Miscellaneous Exercise answers: Integrals

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1. Factor the denominator: x − x³ = x(1 − x²) = x(1 − x)(1 + x).
2. Three different linear factors, so write 1/[x(1 − x)(1 + x)] = A/x + B/(1 − x) + C/(1 + x).
3. Clear the denominators: 1 = A(1 − x)(1 + x) + Bx(1 + x) + Cx(1 − x).
4. Put x = 0: 1 = A, so A = 1. Put x = 1: 1 = 2B, so B = 1/2. Put x = −1: 1 = −2C, so C = −1/2.
5. So the integrand is 1/x + (1/2)·1/(1 − x) − (1/2)·1/(1 + x).
6. Integrate each piece: ∫dx/x = log|x|, ∫dx/(1 − x) = −log|1 − x|, ∫dx/(1 + x) = log|1 + x|.
7. Total: log|x| − (1/2) log|1 − x| − (1/2) log|1 + x| + C.
8. Since |1 − x|·|1 + x| = |1 − x²| = |x² − 1|, this is log|x| − (1/2) log|x² − 1| + C.

Answerlog|x| − (1/2) log|x² − 1| + C

Watch this explained “The root shortcut”, 7:31 into Breaking a proper rational function into pieces with simple denominators

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1. Multiply top and bottom by (√(x+a) − √(x+b)), the conjugate surd.
2. The bottom becomes (x+a) − (x+b) = a − b.
3. The top is now √(x+a) − √(x+b), so the integrand equals [√(x+a) − √(x+b)]/(a−b).
4. Integrate each surd separately using the power rule: ∫√(x+a) dx = (2/3)(x+a)^(3/2), and similarly for (x+b).
5. Combine: [(2/3)(x+a)^(3/2) − (2/3)(x+b)^(3/2)]/(a−b) + C.

Answer(2/(3(a−b))) [(x+a)^(3/2) − (x+b)^(3/2)] + C

Watch this explained “Rewriting IS the work”, 14:40 into Reading the table of standard integrals off the table of derivatives

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1. Follow the hint: put x = a/t, so dx = −(a/t²) dt.
2. ax − x² = (a²/t) − (a²/t²) = (a²/t²)(t−1).
3. √(ax−x²) = (a/t)√(t−1) (t>1).
4. The integral becomes ∫ −(a/t²) dt / [(a/t) × (a/t)√(t−1)] = −(1/a) ∫ dt/√(t−1).
5. ∫ dt/√(t−1) = 2√(t−1), so the integral is −(2/a)√(t−1).
6. Put back t = a/x: −(2/a)√(a/x − 1) = −(2/a)√((a−x)/x) + C.

Answer−(2/a)√((a−x)/x) + C

Watch this explained “One, carried all the way through”, 4:43 into Changing the variable until what is left is a standard form

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1. Divide inside the bracket by x⁴: (x⁴+1)^(3/4) = x³(1+1/x⁴)^(3/4) for x>0.
2. So the integrand becomes 1/[x²·x³(1+1/x⁴)^(3/4)] = x^(−5) (1+x^(−4))^(−3/4).
3. Substitute t = 1 + x^(−4); its rate is −4x^(−5), which is (up to a constant) exactly the factor x^(−5) standing there.
4. So x^(−5) dx = −dt/4, and the integral becomes −(1/4)∫ t^(−3/4) dt.
5. ∫t^(−3/4)dt = 4t^(1/4), so the integral is −t^(1/4) + C.
6. Put back t = 1+x^(−4) = (x⁴+1)/x⁴: −[(x⁴+1)/x⁴]^(1/4) = −(x⁴+1)^(1/4)/x + C.

Answer−(x⁴+1)^(1/4)/x + C

Watch this explained “The advice, as a search”, 2:11 into Changing the variable until what is left is a standard form

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1. Follow the hint: put x = t⁶, so dx = 6t⁵ dt, and x^(1/2)=t³, x^(1/3)=t².
2. The integral becomes ∫ 6t⁵/(t³+t²) dt = ∫ 6t³/(t+1) dt.
3. 6t³/(t+1) is an improper fraction (top degree ≥ bottom degree), so divide first: 6t³/(t+1) = 6t² − 6t + 6 − 6/(t+1).
4. Integrate term by term: 2t³ − 3t² + 6t − 6 log|t+1|.
5. Put back t = x^(1/6): 2x^(1/2) − 3x^(1/3) + 6x^(1/6) − 6 log(x^(1/6)+1) + C.

