin⁻¹((8x − 9)/9) + C
x/((x + 1)(x + 2)) = A/(x + 1) + B/(x + 2), so x = A(x + 2) + B(x + 1).
x = −1 gives A = −1; x = −2 gives B = 2.
∫ = −log|x + 1| + 2 log|x + 2|.
Answer: 2 log(|x + 2|) − log(|x + 1|) + C
Use ∫dx/(x² − a²) = (1/2a) log|(x − a)/(x + a)| with a = 3.
Answer: log(|(x − 3)/(x + 3)|)/6 + C
Write A/(x − 1) + B/(x − 2) + C/(x − 3). Cover-up: A = (3 − 1)/((−1)(−2)) = 1, B = (6 − 1)/((1)(−1)) = −5, C = (9 − 1)/((2)(1)) = 4.
Answer: log(|x − 1|) − 5 log(|x − 2|) + 4 log(|x − 3|) + C
Cover-up: A = 1/((−1)(−2)) = 1/2, B = 2/((1)(−1)) = −2, C = 3/((2)(1)) = 3/2.
Answer: log(|x − 1|)/2 − 2 log(|x − 2|) + 3 log(|x − 3|)/2 + C
x² + 3x + 2 = (x + 1)(x + 2). Then 2x/((x + 1)(x + 2)) = A/(x + 1) + B/(x + 2): A = −2/1 = −2, B = −4/(−1) = 4.
Answer: 4 log(|x + 2|) − 2 log(|x + 1|) + C
The fraction is improper: (1 − x²)/(x − 2x²) = ½ + (1 − x/2)/(x(1 − 2x)).
(1 − x/2)/(x(1 − 2x)) = A/x + B/(1 − 2x), with A = 1 and B = (1 − 1/4)/(1/2) = 3/2.
∫ = x/2 + log|x| + (3/2)·(−½) log|1 − 2x|.
Answer: x/2 + log |x| − 3 log(|1 − 2x|)/4 + C
x/((x − 1)(x² + 1)) = A/(x − 1) + (Bx + C)/(x² + 1).
Compare coefficients of x² + 1 terms: A = 1/2, B = −1/2, C = 1/2.
∫ = ½ log|x − 1| − ¼ log(x² + 1) + ½ tan⁻¹ x.
Answer: log(|x − 1|)/2 − log(x² + 1)/4 + tan⁻¹ x/2 + C
A/(x − 1) + B/(x − 1)² + C/(x + 2). At x = 1: B = 1/3. At x = −2: C = −2/9. Comparing x² terms: A + C = 0, so A = 2/9.
Answer: 2 log(|x − 1|)/9 − 1/(3(x − 1)) − 2 log(|x + 2|)/9 + C
x³ − x² − x + 1 = (x − 1)²(x + 1).
A/(x − 1) + B/(x − 1)² + C/(x + 1): B = 8/2 = 4, C = 2/4 = 1/2, and A + C = 0 so A = −1/2.
Answer: −log(|x − 1|)/2 − 4/(x − 1) + log(|x + 1|)/2 + C
(x − 1)(x + 1)(2x + 3). Cover-up: A = (−1)/((2)(5)) = −1/10, B = (−5)/((−2)(1)) = 5/2, and for 1/(2x + 3): C = (−6)/((9/4 − 1)) = −24/5.
∫C/(2x + 3) dx = (C/2) log|2x + 3| = −(12/5) log|2x + 3|.
Answer: −log(|x − 1|)/10 + 5 log(|x + 1|)/2 − 12 log(|2x + 3|)/5 + C
x² − 4 = (x − 2)(x + 2). Cover-up: A(x + 1) = 5(−1)/((−3)(1)) = 5/3; B(x − 2) = 10/((3)(4)) = 5/6; C(x + 2) = −10/((−1)(−4)) = −5/2.
Answer: 5 log(|x + 1|)/3 + 5 log(|x − 2|)/6 − 5 log(|x + 2|)/2 + C
Divide: (x³ + x + 1)/(x² − 1) = x + (2x + 1)/(x² − 1).
(2x + 1)/((x − 1)(x + 1)) = (3/2)/(x − 1) + (1/2)/(x + 1).
Answer: x²/2 + 3 log(|x − 1|)/2 + log(|x + 1|)/2 + C
2/((1 − x)(1 + x²)) = A/(1 − x) + (Bx + C)/(1 + x²). At x = 1: A = 1; comparing x² terms A − B = 0 so B = 1; constants A + C = 2 so C = 1.
∫ = −log|1 − x| + ½ log(1 + x²) + tan⁻¹ x.
Answer: −log(|1 − x|) + log(1 + x²)/2 + tan⁻¹ x + C
3x − 1 = 3(x + 2) − 7, so the integrand is 3/(x + 2) − 7/(x + 2)².
∫ = 3 log|x + 2| + 7/(x + 2).
Answer: 3 log(|x + 2|) + 7/(x + 2) + C
1/(x⁴ − 1) = ½[1/(x² − 1) − 1/(x² + 1)].
∫1/(x² − 1) dx = ½ log|(x − 1)/(x + 1)| and ∫1/(x² + 1) dx = tan⁻¹ x.
Answer: log(|(x − 1)/(x + 1)|)/4 − tan⁻¹ x/2 + C
Multiply top and bottom by xⁿ⁻¹: xⁿ⁻¹/(xⁿ(xⁿ + 1)). Put t = xⁿ, dt = n xⁿ⁻¹ dx.
(1/n)∫dt/(t(t + 1)) = (1/n) log(t/(t + 1)).
