Questions on integrating standard functions

Question 1

Find ∫cos 2x dx, and ∫sin 4x dx.

Accepted Answer

∫cos 2x dx = ½ sin 2x + c, and ∫sin 4x dx = −¼ cos 4x + c. B1 B1, one for each integral. Where differentiation multiplied by k, integration divides by it. Sine also picks up a minus sign on the way back, so ∫sin x dx = −cos x + c.

Question 2

Find ∫e^3x dx, ∫(1/x) dx and ∫sec² x dx.

Accepted Answer

∫e^3x dx = e^3x/3 + c, ∫(1/x) dx = ln|x| + c, and ∫sec² x dx = tan x + c. B1 B1 B1, one for each integral. The modulus in the logarithm keeps it defined for negative x, where 1/x still has perfectly good areas beneath it.

Question 3

A student writes ∫sin 6x dx = 6 cos 6x + c. State the two errors, and give the correct integral.

Accepted Answer

Differentiating cos 6x gives −6 sin 6x, so reversing that derivative brings a minus sign with it, and the student's answer has kept a plus. B1 for the missing minus sign. The inner coefficient 6 comes down as a divisor when a derivative is reversed, because differentiation multiplied by it; the student has multiplied by 6 where the rule divides. B1 for the 6 dividing rather than multiplying. The integral is −(cos 6x)/6 + c. B1 for the correct answer. Differentiating 6 cos 6x returns −36 sin 6x, thirty-six times the integrand and of the opposite sign, so one line of checking catches both errors at once.

Question 4

Given that ∫₂^k (1/x) dx = ln 3, where k > 2, find the value of k.

Accepted Answer

The integral is [ln x] from 2 to k, and both limits are positive, so no modulus is needed: its value is ln k − ln 2. M1 for the bracketed form with both limits in place. Setting ln k − ln 2 = ln 3 and using the subtraction law gives ln (k/2) = ln 3, so k/2 = 3. M1 for compressing to a single logarithm and removing it. Hence k = 6. A1 for k = 6. The condition k > 2 puts the limits in increasing order and keeps x positive across the interval, which is what allows ln x in place of ln|x| here.

Question 5

Find ∫(e^2x + 3/x + sin 2x) dx.

Accepted Answer

Take the terms one at a time. The integral is e^2x/2 + 3 ln|x| − (cos 2x)/2 + c. B1 B1 B1, one for each term. The danger sits in the last term, where the sine returns a minus sign and the 2 divides at the same time.

Question 6

Find the exact value of ∫₁^e² (3/x) dx.

Accepted Answer

[3 ln x] from 1 to e² = 3 ln e² − 3 ln 1 = 6 − 0 = 6. M1 for 3 ln x, M1 for substituting both limits, A1 for the exact value. The limits are powers of e, so the logarithms evaluate exactly and no calculator is needed.

Question 7

Find the exact value of ∫₀^π/6 cos 3x dx.

Accepted Answer

[⅓ sin 3x] from 0 to π/6 = ⅓ sin (π/2) − 0 = ⅓. M1 for the integrated form, M1 for substituting both limits, A1 for the exact value. The upper limit was chosen so that 3x lands on π/2, where sine is exactly 1. Set the calculator to radians before checking, because degree mode returns a value near 0.0091.

Question 8

Find ∫cos² x dx.

Accepted Answer

Nothing on the standard list covers cos² x, so rewrite it first. The double-angle identity gives cos² x = ½ + ½ cos 2x. Integrating term by term gives x/2 + (sin 2x)/4 + c. M1 for using the double-angle identity, A1 for ½ + ½ cos 2x, M1 for integrating term by term, A1 for the answer. Guessing (sin³ x)/3 or similar is wrong.

Question 9

Find ∫tan² 4x dx.

Accepted Answer

Nothing on the standard list is a squared tangent, so the integrand has to be reshaped first. The identity 1 + tan²A = sec²A, with the angle 4x in place of A, gives tan² 4x = sec² 4x − 1. M1 for using the sec² identity, A1 for sec² 4x − 1. Both terms are now standard: ∫sec² 4x dx = (tan 4x)/4, with the inner coefficient dividing, and ∫1 dx = x. M1 for integrating both terms with the division by 4. The integral is (tan 4x)/4 − x + c. A1 for the answer. The −1 in the identity is a term of the integrand and contributes the −x; an answer of (tan 4x)/4 + c has dropped it. Differentiating the answer gives sec² 4x − 1, which is the integrand.

Question 10

Find the exact value of ∫₀^ln 2 (e^2x + e^−x) dx.

Accepted Answer

Take the terms one at a time, dividing by the inner coefficient each time: ∫e^2x dx = e^2x/2 and ∫e^−x dx = −e^−x, the second divided by −1. M1 for both antiderivatives with their divisors, A1 for [e^2x/2 − e^−x]. At the upper limit, e^2 ln 2 = e^ln 4 = 4 and e^−ln 2 = ½, so the bracket is 4/2 − ½ = 3/2; at the lower limit it is ½ − 1 = −½. M1 for substituting both limits and using the log laws on the exponents. The value is 3/2 − (−½) = 2. A1 for 2.