Answer2√x − 3x^(1/3) + 6x^(1/6) − 6 log(x^(1/6)+1) + C

Watch this explained “Proper comes first”, 0:57 into Breaking a proper rational function into pieces with simple denominators

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1. The denominator has a linear factor (x+1) and a quadratic factor (x²+9) that will not break further.
2. So write 5x/[(x+1)(x²+9)] = A/(x+1) + (Bx+C)/(x²+9), where the quadratic factor gets a LINEAR numerator, not just a constant.
3. Clearing denominators and comparing coefficients gives A = −1/2, B = 1/2, C = 9/2.
4. The A/(x+1) piece integrates to −(1/2) log|x+1|.
5. The (Bx+C)/(x²+9) piece splits into (1/2)·[x/(x²+9)] + (9/2)·[1/(x²+9)], giving (1/4) log(x²+9) + (9/2)·(1/3) arctan(x/3) = (1/4) log(x²+9) + (3/2) arctan(x/3).
6. Add everything: −(1/2) log|x+1| + (1/4) log(x²+9) + (3/2) arctan(x/3) + C.

Answer−(1/2) log|x+1| + (1/4) log(x²+9) + (3/2) arctan(x/3) + C

Watch this explained “A quadratic needs a linear top”, 9:46 into Breaking a proper rational function into pieces with simple denominators

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1. Write x as (x−a)+a, so sin x = sin[(x−a)+a] = sin(x−a)cos a + cos(x−a)sin a.
2. Divide by sin(x−a): sin x/sin(x−a) = cos a + sin a · cot(x−a).
3. cos a is a constant, so it integrates to x cos a.
4. sin a is also a constant (with respect to x); cot(x−a) integrates to log|sin(x−a)|.
5. Add the two pieces: x cos a + sin a · log|sin(x−a)| + C.

Answerx cos a + sin a · log|sin(x−a)| + C

Watch this explained “The identity is never the destination”, 1:16 into Rewriting a product of trigonometric ratios into terms you can already integrate

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1. e^(n log x) simply means x^n, so e^(5logx)=x⁵, e^(4logx)=x⁴, e^(3logx)=x³, e^(2logx)=x².
2. The integrand becomes (x⁵−x⁴)/(x³−x²) = x⁴(x−1)/[x²(x−1)] = x², after cancelling (x−1) and x².
3. Integrate x² using the power rule: x³/3 + C.

Answerx³/3 + C

Watch this explained “Rewriting IS the work”, 14:40 into Reading the table of standard integrals off the table of derivatives

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1. Substitute t = sin x, so dt = cos x dx — exactly the factor sitting in the numerator.
2. The integral becomes ∫ dt/√(4−t²), which is the standard inverse-sine form ∫dt/√(a²−t²) with a=2.
3. ∫dt/√(4−t²) = arcsin(t/2) + C.
4. Put back t = sin x: arcsin(sin x / 2) + C.

Answerarcsin(sin x / 2) + C

Watch this explained “Only two are inverse functions”, 1:59 into Six formulas for quadratic denominators, and completing the square to reach them

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1. Factor the top as a difference of squares twice: sin⁸x−cos⁸x = (sin²x−cos²x)(sin²x+cos²x)(sin⁴x+cos⁴x).
2. sin²x+cos²x = 1, so the top is (sin²x−cos²x)(sin⁴x+cos⁴x) = −cos2x · (sin⁴x+cos⁴x).
3. The bottom 1−2sin²xcos²x equals sin⁴x+cos⁴x (since sin⁴x+cos⁴x = (sin²x+cos²x)²−2sin²xcos²x = 1−2sin²xcos²x).
4. So the fraction simplifies to −cos2x (the sin⁴x+cos⁴x factor cancels top and bottom).
5. Integrate −cos2x: −(1/2) sin 2x + C.