Answer: log(xⁿ/(xⁿ + 1))/n + C
Put t = sin x, dt = cos x dx. 1/((1 − t)(2 − t)) = 1/(1 − t) − 1/(2 − t).
∫ = −log|1 − t| + log|2 − t|.
Answer: log(|(2 − sin x)/(1 − sin x)|) + C
The fraction equals 1 + (−4x² − 10)/((x² + 3)(x² + 4)). With y = x²: (−4y − 10)/((y + 3)(y + 4)) = 2/(y + 3) − 6/(y + 4).
∫dx/(x² + 3) = (1/√3) tan⁻¹(x/√3), ∫dx/(x² + 4) = ½ tan⁻¹(x/2).
Answer: x + 2 tan⁻¹(x/√3)/√3 − 3 tan⁻¹(x/2) + C
Put y = x², dy = 2x dx. 1/((y + 1)(y + 3)) = ½[1/(y + 1) − 1/(y + 3)].
∫ = ½ log((y + 1)/(y + 3)).
Answer: log((x² + 1)/(x² + 3))/2 + C
Multiply top and bottom by x³: x³/(x⁴(x⁴ − 1)). Put t = x⁴, dt = 4x³ dx.
¼∫dt/(t(t − 1)) = ¼ log|(t − 1)/t|.
Answer: log(|(x⁴ − 1)/x⁴|)/4 + C
Put t = eˣ, dt = eˣ dx, dx = dt/t. ∫dt/(t(t − 1)) = log|(t − 1)/t| = log|eˣ − 1| − x.
Answer: log(|eˣ − 1|) − x + C
x/((x − 1)(x − 2)) = −1/(x − 1) + 2/(x − 2) (cover-up: A = 1/(−1), B = 2/1).
∫ = −log|x − 1| + 2 log|x − 2| = log|(x − 2)²/(x − 1)|.
Answer: (B) log|(x − 2)²/(x − 1)| + C
1/(x(x² + 1)) = 1/x − x/(x² + 1).
∫ = log|x| − ½ log(x² + 1).
Answer: (A) log|x| − ½ log(x² + 1) + C
Take u = x, dv = sin x dx. ∫ = −x cos x + ∫cos x dx.
Answer: −x cos x + sin x + C
u = x, dv = sin 3x dx, v = −cos 3x/3. ∫ = −x cos 3x/3 + ⅓∫cos 3x dx.
Answer: −x cos(3x)/3 + sin(3x)/9 + C
u = x², dv = eˣ dx: ∫ = x²eˣ − 2∫x eˣ dx. Apply parts again: ∫x eˣ dx = x eˣ − eˣ.
So ∫ = x²eˣ − 2x eˣ + 2eˣ.
Answer: eˣ(x² − 2x + 2) + C
u = log x, dv = x dx, v = x²/2. ∫ = (x²/2) log x − ∫(x/2) dx.
Answer: x² log x/2 − x²/4 + C
u = log 2x, dv = x dx. ∫ = (x²/2) log 2x − ∫(x²/2)(1/x) dx = (x²/2) log 2x − x²/4.
Answer: x² log(2x)/2 − x²/4 + C
u = log x, dv = x² dx. ∫ = (x³/3) log x − ⅓∫x² dx.
Answer: x³ log x/3 − x³/9 + C
u = sin⁻¹x, dv = x dx. ∫ = (x²/2) sin⁻¹x − ½∫x²/√(1 − x²) dx.
∫x²/√(1 − x²) dx = ½ sin⁻¹x − ½ x√(1 − x²) (put x = sin θ).
So ∫ = (x²/2) sin⁻¹x − ¼ sin⁻¹x + ¼ x√(1 − x²).
Answer: (2x² − 1) sin⁻¹ x/4 + x √(1 − x²)/4 + C
u = tan⁻¹x, dv = x dx. ∫ = (x²/2) tan⁻¹x − ½∫x²/(1 + x²) dx = (x²/2) tan⁻¹x − ½(x − tan⁻¹x).
Answer: (x² + 1) tan⁻¹ x/2 − x/2 + C
u = cos⁻¹x, dv = x dx. ∫ = (x²/2) cos⁻¹x + ½∫x²/√(1 − x²) dx.
∫x²/√(1 − x²) dx = ½ sin⁻¹x − ½ x√(1 − x²), and sin⁻¹x = π/2 − cos⁻¹x (the constant π/2 is absorbed into C).
So ∫ = (x²/2) cos⁻¹x − ¼ cos⁻¹x − ¼ x√(1 − x²).
Answer: (2x² − 1) cos⁻¹ x/4 − x √(1 − x²)/4 + C
u = (sin⁻¹x)², dv = dx. ∫ = x(sin⁻¹x)² − ∫2x sin⁻¹x/√(1 − x²) dx.
∫x sin⁻¹x/√(1 − x²) dx = −√(1 − x²) sin⁻¹x + x (by parts again).
So ∫ = x(sin⁻¹x)² + 2√(1 − x²) sin⁻¹x − 2x.
Answer: x(sin⁻¹ x)² + 2 √(1 − x²) sin⁻¹ x − 2x + C
u = cos⁻¹x, dv = x/√(1 − x²) dx, v = −√(1 − x²).
∫ = −√(1 − x²) cos⁻¹x − ∫(−√(1 − x²))(−1/√(1 − x²)) dx = −√(1 − x²) cos⁻¹x − x.
Answer: −x − √(1 − x²) cos⁻¹ x + C
u = x, dv = sec²x dx. ∫ = x tan x − ∫tan x dx = x tan x + log|cos x|.
