Porisma

Worked solution

Lower even powers with double angles

Find (a) ∫cos⁡2x dx\int\cos^2x\,dx; (b) ∫2sin⁡2(3θ) dθ\int2\sin^2(3\theta)\,d\theta.

CJC · Lecture · Example 12

(a)

cos⁡2x=(1+cos⁡2x)/2\cos^2x=(1+\cos2x)/2, so ∫cos⁡2x dx=12∫(1+cos⁡2x) dx=12(x+12sin⁡2x)+C=12x+14sin⁡2x+C\int\cos^2x\,dx=\frac12\int(1+\cos2x)\,dx=\frac12(x+\frac12\sin2x)+C=\frac12x+\frac14\sin2x+C.

How did this part go?

(b)

2sin⁡2(3θ)=1−cos⁡6θ2\sin^2(3\theta)=1-\cos6\theta. Thus ∫2sin⁡2(3θ) dθ=∫(1−cos⁡6θ) dθ=θ−16sin⁡6θ+C\int2\sin^2(3\theta)\,d\theta=\int(1-\cos6\theta)\,d\theta=\theta-\frac16\sin6\theta+C.

How did this part go?

Common mistakes

The Teacher crosses out the tempting cos cubed x over 3 answer: the power rule cannot be applied to cosine without its inner derivative.

Read the hint