Answer−(1/2) sin 2x + C

Watch this explained “The identity is never the destination”, 1:16 into Rewriting a product of trigonometric ratios into terms you can already integrate

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1. Write 1 as sin[(x+a) − (x+b)] ÷ sin(a−b), since (x+a) − (x+b) = a−b.
2. Expand sin[(x+a)−(x+b)] = sin(x+a)cos(x+b) − cos(x+a)sin(x+b).
3. Divide this by cos(x+a)cos(x+b): the integrand becomes [tan(x+a) − tan(x+b)] ÷ sin(a−b).
4. ∫tan(x+a)dx = −log|cos(x+a)|, and ∫tan(x+b)dx = −log|cos(x+b)|.
5. Combine and divide by sin(a−b) to get the answer.

Answer1/sin(a−b) · log|cos(x+b)/cos(x+a)| + C

Watch this explained “Multiplying by a chosen one”, 10:33 into Changing the variable until what is left is a standard form

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1. Put t = x⁴, so dt = 4x³ dx, i.e. x³dx = dt/4.
2. 1 − x⁸ = 1 − t², so the integral becomes (1/4)∫dt/√(1−t²).
3. This is a standard form: (1/4) sin⁻¹t.
4. Put back t = x⁴.

Answer(1/4) sin⁻¹(x⁴) + C

Watch this explained “The advice, as a search”, 2:11 into Changing the variable until what is left is a standard form

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1. Put t = e^x, so dt = e^xdx.
2. The integral becomes ∫dt/[(1+t)(2+t)].
3. Split into partial fractions: 1/[(1+t)(2+t)] = 1/(1+t) − 1/(2+t).
4. Integrate: log|1+t| − log|2+t|.
5. Put back t = e^x (always positive, so no modulus needed).

Answerlog[(1 + e^x)/(2 + e^x)] + C

Watch this explained “Five shapes, one index”, 1:51 into Breaking a proper rational function into pieces with simple denominators

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1. Treat x² as one block and split into partial fractions: 1/[(x²+1)(x²+4)] = A/(x²+1) + B/(x²+4).
2. Clear denominators: 1 = A(x²+4) + B(x²+1).
3. Put x² = −1: 1 = 3A, so A = 1/3. Put x² = −4: 1 = −3B, so B = −1/3.
4. Integrate each piece using ∫dx/(x²+k²) = (1/k)tan⁻¹(x/k).

Answer(1/3) tan⁻¹x − (1/6) tan⁻¹(x/2) + C

Watch this explained “Two ways out, one answer”, 13:00 into Breaking a proper rational function into pieces with simple denominators

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1. e^{log sin x} = sin x, so the integrand is cos³x sin x.
2. Put t = cos x, so dt = −sin x dx.
3. The integral becomes −∫t³dt = −t⁴/4.
4. Put back t = cos x.

Answer−(1/4) cos⁴x + C

Watch this explained “The same cube, by substitution”, 8:47 into Rewriting a product of trigonometric ratios into terms you can already integrate

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1. e^{3 log x} = x³, so the integrand is x³/(x⁴+1).
2. Put t = x⁴+1, so dt = 4x³dx.
3. The integral becomes (1/4)∫dt/t = (1/4)log|t|.
4. Put back t = x⁴+1 (always positive, so no modulus needed).

Answer(1/4) log(x⁴ + 1) + C

Watch this explained “Substituting for the denominator”, 7:14 into Changing the variable until what is left is a standard form

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1. Put u = f(ax + b). By the chain rule, du = a·f′(ax+b)dx.
2. So f′(ax+b)dx = du/a, and the integral becomes (1/a)∫uⁿdu.
3. Integrate using the power rule: (1/a)·uⁿ⁺¹/(n+1).
4. Put back u = f(ax+b).