Answer: x tan x + log(|cos x|) + C
u = tan⁻¹x, dv = dx. ∫ = x tan⁻¹x − ∫x/(1 + x²) dx = x tan⁻¹x − ½ log(1 + x²).
Answer: x tan⁻¹ x − log(1 + x²)/2 + C
u = (log x)², dv = x dx: ∫ = (x²/2)(log x)² − ∫x log x dx.
∫x log x dx = (x²/2) log x − x²/4, so ∫ = (x²/2)(log x)² − (x²/2) log x + x²/4.
Answer: x²(log x)²/2 − x² log x/2 + x²/4 + C
u = log x, dv = (x² + 1) dx, v = x³/3 + x. ∫ = (x³/3 + x) log x − ∫(x²/3 + 1) dx.
Answer: (x³/3 + x) log x − x³/9 − x + C
Of the form eˣ[f(x) + f′(x)] with f = sin x. So ∫ = eˣ sin x.
Answer: eˣ sin x + C
x/(1 + x)² = 1/(1 + x) − 1/(1 + x)². With f = 1/(1 + x), f′ = −1/(1 + x)², this is eˣ[f + f′].
So ∫ = eˣ/(1 + x).
Answer: eˣ/(1 + x) + C
(1 + sin x)/(1 + cos x) = ½ sec²(x/2) + tan(x/2) (since 1 + sin x = 1 + 2 sin(x/2) cos(x/2) and 1 + cos x = 2 cos²(x/2)).
With f = tan(x/2), f′ = ½ sec²(x/2), this is eˣ[f + f′], so ∫ = eˣ tan(x/2).
Answer: eˣ tan(x/2) + C
With f = 1/x, f′ = −1/x², this is eˣ[f + f′]. So ∫ = eˣ/x.
Answer: eˣ/x + C
(x − 3)/(x − 1)³ = 1/(x − 1)² − 2/(x − 1)³. With f = 1/(x − 1)², f′ = −2/(x − 1)³.
So ∫ = eˣ/(x − 1)².
Answer: eˣ/(x − 1)² + C
Let I = ∫e²ˣ sin x dx. By parts: I = −e²ˣ cos x + 2∫e²ˣ cos x dx.
Again: ∫e²ˣ cos x dx = e²ˣ sin x − 2I. So I = −e²ˣ cos x + 2e²ˣ sin x − 4I.
5I = e²ˣ(2 sin x − cos x), so I = e²ˣ(2 sin x − cos x)/5.
Answer: e^(2x)(2 sin x − cos x)/5 + C
Put x = tan θ: 2x/(1 + x²) = sin 2θ, so sin⁻¹(2x/(1 + x²)) = 2θ = 2 tan⁻¹x (for |x| ≤ 1).
∫2 tan⁻¹x dx = 2[x tan⁻¹x − ½ log(1 + x²)].
Answer: 2x tan⁻¹ x − log(1 + x²) + C
Put t = x³, dt = 3x² dx. ⅓∫eᵗ dt = ⅓ e^(x³).
Answer: (A) (1/3)e^(x³) + C
With f = sec x, f′ = sec x tan x, the integrand is eˣ[f + f′].
So ∫ = eˣ sec x.
Answer: (B) eˣ sec x + C
Use ∫√(a² − x²) dx = (x/2)√(a² − x²) + (a²/2) sin⁻¹(x/a) with a = 2.
Answer: x √(4 − x²)/2 + 2 sin⁻¹(x/2) + C
Put t = 2x, dx = dt/2. ½∫√(1 − t²) dt = ½[(t/2)√(1 − t²) + ½ sin⁻¹t].
Answer: x √(1 − 4x²)/2 + sin⁻¹(2x)/4 + C
x² + 4x + 6 = (x + 2)² + 2. Put t = x + 2 and use ∫√(t² + a²) dt = (t/2)√(t² + a²) + (a²/2) log|t + √(t² + a²)| with a² = 2.
Answer: (x + 2) √(x² + 4x + 6)/2 + log(|x + 2 + √(x² + 4x + 6)|) + C
x² + 4x + 1 = (x + 2)² − 3. Put t = x + 2 and use ∫√(t² − a²) dt = (t/2)√(t² − a²) − (a²/2) log|t + √(t² − a²)| with a² = 3.
Answer: (x + 2) √(x² + 4x + 1)/2 − 3 log(|x + 2 + √(x² + 4x + 1)|)/2 + C
1 − 4x − x² = 5 − (x + 2)². Put t = x + 2 and use ∫√(a² − t²) dt with a² = 5.
Answer: (x + 2) √(1 − 4x − x²)/2 + 5 sin⁻¹((x + 2)/√5)/2 + C
x² + 4x − 5 = (x + 2)² − 9. Put t = x + 2 and use ∫√(t² − a²) dt with a = 3.
Answer: (x + 2) √(x² + 4x − 5)/2 − 9 log(|x + 2 + √(x² + 4x − 5)|)/2 + C
1 + 3x − x² = 13/4 − (x − 3/2)². Put t = x − 3/2 and use ∫√(a² − t²) dt with a² = 13/4:
(t/2)√(…) + (13/8) sin⁻¹(t/a), and t/2 = (2x − 3)/4, t/a = (2x − 3)/√13.
Answer: (2x − 3) √(1 + 3x − x²)/4 + 13 sin⁻¹((2x − 3)/√13)/8 + C
x² + 3x = (x + 3/2)² − 9/4. Put t = x + 3/2 and use ∫√(t² − a²) dt with a² = 9/4: (t/2)√(…) − (a²/2) log|t + √(…)|.
