Questions on integration by substitution and by parts

Question 1

Find ∫2x(x² + 1)³ dx by recognising a reverse chain pattern.

Accepted Answer

The bracket's derivative, 2x, is sitting alongside as a factor, so the integrand is exactly what the chain rule leaves behind. The integral is (x² + 1)⁴/4 + c, the next power up divided by that power. M1 for the recognition, A1 for the answer including the constant.

Question 2

Find ∫2x/(x² + 3) dx.

Accepted Answer

The numerator is exactly the derivative of the denominator, so the f′/f pattern applies and the integral is ln(x² + 3) + c. M1 A1. No modulus signs are needed, since x² + 3 is positive for every real x.

Question 3

Find ∫sin³ x cos x dx.

Accepted Answer

The integrand is a power of sin x multiplied by the derivative of sin x, so it is a reverse chain in sin x: raise the power by one and divide by the new power. The integral is (sin⁴ x)/4 + c. M1 for recognising cos x as the derivative of the function being raised to a power, A1 for the answer including the constant. No substitution is needed once the pattern is seen, and differentiating the answer by the chain rule returns 4 sin³ x cos x ÷ 4.

Question 4

Show that ∫tan x dx = −ln|cos x| + c, and hence write down ∫tan 4x dx.

Accepted Answer

Write tan x as sin x/cos x. The derivative of cos x is −sin x, so the numerator is minus the derivative of the denominator and the integrand is −f′/f with f = cos x. M1 for rewriting as sin x/cos x and identifying the f′/f pattern with its minus sign. An f′/f integrand integrates to ln|f|, so this one integrates to −ln|cos x| + c. A1 for the printed result. The same shape with an inner coefficient of 4 divides by 4: ∫tan 4x dx = −(ln|cos 4x|)/4 + c. B1 for that. The modulus is needed because cos x is negative on half of its cycle.

Question 5

Use the substitution u = x² + 1 to find the exact value of ∫₀² x(x² + 1)³ dx.

Accepted Answer

From u = x² + 1, du = 2x dx, so x dx = du/2. The limits convert as well: x = 0 gives u = 1 and x = 2 gives u = 5. The integral becomes ½∫₁⁵ u³ du = ½[u⁴/4] from 1 to 5 = (625 − 1)/8 = 78. M1 for du/dx, M1 for the substitution, A1 for the integral fully in u with converted limits, M1 for integrating, A1 for 78. Convert the limits with the substitution: carrying 0 and 2 into a u-integral gives (16 − 0)/8 = 2, which integrates the wrong range.

Question 6

Using a suitable substitution, find ∫x√(2x − 1) dx.

Accepted Answer

The awkward part is the root of a linear expression, so take that expression as u: u = 2x − 1. Then x = (u + 1)/2. M1 for the choice with x written in terms of u. Differentiating, du/dx = 2, so dx = du/2. M1 for dx. The integral becomes ∫ ((u + 1)/2) × u^1/2 × (du/2) = ¼∫(u^3/2 + u^1/2) du, once the bracket is multiplied through by the root. A1 for the integral wholly in u with nothing in x left. Integrating the two powers gives ¼[(2/5)u^5/2 + (2/3)u^3/2] = u^5/2/10 + u^3/2/6. M1 for integrating both powers. Since the integral is indefinite there are no limits to convert, but the answer has to go back into x: (2x − 1)^5/2/10 + (2x − 1)^3/2/6 + c. A1 for the answer in x.

Question 7

A student tries to find ∫x e^x² dx by parts, with u = x and dv/dx = e^x², and cannot get past the first line. Explain why parts is the wrong method here, and find the integral.

Accepted Answer

Parts needs v, and v is ∫e^x² dx. That integral is not any combination of the standard functions, so there is no v to write down and the first line of the formula cannot be completed. B1 for v being unavailable. Look at the integrand again instead. The derivative of x² is 2x, and the factor sitting outside the exponential is x, which is half of it. B1 for x as half the derivative of the exponent. So the integrand is ½ × 2x e^x², a reverse chain: something of the form f′(x)e^f(x) integrates to e^f(x). M1 for the reverse chain. The integral is ½e^x² + c. A1 for the answer.

Question 8

Use the substitution u = 1 + √x to show that ∫₁⁴ 1/(1 + √x) dx = 2 − 2 ln (3/2).

Accepted Answer

From u = 1 + √x, √x = u − 1 and so x = (u − 1)², giving dx = 2(u − 1) du. The limits move too: x = 1 gives u = 2 and x = 4 gives u = 3. The integral becomes ∫₂³ 2(u − 1)/u du = 2∫₂³ (1 − 1/u) du = 2[u − ln u] from 2 to 3 = 2[(3 − ln 3) − (2 − ln 2)] = 2[1 − ln (3/2)] = 2 − 2 ln (3/2), as required. M1 for x in terms of u, M1 for dx, A1 for the new limits, M1 for the integrand fully in u, M1 for integrating, A1 for the printed form.

Question 9

Use the substitution x = 2 sin θ to show that ∫₀² √(4 − x²) dx = π, and explain how the value can be checked from the graph of y = √(4 − x²).

