Porisma

Worked solution

Apply double-angle and product-to-sum identities before integrating

Find the following integrals:

(a) ∫11+cos⁡4θ dθ\displaystyle\int\dfrac1{1+\cos4\theta}\,d\theta, marks

(b) ∫cos⁡2xsin⁡3x dx\displaystyle\int\cos2x\sin3x\,dx, given 2sin⁡Acos⁡B=sin⁡(A+B)+sin⁡(A−B)2\sin A\cos B=\sin(A+B)+\sin(A-B). marks

[2016/NYJC/Promo/3 modified]

CJC · Review Problems · Question 7

Answer

(a) 14tan⁡2θ+C\frac14\tan2\theta+C; (b) −110cos⁡5x−12cos⁡x+C-\frac1{10}\cos5x-\frac12\cos x+C.

(a)

Use the source double-angle guidance: 1+cos⁡4θ=2cos⁡22θ1+\cos4\theta=2\cos^22\theta. Then 12∫sec⁡22θ dθ=14tan⁡2θ+C\frac12\int\sec^22\theta\,d\theta=\frac14\tan2\theta+C.

How did this part go?

(b)

Use the supplied factor formula with A=3xA=3x, B=2xB=2x: cos⁡2xsin⁡3x=12(sin⁡5x+sin⁡x)\cos2x\sin3x=\frac12(\sin5x+\sin x). Integrating gives −110cos⁡5x−12cos⁡x+C-\frac1{10}\cos5x-\frac12\cos x+C.

How did this part go?

Common mistakes

The doubled angle in sec²(2θ) supplies another factor 1/2 in its antiderivative.

Read the hint