Answer: (2x + 3) √(x² + 3x)/4 − 9 log(|x + 3/2 + √(x² + 3x)|)/8 + C
√(1 + x²/9) = (1/3)√(x² + 9).
⅓∫√(x² + 9) dx = ⅓[(x/2)√(x² + 9) + (9/2) log|x + √(x² + 9)|].
Answer: x √(x² + 9)/6 + 3 log(|x + √(x² + 9)|)/2 + C
Use ∫√(x² + a²) dx = (x/2)√(x² + a²) + (a²/2) log|x + √(x² + a²)| with a = 1.
Answer: (A) (x/2)√(1 + x²) + ½ log|x + √(1 + x²)| + C
x² − 8x + 7 = (x − 4)² − 9. Put t = x − 4 and use ∫√(t² − a²) dt = (t/2)√(t² − a²) − (a²/2) log|t + √(t² − a²)| with a = 3.
The log coefficient is 9/2 with a minus sign.
Answer: (D) ½(x − 4)√(x² − 8x + 7) − (9/2) log|x − 4 + √(x² − 8x + 7)| + C
F(x) = x²/2 + x. F(1) − F(−1) = 3/2 − (1/2 − 1) = 3/2 + 1/2 = 2.
Answer: 2
F(x) = log x. F(3) − F(2) = log 3 − log 2 = log(3/2).
Answer: log(3/2)
F(x) = x⁴ − (5/3)x³ + 3x² + 9x.
F(2) = 16 − 40/3 + 12 + 18 = 46 − 40/3 = 98/3.
F(1) = 1 − 5/3 + 3 + 9 = 13 − 5/3 = 34/3. Difference = 64/3.
Answer: 64/3
F(x) = −½ cos 2x. F(π/4) − F(0) = 0 − (−½) = 1/2.
Answer: 1/2
F(x) = ½ sin 2x. F(π/2) − F(0) = 0 − 0 = 0.
Answer: 0
F(x) = eˣ. F(5) − F(4) = e⁵ − e⁴.
Answer: e^(5) − e^(4)
F(x) = log|sec x|. F(π/4) = log √2 = ½ log 2 and F(0) = 0.
Answer: log 2/2
F(x) = log|cosec x − cot x|.
F(π/4) = log(√2 − 1), F(π/6) = log(2 − √3). Difference = log[(√2 − 1)/(2 − √3)].
Answer: log(√2 − 1) − log(2 − √3)
F(x) = sin⁻¹x. F(1) − F(0) = π/2 − 0.
Answer: π/2
F(x) = tan⁻¹x. F(1) − F(0) = π/4.
Answer: π/4
F(x) = ½ log|(x − 1)/(x + 1)|. F(3) = ½ log(1/2), F(2) = ½ log(1/3).
Difference = ½[log(1/2) − log(1/3)] = ½ log(3/2).
Answer: log(3/2)/2
cos²x = (1 + cos 2x)/2, F(x) = x/2 + sin 2x/4. F(π/2) − F(0) = π/4.
Answer: π/4
F(x) = ½ log(x² + 1). F(3) − F(2) = ½(log 10 − log 5) = ½ log 2.
Answer: log 2/2
Split: 2x/(5x² + 1) + 3/(5x² + 1).
∫2x/(5x² + 1) dx = (1/5) log(5x² + 1) and ∫3/(5x² + 1) dx = (3/√5) tan⁻¹(√5 x).
Evaluate from 0 to 1: (1/5) log 6 + (3/√5) tan⁻¹√5.
Answer: log 6/5 + 3 tan⁻¹(√5)/√5
F(x) = ½ e^(x²). F(1) − F(0) = (e − 1)/2.
Answer: (e^(1) − 1)/2
5x²/((x + 1)(x + 3)) = 5 + (−20x − 15)/((x + 1)(x + 3)) = 5 + (5/2)/(x + 1) − (45/2)/(x + 3).
F(x) = 5x + (5/2) log|x + 1| − (45/2) log|x + 3|.
F(2) − F(1) = 5 + (5/2) log(3/2) − (45/2) log(5/4).
Answer: 5 + 5 log(3/2)/2 − 45 log(5/4)/2
F(x) = 2 tan x + x⁴/4 + 2x. F(π/4) = 2 + π⁴/1024 + π/2 and F(0) = 0.
Answer: 2 + π⁴/1024 + π/2
sin²(x/2) − cos²(x/2) = −cos x.
F(x) = −sin x, and F(π) − F(0) = 0.
Answer: 0
Split: 6x/(x² + 4) + 3/(x² + 4).
F(x) = 3 log(x² + 4) + (3/2) tan⁻¹(x/2). F(2) − F(0) = 3 log 8 − 3 log 4 + (3/2)(π/4) = 3 log 2 + 3π/8.
Answer: 3 log 2 + 3π/8
∫x eˣ dx = x eˣ − eˣ, from 0 to 1: (e − e) − (0 − 1) = 1.
∫sin(πx/4) dx = −(4/π) cos(πx/4), from 0 to 1: (4/π)(1 − cos(π/4)) = (4/π)(1 − √2/2).
Sum = 1 + (4/π)(1 − √2/2).
Answer: 1 + 4(1 − √2/2)/π
F(x) = tan⁻¹x. F(√3) − F(1) = π/3 − π/4 = π/12.
Answer: (D) π/12
4 + 9x² = 9(x² + 4/9). F(x) = (1/6) tan⁻¹(3x/2).
