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Integrating standard functions questions
Every derivative fact from the differentiation unit now runs backwards: exponentials, sines, cosines, and at long last 1 over x, whose integral is the logarithm the power rule could never produce. Where a function does not integrate directly, a trig identity can reshape it into a form that does.
7 original questions · 23 marks · the integrating standard functions notes · Integration
Every question here is written for this library rather than taken from a past paper. Write your answer out before opening the worked one: the answers award marks point by point, and the marks are easier to see when you have something of your own to compare against.
Find ∫cos 2x dx, and ∫sin 4x dx.
Worked 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, and dropping it is the single commonest error in this topic.Find ∫e3x dx, ∫(1/x) dx and ∫sec2 x dx.
Worked answer
∫e3x dx = e3x/3 + c, ∫(1/x) dx = ln|x| + c, and ∫sec2 x dx = tan x + c. B1 B1 B1, one for each integral. Each one is a derivative fact read backwards. The modulus in the logarithm keeps it defined for negative x, where 1/x still has perfectly good areas beneath it, and a missing modulus quietly narrows the answer to positive x alone.Find ∫(e2x + 3/x + sin 2x) dx.
Worked answer
Take the terms one at a time. The integral is e2x/2 + 3 ln|x| − (cos 2x)/2 + c. B1 B1 B1, one for each term. Three standard results, one line each. The danger sits in the last term, where the sine returns a minus sign and the 2 divides at the same time.Find the exact value of ∫1e² (3/x) dx.
Worked 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. Writing 5.999 throws away the accuracy mark on an exact-value question.Find the exact value of ∫0π/6 cos 3x dx.
Worked 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 and no marks.Find ∫cos2 x dx.
Worked 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. Identities come before integrals. The algebra turns an impossible integrand into two easy ones, and the alternative of guessing (sin3 x)/3 or similar is simply wrong.Show that ∫0π/8 sin2 2x dx = π/16 − 1/8.
Worked 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. Two things have to go right at once. 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. Candidates who write ½ − ½ cos 2x reach π/16 − √2/8, which is wrong and unrecoverable.
The same practice on paper: the printable workbook for this topic, questions and a worked answer book.
Practise integrating standard functions one question at a time
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