Accepted Answer

A root of a difference of two squares is what x = a sin θ is for, with a² = 4 here, so a = 2. Differentiating, dx = 2 cos θ dθ. M1 for dx. Under the substitution 4 − x² = 4 − 4 sin²θ = 4 cos²θ by the Pythagorean identity, so the root is 2 cos θ; θ runs over [0, π/2], where cos θ is not negative, so no modulus is needed. M1 for the root becoming 2 cos θ, with the sign settled. The limits move with the substitution: x = 0 gives sin θ = 0 and θ = 0, and x = 2 gives sin θ = 1 and θ = π/2. A1 for the new limits. The integral is now ∫₀^π/2 2 cos θ × 2 cos θ dθ = 4∫₀^π/2 cos²θ dθ, and cos²θ = ½ + ½ cos 2θ. M1 for the double-angle identity. So the value is 4[θ/2 + (sin 2θ)/4] from 0 to π/2 = 4(π/4 + 0) − 0 = π, as required. A1 for reaching π. For the check, y = √(4 − x²) is the upper half of the circle x² + y² = 4, so the region for 0 ≤ x ≤ 2 is a quarter of a disc of radius 2, of area ¼ × π × 2² = π. B1 for the quarter-circle check.

Question 10

Use integration by parts to find ∫xe^2x dx.

Accepted Answer

Choose the factor that simplifies when differentiated, so u = x and dv/dx = e^2x. Then ∫xe^2x dx = (x/2)e^2x − ∫(1/2)e^2x dx = (x/2)e^2x − e^2x/4 + c. M1 for a correct application of the formula, A1 for the first stage, M1 for integrating the remaining term, A1 for the answer with its constant.

Question 11

Use integration by parts to find ∫x ln x dx.

Accepted Answer

Here the logarithm has to be the differentiated factor, because ∫ln x dx is not a standard result you can quote. Take u = ln x and dv/dx = x. Then ∫x ln x dx = (x²/2) ln x − ∫(x²/2)(1/x) dx = (x²/2) ln x − x²/4 + c. M1 for the choice and the formula, A1 for the first stage, M1 for simplifying x²/2 × 1/x to x/2, A1 for the answer.

Question 12

Find the exact value of ∫₀^π/2 x sin x dx.

Accepted Answer

Parts with u = x and dv/dx = sin x gives [−x cos x]₀^π/2 + ∫₀^π/2 cos x dx = [−x cos x + sin x] from 0 to π/2. At the upper limit that is −(π/2)(0) + 1 = 1, and at the lower limit it is 0 + 0 = 0, so the value is exactly 1. M1 for the parts, A1 for −x cos x + sin x, M1 for substituting the limits, A1 for 1. The boundary term vanishes at both ends for different reasons, cos (π/2) = 0 at the top and x = 0 at the bottom, and the two minus signs in −∫(−cos x) dx cancel.

Question 13

Find ∫ln x dx, and hence find the exact value of ∫₁^e ln x dx.

Accepted Answer

There is only one factor, so supply a second: write the integrand as ln x × 1, take u = ln x and dv/dx = 1, giving v = x. M1 for taking dv/dx = 1. Then ∫ln x dx = x ln x − ∫x × (1/x) dx = x ln x − ∫1 dx. A1 for that first stage. The remaining integral is x, so ∫ln x dx = x ln x − x + c. A1 for the antiderivative. Now the definite value. At x = e the bracket is e ln e − e = e − e = 0, and at x = 1 it is 1 × 0 − 1 = −1. M1 for substituting both limits, A1 for the two bracket values. The value is 0 − (−1) = 1. A1 for 1. Differentiating x ln x − x gives ln x + 1 − 1, which is the check.

Question 14

Find the exact value of ∫₀¹ x² e^x dx, giving your answer in terms of e.

Accepted Answer

Parts with u = x² and dv/dx = e^x gives x²e^x − 2∫x e^x dx. The remaining integral needs parts a second time, giving ∫x e^x dx = x e^x − e^x. So the antiderivative is x²e^x − 2x e^x + 2e^x. At x = 1 that is e − 2e + 2e = e, and at x = 0 it is 0 − 0 + 2 = 2, so the value is e − 2. M1 for the first parts, A1 for the first stage, M1 for the second parts, A1 for the full antiderivative, A1 for e − 2. Each application drops the power on x by one, so a polynomial of degree n needs n rounds and the sign alternates as you go.

Question 15

Find ∫e^2x sin x dx.

Accepted Answer

Neither factor simplifies when differentiated, so parts will not terminate; it comes back to where it started instead, and that is what makes the answer available. Write I for the integral. With u = sin x and dv/dx = e^2x, so v = e^2x/2, I = (e^2x sin x)/2 − ½∫e^2x cos x dx. M1 for the first application, A1 for that stage. Apply parts again to the new integral, with u = cos x and the same dv/dx: ∫e^2x cos x dx = (e^2x cos x)/2 + ½∫e^2x sin x dx, the plus coming from the minus in the derivative of cos x meeting the minus in the formula. M1 for the second application, A1 for the second stage. The last integral is I again, so substituting gives I = (e^2x sin x)/2 − (e^2x cos x)/4 − I/4, and collecting, (5/4)I = e^2x(2 sin x − cos x)/4. M1 for recognising I and solving for it. Hence I = e^2x(2 sin x − cos x)/5 + c. A1 for the answer with its constant. Both applications must differentiate the trigonometric factor; swapping roles halfway undoes the first application and returns the identity I = I. The constant is added at the end, after the equation for I has been solved, and differentiating the answer by the product rule returns e^2x sin x.

Questions on integration by substitution and by parts

Reverse chain patterns and substitution

Integration by parts