F(2/3) − F(0) = (1/6) tan⁻¹ 1 = π/24.
Answer: (C) π/24
Put t = x² + 1, dt = 2x dx. Limits: x = 0 gives t = 1; x = 1 gives t = 2.
½∫₁² dt/t = ½ log 2.
Answer: log 2/2
Put t = sin φ, dt = cos φ dφ, and cos⁴φ = (1 − t²)². Limits 0 to 1.
∫₀¹ t^(1/2)(1 − 2t² + t⁴) dt = 2/3 − 4/7 + 2/11 = 64/231.
Answer: 64/231
For 0 ≤ x ≤ 1, put x = tan θ: sin⁻¹(2x/(1 + x²)) = 2θ = 2 tan⁻¹x.
∫₀¹ 2 tan⁻¹x dx = 2[x tan⁻¹x − ½ log(1 + x²)]₀¹ = 2(π/4 − ½ log 2) = π/2 − log 2.
Answer: π/2 − log 2
Put x + 2 = t², dx = 2t dt. Limits: x = 0 gives t = √2; x = 2 gives t = 2.
∫(t² − 2)(t)(2t) dt = 2∫(t⁴ − 2t²) dt = 2[t⁵/5 − 2t³/3] from √2 to 2.
At 2: 32/5 − 16/3 = 16/15. At √2: 4√2/5 − 4√2/3 = −8√2/15. So 2(16/15 + 8√2/15) = (32 + 16√2)/15.
Answer: (32 + 16 √2)/15
Put t = cos x, dt = −sin x dx. Limits: π/2 → 0 and 0 → 1.
∫₀¹ dt/(1 + t²) = tan⁻¹ 1 = π/4.
Answer: π/4
x + 4 − x² = 17/4 − (x − 1/2)². Put t = x − 1/2, a = √17/2.
∫dt/(a² − t²) = (1/2a) log|(a + t)/(a − t)| = (1/√17) log|(√17 + 2t)/(√17 − 2t)|.
t runs from −1/2 to 3/2. The ratio at 3/2 is (√17 + 3)/(√17 − 3), and at −1/2 it is (√17 − 1)/(√17 + 1).
Divide: (√17 + 3)(√17 + 1)/((√17 − 3)(√17 − 1)) = (20 + 4√17)/(20 − 4√17) = (5 + √17)/(5 − √17).
Answer: log((5 + √17)/(5 − √17))/√17
x² + 2x + 5 = (x + 1)² + 4. ∫ = ½ tan⁻¹((x + 1)/2).
At 1: ½ tan⁻¹1 = π/8. At −1: 0.
Answer: π/8
d/dx(e²ˣ/x) = 2e²ˣ/x − e²ˣ/x² = 2e²ˣ(1/x − 1/(2x²)).
So the integrand is ½ d/dx(e²ˣ/x). ∫ = ½[e²ˣ/x]₁² = ½(e⁴/2 − e²) = e⁴/4 − e²/2.
Answer: e^(4)/4 − e^(2)/2
Write (x − x³)^(1/3) = x(1/x² − 1)^(1/3), so the integrand is (1/x² − 1)^(1/3)/x³.
Put t = 1/x² − 1, dt = −2/x³ dx. Limits: x = 1/3 gives t = 8; x = 1 gives t = 0.
(1/2)∫₀⁸ t^(1/3) dt = (1/2)(3/4)·8^(4/3) = (3/8)(16) = 6.
Answer: (A) 6
By the Fundamental Theorem of Calculus, d/dx ∫₀ˣ g(t) dt = g(x). Here g(t) = t sin t.
So f′(x) = x sin x.
Answer: (B) x sin x
Let I = ∫₀^(π/2) cos²x dx. By P₃, I = ∫₀^(π/2) sin²x dx.
Add: 2I = ∫₀^(π/2) 1 dx = π/2, so I = π/4.
Answer: π/4
By P₃, I = ∫√(cos x)/(√(cos x) + √(sin x)) dx. Add the two forms: 2I = ∫₀^(π/2) 1 dx = π/2.
So I = π/4.
Answer: π/4
By P₃, I = ∫cos^(3/2)x/(cos^(3/2)x + sin^(3/2)x) dx. Add: 2I = π/2.
So I = π/4.
Answer: π/4
By P₃, I = ∫sin⁵x/(sin⁵x + cos⁵x) dx. Add: 2I = π/2, so I = π/4.
Answer: π/4
|x + 2| = −(x + 2) for x < −2 and x + 2 for x ≥ −2. Split at −2 (P₂).
∫₋₅⁻² −(x + 2) dx = 9/2 and ∫₋₂⁵ (x + 2) dx = 49/2. Sum = 29.
Answer: 29
|x − 5| changes sign at x = 5. Split into [2, 5] and [5, 8].
Each triangle has base 3 and height 3, giving 9/2 each. Sum = 9.
Answer: 9
By P₃, I = ∫₀¹ (1 − x) xⁿ dx = ∫₀¹ (xⁿ − xⁿ⁺¹) dx.
= 1/(n + 1) − 1/(n + 2) = 1/((n + 1)(n + 2)).
Answer: 1/((n + 1)(n + 2))
By P₃, I = ∫₀^(π/4) log(1 + tan(π/4 − x)) dx = ∫ log(2/(1 + tan x)) dx = ∫(log 2 − log(1 + tan x)) dx.
So 2I = (π/4) log 2, hence I = (π/8) log 2.
