Distinguish cos(x²) from cos²x · source reference

Printed question

(a) Find $\int x\cos x^2 dx$. [1]

(b) Use integration by parts to find $\int x\cos2x dx$. [3]

Hence or otherwise find $\int_{\pi/4}^{\pi/2}x\cos^2x dx$. [3]

[2020/RI/Prelim/I/5]

Printed working

(a)

$$\int x\cos x^2dx=\frac12\sin x^2+c.$$

(b)

$u=x$, $dv/dx=\cos2x$, $du/dx=1$, $v=\frac12\sin2x$.

$$\int x\cos2x dx=\frac x2\sin2x-\frac12\int\sin2x dx=\frac x2\sin2x+\frac{\cos2x}4+c.$$ $$\begin{aligned}\int_{\pi/4}^{\pi/2}x\cos^2x dx&=\int_{\pi/4}^{\pi/2}x\frac{\cos2x+1}2dx\\&=\frac12\int_{\pi/4}^{\pi/2}x\cos2x dx+\frac12\int_{\pi/4}^{\pi/2}x dx\\&=\frac12\left[\frac x2\sin2x+\frac{\cos2x}4\right]_{\pi/4}^{\pi/2}+\frac12[x^2/2]_{\pi/4}^{\pi/2}\\&=\frac12\left[(0-1/4)-(\pi/8+0)\right]+\frac14(\pi^2/4-\pi^2/16)\\&=\frac{3\pi^2}{64}-\frac18-\frac\pi{16}.\end{aligned}$$

Teacher notes

Use $\int f'(x)\cos f(x)dx=\sin f(x)+c$ for (a); and integration by parts for (b). Use $\int f'(x)\sin f(x)dx=-\cos f(x)+c$ on the remaining sine integral.

From MF27: $\cos2A=2\cos^2A-1$. Need to re-express $\cos^2x/\sin^2x$ in terms of multiple angles, i.e. $\cos2x$, in order to integrate directly. The starred integral is integrated by parts with $u=x$, $dv/dx=\cos2x$.

Printed answer key

(a) $\frac12\sin x^2+c$; (b) $(x/2)\sin2x+\cos2x/4+c$; $3\pi^2/64-1/8-\pi/16$.