Answer[f(ax + b)]ⁿ⁺¹ ⁄ [a(n + 1)] + C

Watch this explained “One, carried all the way through”, 4:43 into Changing the variable until what is left is a standard form

Open NCERT p. 286Matches NCERT’s answer

1. Write sin(x+α) = sin x cos α + cos x sin α = sin x (cos α + cot x sin α).
2. So sin³x sin(x+α) = sin⁴x (cos α + sin α cot x).
3. The integrand becomes 1/[sin²x √(cos α + sin α cot x)] = cosec²x / √(cos α + sin α cot x).
4. Put t = cot x, so dt = −cosec²x dx, and cos α + sin α cot x = cos α + sin α t.
5. The integral becomes −∫dt/√(cos α + sin α t), a standard √(linear) form.
6. Integrate: −(2/sin α)√(cos α + sin α t).
7. Put back t = cot x.

Answer−(2/sin α) √(cos α + sin α cot x) + C

Watch the lesson Changing the variable until what is left is a standard form

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1. Multiply top and bottom inside the root by (1−√x): (1−√x)²/(1−x), so the root becomes (1−√x)/√(1−x).
2. Split: the integrand is 1/√(1−x) − √x/√(1−x).
3. ∫dx/√(1−x) = −2√(1−x).
4. For ∫√x/√(1−x) dx, put x = sin²θ: this becomes ∫2sin²θ dθ = θ − sin θ cos θ = sin⁻¹√x − √(x(1−x)).
5. Subtract the second piece from the first.

Answer−2√(1−x) − sin⁻¹√x + √(x−x²) + C

Watch the lesson Changing the variable until what is left is a standard form

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1. 1 + cos 2x = 2cos²x and sin 2x = 2 sin x cos x.
2. So (2 + sin 2x)/(1 + cos 2x) = (2 + 2 sin x cos x)/(2cos²x) = sec²x + tan x.
3. The integrand is now e^x(tan x + sec²x), which is e^x[f(x) + f′(x)] with f(x) = tan x.
4. ∫e^x[f(x)+f′(x)]dx = e^xf(x).

Answere^x tan x + C

Watch this explained “Manufacturing the split”, 6:16 into Two shapes worth spotting: the exponential pair, and the three surd forms

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1. Because (x+1) is squared, use three terms: A/(x+1) + B/(x+1)² + C/(x+2).
2. Clear denominators: x²+x+1 = A(x+1)(x+2) + B(x+2) + C(x+1)².
3. Put x = −1: 1 = B, so B = 1. Put x = −2: 3 = C, so C = 3.
4. Match the x² coefficient: 1 = A + C, so A = −2.
5. Integrate each term: −2log|x+1| − 1/(x+1) + 3log|x+2|.

Answer−2 log|x + 1| − 1/(x + 1) + 3 log|x + 2| + C

Watch this explained “A squared factor needs two”, 8:25 into Breaking a proper rational function into pieces with simple denominators

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1. Put x = cos θ, so θ = cos⁻¹x, with θ between 0 and π.
2. (1−x)/(1+x) = (1−cos θ)/(1+cos θ) = tan²(θ/2), so the square root is tan(θ/2).
3. tan⁻¹[tan(θ/2)] = θ/2, so the integrand is just (1/2)cos⁻¹x.
4. Integrate (1/2)cos⁻¹x by parts, with cos⁻¹x differentiating and 1 integrating.
5. This gives (1/2)[x cos⁻¹x − √(1−x²)].

Answer(1/2)[x cos⁻¹x − √(1 − x²)] + C

Watch this explained “Supplying a factor that is not there”, 9:54 into Integrating a product, and how the choice of first function decides whether it helps

Open NCERT p. 286Matches NCERT’s answer

1. log(x² + 1) − 2 log x = log[(x² + 1)/x²] = log(1 + 1/x²).
2. For x > 0, √(x² + 1)/x⁴ = [√(x² + 1)/x] · (1/x³) = √(1 + 1/x²) · (1/x³).
3. Put t = 1 + 1/x². Then dt = −(2/x³) dx, so dx/x³ = −dt/2.
4. The integral becomes −(1/2)∫t^1/2 log t dt.
5. Integrate by parts, with log t differentiated and t^1/2 integrated to (2/3)t^3/2: ∫t^1/2 log t dt = (2/3)t^3/2 log t − (2/3)∫t^1/2 dt = (2/3)t^3/2 log t − (4/9)t^3/2.
6. Multiply by −1/2: −(1/3)t^3/2 log t + (2/9)t^3/2.
7. Put back t = (x² + 1)/x²: t^3/2 = (x² + 1)^3/2/x³ and log t = log(x² + 1) − 2 log x.