Answer: π log 2/8
By P₃, I = ∫₀²(2 − x)√x dx = ∫(2x^(1/2) − x^(3/2)) dx.
= (4/3)x^(3/2) − (2/5)x^(5/2) from 0 to 2 = (4/3)(2√2) − (2/5)(4√2) = 8√2/3 − 8√2/5 = 16√2/15.
Answer: 16 √2/15
2 log sin x − log sin 2x = 2 log sin x − log 2 − log sin x − log cos x = log sin x − log cos x − log 2.
∫₀^(π/2) log sin x dx = ∫₀^(π/2) log cos x dx (P₃), so those two cancel.
I = −(π/2) log 2.
Answer: −π log 2/2
sin²x is even, so I = 2∫₀^(π/2) sin²x dx = 2(π/4) = π/2.
Answer: π/2
By P₃, I = ∫(π − x)/(1 + sin x) dx. Add: 2I = π∫₀^π dx/(1 + sin x).
∫₀^π dx/(1 + sin x) = ∫(1 − sin x)/cos²x dx = [tan x − sec x]₀^π = 2. So 2I = 2π and I = π.
Answer: π
sin⁷x is an odd function, so the integral over a symmetric interval is 0.
Answer: 0
By P₆, ∫₀^(2π) cos⁵x dx = 2∫₀^π cos⁵x dx when f(2π − x) = f(x), and cos⁵(π − x) = −cos⁵x makes ∫₀^π cos⁵x dx = 0.
So I = 0.
Answer: 0
By P₃, I = ∫(cos x − sin x)/(1 + sin x cos x) dx = −I.
So 2I = 0 and I = 0.
Answer: 0
1 + cos x = 2cos²(x/2), so I = π log 2 + 2∫₀^π log cos(x/2) dx = π log 2 + 4∫₀^(π/2) log cos t dt.
Using ∫₀^(π/2) log cos t dt = −(π/2) log 2: I = π log 2 − 2π log 2 = −π log 2.
Answer: −π log 2
By P₃, I = ∫√(a − x)/(√(a − x) + √x) dx. Add: 2I = ∫₀^a dx = a.
So I = a/2.
Answer: a/2
Split at x = 1 (P₂): ∫₀¹ (1 − x) dx = 1/2 and ∫₁⁴ (x − 1) dx = 9/2.
Sum = 5.
Answer: 5
Let I = ∫₀^a f(x)g(x) dx. By P₃, I = ∫₀^a f(a − x)g(a − x) dx = ∫₀^a f(x)g(a − x) dx, using f(a − x) = f(x).
Add the two forms: 2I = ∫₀^a f(x)[g(x) + g(a − x)] dx = 4∫₀^a f(x) dx.
So I = 2∫₀^a f(x) dx.
Answer: Hence proved.
x³, x cos x and tan⁵x are all odd, so their integrals over [−π/2, π/2] vanish (P₇).
What remains is ∫ 1 dx = π.
Answer: (C) π
By P₃, I = ∫log((4 + 3 cos x)/(4 + 3 sin x)) dx = −I.
So 2I = 0 and I = 0.
Answer: (C) 0
1/(x(1 − x²)) = 1/x + x/(1 − x²) (cover-up: 1/x at x = 0; the remaining part x/(1 − x²)).
∫ = log|x| − ½ log|1 − x²| = ½ log|x²/(1 − x²)|.
Answer: log(|x²/(1 − x²)|)/2 + C
Rationalize: 1/(√(x + a) + √(x + b)) = (√(x + a) − √(x + b))/(a − b).
Integrate each power: (2/(3(a − b)))[(x + a)^(3/2) − (x + b)^(3/2)].
Answer: 2((x + a)^(3/2) − (x + b)^(3/2))/(3(a − b)) + C
Put x = a/t, dx = −a/t² dt. Then ax − x² = (a²/t)(1 − 1/t)·... = a²(t − 1)/t², so √(ax − x²) = a√(t − 1)/t.
The integrand becomes (t/a)(t/(a√(t − 1)))(−a/t²) dt = −dt/(a√(t − 1)).
∫ = −(2/a)√(t − 1) = −(2/a)√((a − x)/x).
Answer: −2 √((a − x)/x)/a + C
Divide top and bottom: 1/(x²(x⁴ + 1)^(3/4)) = 1/(x⁵(1 + x⁻⁴)^(3/4)).
Put t = 1 + x⁻⁴, dt = −4x⁻⁵ dx. ∫ = −¼∫t^(−3/4) dt = −t^(1/4) = −(1 + x⁻⁴)^(1/4) = −(x⁴ + 1)^(1/4)/x (for x > 0).
Answer: −(x⁴ + 1)^(1/4)/x + C
Put x = t⁶, dx = 6t⁵ dt. The integrand becomes 6t⁵/(t³ + t²) = 6t³/(t + 1).
6t³/(t + 1) = 6(t² − t + 1) − 6/(t + 1). Integrate: 2t³ − 3t² + 6t − 6 log(1 + t).
Replace t = x^(1/6).
Answer: 2 √x − 3x^(1/3) + 6x^(1/6) − 6 log(1 + x^(1/6)) + C
5x/((x + 1)(x² + 9)) = A/(x + 1) + (Bx + C)/(x² + 9). At x = −1: A = −5/10 = −1/2.
Comparing x²: A + B = 0, so B = 1/2. Comparing constants: 9A + C = 0, so C = 9/2.
∫ = −½ log|x + 1| + ¼ log(x² + 9) + (9/2)(1/3) tan⁻¹(x/3).
