Q15 (c)

Criteria	Marks
Provides correct solution	4
Attempts to use integration by parts with the correct integral, or equivalent merit	3
Obtains integral in terms of $\theta$, or equivalent merit	2
Attempts to use the given substitution, or equivalent merit	1

Sample answer:
Let $x=\tan^2\theta$.
$x=0\Rightarrow\theta=0,\qquad x=1\Rightarrow\theta=\frac\pi4$
$dx=2\tan\theta\sec^2\theta\,d\theta$

$$\sin^{-1}\sqrt{\frac{x}{1+x}} =\sin^{-1}\sqrt{\frac{\tan^2\theta}{\sec^2\theta}} =\sin^{-1}\sqrt{\sin^2\theta}=\theta$$

for $0\le\theta\le\frac\pi4$, as $\sin\theta\ge0$.

$$\begin{aligned} \int_0^1\sin^{-1}\sqrt{\frac{x}{1+x}}\,dx &=\int_0^{\pi/4}2\theta\tan\theta\sec^2\theta\,d\theta\\ &=\left[\theta\tan^2\theta\right]_0^{\pi/4}-\int_0^{\pi/4}\tan^2\theta\,d\theta\\ &=\frac\pi4-\int_0^{\pi/4}(-1+\sec^2\theta)\,d\theta\\ &=\frac\pi4-[-\theta+\tan\theta]_0^{\pi/4}\\ &=\frac\pi4+\frac\pi4-1=\frac\pi2-1 \end{aligned}$$