Evaluate the following integrals: ^1_0 1-x 1+x dx

Question

Evaluate the following integrals:
$\int\limits^1_0\sqrt{\frac{1-\text{x}}{1+\text{x}}}\text{ dx}$

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Answer

Let $\text{x}=\cos2\theta$
Differentiating w.r.t. x, we get
$\text{dx}=-2\sin2\theta\text{ d}\theta$
Now, $\text{x}=0\Rightarrow\theta=\frac{\pi}{4}$
$\text{x}=1\Rightarrow\theta=0$
$\therefore\ \int^\limits1_0\sqrt{\frac{1-\text{x}}{1+\text{x}}}\text{ dx}=\int^0\limits_\frac{\pi}{4}\sqrt{\frac{1-\cos2\theta}{1+\cos2\theta}}\big(-2\sin2\theta\big)\text{d}\theta$
$=\int_0\limits^\frac{\pi}{4}\sqrt{\frac{1-\cos2\theta}{1+\cos2\theta}}\big(2\sin2\theta\big)\text{d}\theta$ $\Big[\because\sin2\theta=2\sin\theta\cos\theta;\text{ and }\sin^2\theta=\frac{1-\cos2\theta}{2}\Big]$
$=2=\int_0\limits^\frac{\pi}{4}\frac{\sin\theta}{\cos\theta}\cdot\sin2\theta\text{ d}\theta$
$=4\int_0\limits^\frac{\pi}{4}\sin^2\theta\text{ d}\theta$
$=2\int_0\limits^\frac{\pi}{4}\big(1-\cos2\theta\big)\text{d}\theta$
$=2\Big[\theta-\frac{\sin^2\theta}{2}\Big]^{\frac{\pi}{4}}_0$
$=2\Big[\frac{\pi}{4}-\frac{1}{2}\Big]$
$=\frac{\pi}{2}-1$

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