Answer: −log(|x + 1|)/2 + log(x² + 9)/4 + 3 tan⁻¹(x/3)/2 + C
Write x = (x − a) + a: sin x = sin(x − a) cos a + cos(x − a) sin a.
So the integrand is cos a + sin a · cot(x − a).
∫ = x cos a + sin a · log|sin(x − a)|.
Answer: x cos a + sin a log(|sin(x − a)|) + C
e^(n log x) = xⁿ. So the integrand is (x⁵ − x⁴)/(x³ − x²) = x⁴(x − 1)/(x²(x − 1)) = x².
∫x² dx = x³/3.
Answer: x³/3 + C
Put t = sin x, dt = cos x dx. ∫dt/√(4 − t²) = sin⁻¹(t/2).
Answer: sin⁻¹(sin x/2) + C
sin⁸x − cos⁸x = (sin⁴x − cos⁴x)(sin⁴x + cos⁴x) = (sin²x − cos²x)(sin⁴x + cos⁴x).
Denominator: 1 − 2 sin²x cos²x = (sin²x + cos²x)² − 2 sin²x cos²x = sin⁴x + cos⁴x.
So the integrand is sin²x − cos²x = −cos 2x, and ∫ = −½ sin 2x.
Answer: −sin(2x)/2 + C
tan(x + a) − tan(x + b) = sin(a − b)/(cos(x + a) cos(x + b)).
So 1/(cos(x + a) cos(x + b)) = [tan(x + a) − tan(x + b)]/sin(a − b).
∫tan(x + a) dx = log|sec(x + a)| = −log|cos(x + a)|, so ∫ = [log|cos(x + b)| − log|cos(x + a)|]/sin(a − b).
Answer: (log(|cos(x + b)|) − log(|cos(x + a)|))/sin(a − b) + C
Put t = x⁴, dt = 4x³ dx. ¼∫dt/√(1 − t²) = ¼ sin⁻¹t.
Answer: sin⁻¹(x⁴)/4 + C
Put t = eˣ, dt = eˣ dx. 1/((1 + t)(2 + t)) = 1/(1 + t) − 1/(2 + t).
∫ = log((1 + t)/(2 + t)).
Answer: log((1 + eˣ)/(2 + eˣ)) + C
1/((x² + 1)(x² + 4)) = ⅓[1/(x² + 1) − 1/(x² + 4)].
∫ = ⅓ tan⁻¹x − ⅓·½ tan⁻¹(x/2).
Answer: tan⁻¹ x/3 − tan⁻¹(x/2)/6 + C
e^(log sin x) = sin x, so the integrand is cos³x sin x.
Put t = cos x, dt = −sin x dx: −∫t³ dt = −t⁴/4.
Answer: −cos⁴x/4 + C
e^(3 log x) = x³. Put t = x⁴ + 1, dt = 4x³ dx: ¼∫dt/t = ¼ log(x⁴ + 1).
Answer: log(x⁴ + 1)/4 + C
Put t = f(ax + b). Then dt = a f′(ax + b) dx, so f′(ax + b) dx = dt/a.
∫ = (1/a)∫tⁿ dt = tⁿ⁺¹/(a(n + 1)) = [f(ax + b)]ⁿ⁺¹/(a(n + 1)).
Answer: [f(ax + b)]ⁿ⁺¹ / (a(n + 1)) + C
sin³x sin(x + α) = sin⁴x (cos α + cot x sin α), so the integrand is 1/(sin²x √(cos α + sin α cot x)).
Put t = cos α + sin α cot x, dt = −sin α cosec²x dx. ∫ = −(1/sin α)∫dt/√t = −(2/sin α)√t.
Answer: −2 √(cos α + sin α/tan x)/sin α + C
Put √x = cos θ, so x = cos²θ and dx = −2 sin θ cos θ dθ. Then √((1 − cos θ)/(1 + cos θ)) = tan(θ/2).
tan(θ/2) sin θ = 2 sin²(θ/2) = 1 − cos θ, so the integral is −2∫(1 − cos θ) cos θ dθ = −2∫(cos θ − cos²θ) dθ = −2 sin θ + θ + sin θ cos θ.
Replace cos θ = √x and sin θ = √(1 − x).
Answer: cos⁻¹(√x) + (√x − 2) √(1 − x) + C
(2 + sin 2x)/(1 + cos 2x) = (2 + 2 sin x cos x)/(2cos²x) = sec²x + tan x.
With f = tan x, f′ = sec²x, the integrand is eˣ[f + f′]. So ∫ = eˣ tan x.
Answer: eˣ tan x + C
A/(x + 1) + B/(x + 1)² + C/(x + 2). At x = −1: B = 1. At x = −2: C = 3/1 = 3.
Comparing x² terms: A + C = 1, so A = −2.
Answer: −2 log(|x + 1|) − 1/(x + 1) + 3 log(|x + 2|) + C
Put x = cos 2θ. Then √((1 − x)/(1 + x)) = tan θ, so the integrand is θ = ½ cos⁻¹x.
∫½ cos⁻¹x dx = ½[x cos⁻¹x − √(1 − x²)].
Answer: (x cos⁻¹ x − √(1 − x²))/2 + C
log(x² + 1) − 2 log x = log(1 + 1/x²). Put t = 1 + 1/x², dt = −2/x³ dx, and √(x² + 1)/x⁴ dx = √t/x³ dx = −(√t/2) dt (for x > 0).
−½∫t^(1/2) log t dt = −½[(2/3)t^(3/2) log t − (4/9)t^(3/2)] = −⅓ t^(3/2) log t + (2/9) t^(3/2).