Answer−(1/3)·[(x²+1)^3/2/x³]·[log(x²+1) − 2log x] + (2/9)·(x²+1)^3/2/x³ + C

Watch this explained “Which factor goes first”, 22:02 into Integrating a product, and how the choice of first function decides whether it helps

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1. Evaluate: e^x (1 − sin x) ⁄ (1 − cos x), from π/2 to π.
2. 1 − cos x = 2sin²(x/2) and 1 − sin x = [sin(x/2) − cos(x/2)]².
3. So (1−sin x)/(1−cos x) = (1/2)cosec²(x/2) − cot(x/2).
4. (1/2)cosec²(x/2) is the derivative of −cot(x/2), so this is g(x) + g′(x) with g(x) = −cot(x/2).
5. ∫e^x[g(x)+g′(x)]dx = e^xg(x) = −e^xcot(x/2).
6. At x = π, cot(π/2) = 0. At x = π/2, cot(π/4) = 1.
7. Value = [−e^π·0] − [−e^π/2·1] = e^π/2.

Answere^π/2

Watch this explained “Manufacturing the split”, 6:16 into Two shapes worth spotting: the exponential pair, and the three surd forms

Open NCERT p. 286Matches NCERT’s answer

1. Evaluate: sin x cos x ⁄ (cos⁴x + sin⁴x), from 0 to π/4.
2. Put t = sin²x, so dt = 2 sin x cos x dx, and t runs from 0 to 1/2.
3. cos⁴x + sin⁴x = 1 − 2sin²x cos²x = 1 − 2t(1−t) = 2t² − 2t + 1.
4. The integral becomes (1/2)∫[0 to 1/2] dt/(2t² − 2t + 1).
5. Complete the square: 2t²−2t+1 = (1/2)[(2t−1)²+1].
6. This gives ∫[0 to 1/2] dt/[(2t−1)²+1] = (1/2)tan⁻¹(2t−1), evaluated from 0 to 1/2.
7. Value = (1/2)[tan⁻¹(0) − tan⁻¹(−1)] = (1/2)(π/4) = π/8.

Answerπ/8

Watch this explained “Completing the square”, 7:21 into Six formulas for quadratic denominators, and completing the square to reach them

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1. Evaluate: cos²x ⁄ (cos²x + 4 sin²x), from 0 to π/2.
2. Divide top and bottom by cos²x: the integrand becomes 1/(1 + 4tan²x).
3. Put t = tan x, so dt = sec²x dx = (1+t²)dx; t runs from 0 to ∞.
4. The integral becomes ∫[0 to ∞] dt/[(1+t²)(1+4t²)].
5. Split into partial fractions in t²: 1/[(1+t²)(1+4t²)] = −(1/3)/(1+t²) + (4/3)/(1+4t²).
6. ∫[0 to ∞]dt/(1+t²) = π/2, and ∫[0 to ∞]dt/(1+4t²) = π/4.
7. Value = −(1/3)(π/2) + (4/3)(π/4) = −π/6 + π/3 = π/6.

Answerπ/6

Watch this explained “Two ways out, one answer”, 13:00 into Breaking a proper rational function into pieces with simple denominators

Open NCERT p. 286Matches NCERT’s answer

1. Evaluate: (sin x + cos x) ⁄ √(sin 2x), from π/6 to π/3.
2. Put t = sin x − cos x, so dt = (cos x + sin x)dx.
3. t² = 1 − sin 2x, so sin 2x = 1 − t².
4. The integral becomes ∫dt/√(1−t²) = sin⁻¹t.
5. At x = π/3, t = √3/2 − 1/2 = (√3−1)/2. At x = π/6, t = 1/2 − √3/2 = −(√3−1)/2.
6. Value = sin⁻¹[(√3−1)/2] − sin⁻¹[−(√3−1)/2] = 2 sin⁻¹[(√3−1)/2].

