Turn the radical into a trigonometric identity · source reference

Printed question

(a) Use $x=\tan t$, where $0<t<\pi/2$, to find $\int\frac1{x^2\sqrt{x^2+1}}dx$. [4]

(b)(i) Write down $\frac d{dx}(\cos x^3)$. [1]

(ii) Hence find $\int x^5\sin x^3dx$. [4]

[2023/SAJC/Prelim/II/3]

Printed working

(a)

$x=\tan t$, $dx/dt=\sec^2t$.

$$\begin{aligned}\int\frac{dx}{x^2\sqrt{x^2+1}}&=\int\frac{\sec^2t}{\tan^2t\sqrt{\tan^2t+1}}dt=\int\frac{\sec^2t}{\tan^2t(\sec t)}dt\\&=\int\frac{\sec t}{\tan^2t}dt=\int\frac{\cos t}{\sin^2t}dt=\int(\cos t)(\sin t)^{-2}dt\\&=-\frac1{\sin t}+C=-\frac{\sqrt{1+x^2}}x+C.\end{aligned}$$

$C$ is arbitrary. Alternatively $\int\cos t/\sin^2t\,dt=\int\csc t\cot t\,dt=-\csc t+C$.

(b)(i)

$$\frac d{dx}(\cos x^3)=-3x^2\sin x^3.$$

(b)(ii)

$u=x^3$, $dv/dx=x^2\sin x^3$, $du/dx=3x^2$, $v=-\frac13\cos x^3$.

$$\int x^5\sin x^3dx=\int x^3\cdot x^2\sin x^3dx=-\frac{x^3}3\cos x^3+\int x^2\cos x^3dx=-\frac{x^3}3\cos x^3+\frac13\int3x^2\cos x^3dx=-\frac{x^3}3\cos x^3+\frac13\sin x^3+C.$$

Teacher notes

Substitution: let $x=g(t)$, $dx/dt=g'(t)$; $\int f(x)dx=\int f(g(t))g'(t)dt$. Express the integrand in $t$ and the final indefinite answer in $x$.

Power formula: $\int f'(t)(f(t))^n dt=(f(t))^{n+1}/(n+1)+c$, with $f(t)=\sin t,n=-2$.

(b)(ii) integration by parts $\int u\,dv/dx\,dx=uv-\int v\,du/dx\,dx$.

Printed answer key

(a) $-\sqrt{1+x^2}/x+C$; (b)(i) $-3x^2\sin x^3$; (ii) $-(x^3/3)\cos x^3+\frac13\sin x^3+C$.

Source diagrams in native form