Question 11

Find ∫2 sin 4x cos 4x dx, and check your answer by differentiating it.

Accepted Answer

No standard result covers a sine multiplied by a cosine, but the double-angle identity sin 2A = 2 sin A cos A does, with the angle 4x in place of A: 2 sin 4x cos 4x = sin 8x. M1 for the identity, A1 for sin 8x. That is now a single standard integral, and the inner coefficient 8 divides: −(cos 8x)/8 + c. A1 for the integral. Checking, the derivative of cos 8x is −8 sin 8x, so the derivative of −(cos 8x)/8 is sin 8x, which is the integrand in its rewritten form. B1 for the check.

Question 12

Show that ∫₀^π/8 sin² 2x dx = π/16 − 1/8.

Accepted Answer

Start from cos 4x = 1 − 2 sin² 2x, which rearranges to sin² 2x = ½ − ½ cos 4x. Integrating gives [x/2 − (sin 4x)/8]. At the upper limit, x/2 = π/16 and sin 4x = sin (π/2) = 1, so the bracket is π/16 − 1/8. At the lower limit it is 0. The value is therefore π/16 − 1/8, as required. M1 for using cos 4x, A1 for ½ − ½ cos 4x, M1 for integrating, A1 for x/2 − (sin 4x)/8, A1 for reaching the printed answer. The identity needs the double of the angle already present, so sin² 2x calls for cos 4x rather than cos 2x, and the k-factor of 4 then divides the sine term to give the 8 in the denominator. Writing ½ − ½ cos 2x uses cos 2x in place of cos 4x and reaches π/16 − √2/8.

Question 13

Show that ∫₋₄⁻¹ (1/x) dx = −ln 4, and explain why the same method cannot be used to find ∫₋₁² (1/x) dx.

Accepted Answer

Every x between −4 and −1 is negative, which is what the modulus in ∫(1/x) dx = ln|x| + c is there for. The value is [ln|x|] from −4 to −1 = ln 1 − ln 4. M1 for the bracketed form with the modulus and both limits. That is 0 − ln 4 = −ln 4, which may also be written −2 ln 2 and is about −1.386. A1 for ln 1 − ln 4, A1 for −ln 4. The value is negative because 1/x is negative right across [−4, −1], so the region lies below the axis; the size of that region is ln 4. B1 for reading the sign. The second integral is a different matter. Its interval contains x = 0, where 1/x is not defined and grows without bound on either side, so no single antiderivative is valid across the whole of [−1, 2] and substituting the limits into ln|x| would require ln 0. B1 for the point x = 0 lying inside the interval.

Question 14

Show, without evaluating either integral, that ∫₀^π/2 sin² x dx and ∫₀^π/2 cos² x dx are equal, and hence find their common value.

Accepted Answer

Write S for the first and T for the second. Adding them, S + T = ∫₀^π/2 (sin² x + cos² x) dx = ∫₀^π/2 1 dx = π/2, since the Pythagorean identity turns the integrand into 1. B1 for S + T = π/2. For the equality, cos x = sin (π/2 − x), so cos² x = sin² (π/2 − x): on [0, π/2] the graph of cos² x is the graph of sin² x reflected in the line x = π/4, and reflecting a region does not change its area. M1 for the reflection, A1 for S = T. Two equal numbers whose sum is π/2 are each half of it. M1 for halving. So S = T = π/4. A1 for π/4. Integrating ½ − ½ cos 2x confirms the value, since the sine term happens to vanish at both ends of this interval, but that route is not needed.

Question 15

Given that ∫f(x) dx = F(x) + c, prove that ∫f(kx) dx = F(kx)/k + c for any non-zero constant k. Hence write down ∫cos 5x dx and ∫e^x/3 dx, and explain why the same argument does not give ∫f(x²) dx = F(x²)/(2x) + c.

Accepted Answer

Differentiate the proposed answer. By the chain rule the derivative of F(kx) is F′(kx) × k, so the derivative of F(kx)/k is F′(kx), and F′ = f, so it is f(kx). M1 for differentiating F(kx) by the chain rule. The proposal therefore differentiates to the integrand, which is what it means to be an antiderivative, and the constant k the chain rule introduces is cancelled by the k already dividing. A1 for the cancellation, which is the proof. Taking f = cos and F = sin with k = 5 gives ∫cos 5x dx = (sin 5x)/5 + c. B1. Taking f and F both to be the exponential with k = 1/3, dividing by 1/3 means multiplying by 3, so ∫e^x/3 dx = 3e^x/3 + c. B1. For F(x²) the chain rule produces F′(x²) × 2x, and 2x is not a constant, so it cannot be taken outside the integral sign and cancelled; differentiating F(x²)/(2x) needs the quotient rule and returns f(x²) − F(x²)/(2x²), which is not the integrand. B1 for the multiplier not being constant.