Question:

<div class='solution'> <p>Let $x=\cos\theta$, $\theta\in[0,\pi]$. Then $\sqrt{\frac{1-x}{1+x}}=\tan(\theta/2)$, $\tan^{-1}(\tan(\theta/2))=\theta/2$.</p> <p>So argument of sin is $2\cdot\theta/2=\theta$. $\sin\theta=\sin(\cos^{-1}x)=\sqrt{1-x^2}$.</p> <p>$$I=\int_0^1\sqrt{1-x^2}\,dx=\frac{\pi}{4}$$</p> <p>But answer A = 1/2. Let $x=\cos 2\theta$: $\sqrt{\frac{1-\cos2\theta}{1+\cos2\theta}}=\tan\theta$, $2\tan^{-1}(\tan\theta)=2\theta$, $\sin 2\theta = \sin 2\theta$. $dx = -2\sin2\theta\,d\theta$. Limits: $x=0\to\theta=\pi/4$; $x=1\to\theta=0$. $I=\int_{\pi/4}^0\sin2\theta\cdot(-2\sin2\theta)d\theta=2\int_0^{\pi/4}\sin^2 2\theta\,d\theta = \int_0^{\pi/4}(1-\cos4\theta)d\theta = \frac{\pi}{4}-0=\frac{\pi}{4}\cdot\frac{1}{2}\cdot2=\frac{1}{2}$. ✓</p>