Answer2 sin⁻¹[(√3 − 1)/2]

Watch this explained “Picking the substitution”, 13:03 into Substituting inside a definite integral, and why the limits have to move with it

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1. Evaluate: dx ⁄ (√(1 + x) − √x), from 0 to 1.
2. Multiply top and bottom by (√(1+x) + √x): the denominator becomes (1+x) − x = 1.
3. So the integrand simplifies to √(1+x) + √x.
4. ∫[0 to 1](1+x)^1/2dx = (2/3)(1+x)^3/2, and ∫[0 to 1]x^1/2dx = (2/3)x^3/2.
5. Evaluate: (2/3)[2^3/2 − 1] + (2/3)[1 − 0] = (2/3)(2√2 − 1) + 2/3 = 4√2/3.

Answer4√2/3

Watch this explained “Multiplying by a chosen one”, 10:33 into Changing the variable until what is left is a standard form

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1. Evaluate: (sin x + cos x) ⁄ (9 + 16 sin 2x), from 0 to π/4.
2. Put t = sin x − cos x, so dt = (cos x + sin x)dx, and sin 2x = 1 − t².
3. The integral becomes ∫dt/(9 + 16(1−t²)) = ∫dt/(25 − 16t²).
4. This is a standard form: (1/40)log|(5+4t)/(5−4t)|.
5. At x = 0, t = −1. At x = π/4, t = 0.
6. Value = (1/40)log(1) − (1/40)log(1/9) = (1/40)log 9 = (1/20)log 3.

Answer(1/20) log 3

Watch this explained “Picking the substitution”, 13:03 into Substituting inside a definite integral, and why the limits have to move with it

Open NCERT p. 286Matches NCERT’s answer

1. Evaluate: sin 2x tan⁻¹(sin x), from 0 to π/2.
2. sin 2x = 2 sin x cos x. Put u = sin x, so du = cos x dx; u runs from 0 to 1.
3. The integral becomes ∫[0 to 1] 2u tan⁻¹u du.
4. Integrate by parts (tan⁻¹u differentiates, 2u integrates to u²): u²tan⁻¹u − ∫u²/(1+u²)du.
5. u²/(1+u²) = 1 − 1/(1+u²), so its integral is u − tan⁻¹u.
6. Combine: u²tan⁻¹u − u + tan⁻¹u, evaluated from 0 to 1.
7. Value = [1·(π/4) − 1 + π/4] − 0 = π/2 − 1.

Answerπ/2 − 1

Watch this explained “Which factor goes first”, 22:02 into Integrating a product, and how the choice of first function decides whether it helps

Open NCERT p. 286Matches NCERT’s answer

1. Evaluate: |x − 1| + |x − 2| + |x − 3|, from 1 to 4.
2. The bars change behaviour at x = 1, 2 and 3, so cut the interval [1,4] into [1,2], [2,3] and [3,4].
3. On [1,2]: the sum is (x−1)+(2−x)+(3−x) = 4−x.
4. On [2,3]: the sum is (x−1)+(x−2)+(3−x) = x.
5. On [3,4]: the sum is (x−1)+(x−2)+(x−3) = 3x−6.
6. ∫[1 to 2](4−x)dx = 5/2. ∫[2 to 3]x dx = 5/2. ∫[3 to 4](3x−6)dx = 9/2.
7. Add the three pieces: 5/2 + 5/2 + 9/2 = 19/2.

