Evaluate the following definite integrals: _ 0 ^ 1 1-x 1+x dx

Question

Evaluate the following definite integrals:
$\int\limits_{0}^{1}\frac{1-\text{x}}{1+\text{x}}\text{ dx}$

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Answer

We have,
$\int_{0}^\limits{1}\frac{1-\text{x}}{1+\text{x}}\text{ dx}$
Let $\text{x}=\cos2\theta$
$\text{dx}=-2\sin2\theta\text{ d}\theta$
Now,
$\int_{0}^\limits{1}\frac{1-\text{x}}{1+\text{x}}\text{ dx}$
$=\int_{\frac{\pi}{4}}^\limits{0}\frac{1-\cos2\theta}{1+\cos2\theta}\times(-2\sin2\theta)\text{d}\theta$
$\int_{\frac{\pi}{4}}^\limits{0}\frac{2\sin^2\theta}{2\cos^2\theta}\times2\sin2\theta\text{ d}\theta$ $\bigg[\because\ -\int_{\text{a}}^\limits{\text{b}}\text{f(x)}\text{dx}=\int_{\text{b}}^\limits{\text{a}}\text{f(x)}\text{dx}\bigg]$
$=\int_{\frac{\pi}{4}}^\limits{0}\frac{4\sin^3\theta}{\cos\theta}\text{ d}\theta$
Let $\cos\theta=\text{t}$
$-\sin\theta\text{ d}\theta=\text{dt}$
Now,
$\theta=0\Rightarrow\text{t}=1$
$\theta=\frac{\pi}{4}\Rightarrow\text{t}=\frac{1}{\sqrt{2}}$
$\therefore\ \int_{0}^\limits{\frac{\pi}{4}}\frac{4\sin^3\theta}{\cos\theta}\text{ d}\theta$
$=-4\int_{1}^\limits{\frac{1}{\sqrt{2}}}\frac{\big(1-\text{t}^2\big)}{\text{t}}\text{ dt}$
$=-4\Big[\log\text{t}-\frac{\text{t}^2}{2}\Big]^{\frac{1}{\sqrt{2}}}_1$
$=-4\Big[\log\Big(\frac{1}{\sqrt{2}}\Big)-\frac{1}{4}-0+\frac{1}{2}\Big]$
$=-4\Big[\log\sqrt{2}+\frac{1}{4}\Big]$
$\therefore\ \int_{0}^\limits{1}\frac{1-\text{x}}{1+\text{x}}\text{ dx}=2\log2-1$

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