Answer: −(1 + 1/x²)^(3/2) log(1 + 1/x²)/3 + 2(1 + 1/x²)^(3/2)/9 + C
(1 − sin x)/(1 − cos x) = ½cosec²(x/2) − cot(x/2). With f = −cot(x/2), f′ = ½cosec²(x/2), the integrand is eˣ[f + f′].
∫ = −eˣ cot(x/2). At π: 0. At π/2: −e^(π/2) · 1. Value = 0 − (−e^(π/2)) = e^(π/2).
Answer: e^(π/2)
Divide top and bottom by cos⁴x: tan x sec²x/(1 + tan⁴x). Put t = tan²x, dt = 2 tan x sec²x dx.
½∫₀¹ dt/(1 + t²) = ½ · π/4 = π/8.
Answer: π/8
Divide by cos²x: 1/(1 + 4 tan²x). Write 1/(1 + 4t²) with t = tan x and dx = dt/(1 + t²).
1/((1 + 4t²)(1 + t²)) = (1/3)[4/(1 + 4t²) − 1/(1 + t²)]. ∫₀^∞ = (1/3)[2 · π/2 − π/2] = π/6.
Answer: π/6
Put t = sin x − cos x, dt = (cos x + sin x) dx, and sin 2x = 1 − t².
t runs from (1 − √3)/2 to (√3 − 1)/2. ∫dt/√(1 − t²) = sin⁻¹t. By symmetry the value is 2 sin⁻¹((√3 − 1)/2).
Answer: 2 sin⁻¹((√3 − 1)/2)
Rationalize: 1/(√(1 + x) − √x) = √(1 + x) + √x.
∫ = (2/3)[(1 + x)^(3/2) + x^(3/2)] from 0 to 1 = (2/3)(2√2 + 1 − 1) = 4√2/3.
Answer: 4 √2/3
Put t = sin x − cos x, dt = (cos x + sin x) dx, and sin 2x = 1 − t². Then 9 + 16 sin 2x = 25 − 16t².
t runs from −1 (at x = 0) to 0 (at x = π/4). ∫dt/(25 − 16t²) = (1/40) log|(5 + 4t)/(5 − 4t)|.
At t = 0 this is 0; at t = −1 it is (1/40) log(1/9). So I = −(1/40) log(1/9) = (1/40) log 9 = (1/20) log 3.
Answer: log 3/20
sin 2x = 2 sin x cos x. Put t = sin x, dt = cos x dx: I = 2∫₀¹ t tan⁻¹t dt.
∫t tan⁻¹t dt = ½(t² + 1) tan⁻¹t − t/2. From 0 to 1: ½(2)(π/4) − ½ = π/4 − ½.
I = 2(π/4 − ½) = π/2 − 1.
Answer: π/2 − 1
Integrate each term separately on [1, 4].
∫₁⁴|x − 1| dx = 9/2. ∫₁⁴|x − 2| dx = 1/2 + 2 = 5/2. ∫₁⁴|x − 3| dx = 2 + 1/2 = 5/2.
Sum = 9/2 + 5/2 + 5/2 = 19/2.
Answer: 19/2
1/(x²(x + 1)) = 1/x² − 1/x + 1/(x + 1).
∫ = −1/x − log x + log(x + 1) = −1/x + log((x + 1)/x).
At 3: −1/3 + log(4/3). At 1: −1 + log 2. Difference = 2/3 + log(4/3) − log 2 = 2/3 + log(2/3).
Answer: Hence proved.
By parts: ∫x eˣ dx = x eˣ − eˣ.
From 0 to 1: (e − e) − (0 − 1) = 1.
Answer: Hence proved.
f(x) = x¹⁷cos⁴x satisfies f(−x) = −f(x), so f is odd.
The integral of an odd function over [−1, 1] is 0.
Answer: Hence proved.
sin³x = sin x (1 − cos²x). Put t = cos x: ∫₀¹ (1 − t²) dt = 1 − 1/3 = 2/3.
Answer: Hence proved.
2 tan³x = 2 tan x (sec²x − 1). ∫2 tan x sec²x dx = tan²x and ∫2 tan x dx = −2 log|cos x|... so the primitive is tan²x + 2 log|cos x|.
From 0 to π/4: 1 + 2 log(1/√2) = 1 − log 2.
Answer: Hence proved.
By parts: ∫sin⁻¹x dx = x sin⁻¹x + √(1 − x²).
From 0 to 1: (π/2 + 0) − (0 + 1) = π/2 − 1.
Answer: Hence proved.
Multiply top and bottom by eˣ: eˣ/(e²ˣ + 1). Put t = eˣ, dt = eˣ dx.
∫dt/(t² + 1) = tan⁻¹t = tan⁻¹(eˣ).
Answer: (A) tan⁻¹(eˣ) + C
cos 2x = (cos x − sin x)(cos x + sin x), so the integrand is (cos x − sin x)/(cos x + sin x).
This is the derivative of log|sin x + cos x| over itself, so ∫ = log|sin x + cos x|.
Answer: (B) log|sin x + cos x| + C
Let I = ∫ₐᵇ x f(x) dx. By P₃, I = ∫ₐᵇ (a + b − x) f(a + b − x) dx = ∫ₐᵇ (a + b − x) f(x) dx.
Add: 2I = (a + b)∫ₐᵇ f(x) dx, so I = ((a + b)/2) ∫ₐᵇ f(x) dx.
Answer: (D) ((a + b)/2) ∫ₐᵇ f(x) dx