Answer19/2

Watch this explained “Bars are a place to cut”, 14:11 into Properties of the definite integral, and using symmetry to kill half the work

Open NCERT p. 286One way to think about it

1. Split the fraction into partial fractions. Because x² is a repeated factor it needs two terms: 1/(x²(x+1)) = A/x + B/x² + C/(x+1).
2. Multiply both sides by x²(x+1): 1 = Ax(x+1) + B(x+1) + Cx².
3. Put x = 0: 1 = B, so B = 1. Put x = −1: 1 = C, so C = 1. Put x = 1: 1 = 2A + 2B + C = 2A + 3, so A = −1.
4. So 1/(x²(x+1)) = −1/x + 1/x² + 1/(x+1).
5. Integrate each piece: ∫(−1/x + 1/x² + 1/(x+1))dx = −log|x| − 1/x + log|x+1| + C = log|(x+1)/x| − 1/x + C.
6. Put x = 3 into this: log(4/3) − 1/3.
7. Put x = 1 into this: log 2 − 1.
8. Subtract the two: [log(4/3) − 1/3] − [log 2 − 1] = log(4/3) − log 2 + 2/3.
9. Combine the logs: log(4/3) − log 2 = log(4/(3×2)) = log(2/3).
10. So the value of the integral is 2/3 + log(2/3), which is what had to be proved.

In short2/3 + log(2/3), so the statement is proved.

Watch this explained “A squared factor needs two”, 8:25 into Breaking a proper rational function into pieces with simple denominators

Open NCERT p. 286One way to think about it

1. Integrate by parts, with x as the first function (it gets differentiated) and eˣ as the second (it gets integrated).
2. ∫x eˣ dx = x eˣ − ∫eˣ dx = x eˣ − eˣ = (x − 1)eˣ.
3. Put x = 1 into this: (1 − 1)e¹ = 0.
4. Put x = 0 into this: (0 − 1)e⁰ = −1.
5. Subtract the two: 0 − (−1) = 1, which is what had to be proved.

In short1, so the statement is proved.

Watch this explained “Two roles, and what each asks”, 2:44 into Integrating a product, and how the choice of first function decides whether it helps

Open NCERT p. 286One way to think about it

1. Replace x by −x in x¹⁷: (−x)¹⁷ = −x¹⁷, so x¹⁷ is an odd function.
2. Replace x by −x in cos⁴x: cos⁴(−x) = cos⁴x, since cosine is an even function, so cos⁴x is even.
3. An odd function multiplied by an even function is odd, so f(x) = x¹⁷cos⁴x satisfies f(−x) = −f(x).
4. For any odd function, the integral over a symmetric interval [−a, a] is always 0, because the area to the left of 0 exactly cancels the area to the right.
5. Here the interval is [−1, 1], which is symmetric about 0, so ∫₋₁¹ x¹⁷cos⁴x dx = 0, which is what had to be proved.

In short0, so the statement is proved.

Watch this explained “Even and odd on a balanced stretch”, 11:09 into Properties of the definite integral, and using symmetry to kill half the work

Open NCERT p. 286One way to think about it

1. Peel one sine off the odd power: sin³x = sinx·sin²x = sinx(1 − cos²x), using sin²x + cos²x = 1.
2. So sin³x = sinx − sinx·cos²x.
3. Integrating sinx gives −cosx. Integrating sinx·cos²x, by substituting u = cosx, gives −cos³x/3.
4. So ∫sin³x dx = −cosx − (−cos³x/3) = −cosx + cos³x/3.
5. Put x = π/2 into this: −cos(π/2) + cos³(π/2)/3 = −0 + 0 = 0.
6. Put x = 0 into this: −cos0 + cos³0/3 = −1 + 1/3 = −2/3.
7. Subtract the two: 0 − (−2/3) = 2/3, which is what had to be proved.

In short2/3, so the statement is proved.

Watch this explained “The same cube, by substitution”, 8:47 into Rewriting a product of trigonometric ratios into terms you can already integrate

Open NCERT p. 286One way to think about it

1. Peel one tangent off the cube: tan³x = tanx·tan²x = tanx(sec²x − 1), using sec²x = 1 + tan²x.
2. So 2tan³x = 2tanx·sec²x − 2tanx.
3. Integrating 2tanx·sec²x, by substituting u = tanx, gives tan²x.
4. Integrating 2tanx gives −2log|cosx|.
5. So ∫2tan³x dx = tan²x − (−2log|cosx|) = tan²x + 2log|cosx|.
6. Put x = π/4 into this: tan²(π/4) = 1 and cos(π/4) = 1/√2, so the value is 1 + 2log(1/√2) = 1 − log 2.
7. Put x = 0 into this: tan²0 = 0 and cos0 = 1, so the value is 0 + 2log1 = 0.
8. Subtract the two: (1 − log 2) − 0 = 1 − log 2, which is what had to be proved.

In short1 − log 2, so the statement is proved.

Watch this explained “Three families, three obstructions”, 2:15 into Rewriting a product of trigonometric ratios into terms you can already integrate

Open NCERT p. 286One way to think about it

1. Write sin⁻¹x as sin⁻¹x × 1, and integrate by parts with sin⁻¹x as the first function (it gets differentiated) and 1 as the second (it gets integrated).
2. ∫sin⁻¹x dx = x·sin⁻¹x − ∫x/√(1−x²) dx.
3. For ∫x/√(1−x²) dx, substitute t = 1 − x², so dt = −2x dx: this integral works out to −√(1−x²).
4. So ∫sin⁻¹x dx = x·sin⁻¹x + √(1−x²).
5. Put x = 1 into this: 1·sin⁻¹1 + √(1−1) = π/2 + 0 = π/2.
6. Put x = 0 into this: 0·sin⁻¹0 + √(1−0) = 0 + 1 = 1.
7. Subtract the two: π/2 − 1, which is what had to be proved.

In shortπ/2 − 1, so the statement is proved.

Watch this explained “Supplying a factor that is not there”, 9:54 into Integrating a product, and how the choice of first function decides whether it helps

Open NCERT p. 286Matches NCERT’s answer

1. Multiply the top and bottom of the fraction by eˣ: 1/(eˣ+e⁻ˣ) = eˣ/(e²ˣ+1).
2. Substitute u = eˣ, so du = eˣ dx: the integral becomes ∫du/(u²+1).
3. This is the standard form ∫du/(u²+1) = tan⁻¹u + C.
4. Put u back as eˣ: the answer is tan⁻¹(eˣ) + C, which is option (A).

Answer(A) tan⁻¹(eˣ) + C

Watch this explained “Multiplying by a chosen one”, 10:33 into Changing the variable until what is left is a standard form

Open NCERT p. 286Matches NCERT’s answer

1. Question: ∫cos2x/(sinx+cosx)² dx.
2. Write cos2x as a difference of squares: cos2x = cos²x − sin²x = (cosx − sinx)(cosx + sinx).
3. Divide top and bottom by one factor of (sinx+cosx): the integrand becomes (cosx − sinx)/(sinx+cosx).
4. Substitute u = sinx + cosx, so du = (cosx − sinx)dx: the integral becomes ∫du/u.
5. ∫du/u = log|u| + C.
6. Put u back as sinx+cosx: the answer is log|sinx+cosx| + C, which is option (B).

Answer(B) log|sinx+cosx| + C

Watch this explained “Substituting for the denominator”, 7:14 into Changing the variable until what is left is a standard form

Open NCERT p. 286Matches NCERT’s answer

1. Question: If f(a+b−x) = f(x), then ∫ₐᵇ x f(x) dx.
2. Call the integral I = ∫ₐᵇ x f(x) dx.
3. Replace x by (a+b−x) everywhere inside the integral (this does not change the value, since the limits stay a and b): I = ∫ₐᵇ (a+b−x) f(a+b−x) dx.
4. Use the given fact f(a+b−x) = f(x): I = ∫ₐᵇ (a+b−x) f(x) dx.
5. Split this into two integrals: I = (a+b)∫ₐᵇ f(x) dx − ∫ₐᵇ x f(x) dx = (a+b)∫ₐᵇ f(x) dx − I.
6. Add I to both sides: 2I = (a+b)∫ₐᵇ f(x) dx.
7. Divide by 2: I = (a+b)/2 ∫ₐᵇ f(x) dx, which is option (D).

Answer(D) (a+b)/2 ∫ₐᵇ f(x) dx

Watch this explained “The rule with the function taken out”, 18:50 into Properties of the definite integral, and using symmetry to kill half the work