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

Question

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

✓

Answer

Let $\text{I =}\int\tan^{-1}\sqrt{\frac{1-\text{x}}{1+\text{x}}}\text{dx}$
Putting $\text{x}=\cos\theta$
$\Rightarrow\text{dx}=-\sin\theta\text{d}\theta$
$\&\theta=\cos^{-1}\text{x}$
$\therefore\text{I}=\int\tan^{-1}\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}(-\sin\theta)\text{d}\theta$
$=\int\tan^{-1}\sqrt{\frac{2\sin^2\frac{\theta}{2}}{2\cos^2\frac{\theta}{2}}}(-\sin\theta)\text{d}\theta$
$=\int\tan^{-1}\Big(\tan\frac{\theta}{2}\Big)(-\sin\theta)\text{d}\theta$
$=-\frac{1}{2}\int\theta\sin\theta\text{d}\theta$
$=-\frac{1}{2}\Big[\theta\int\sin\theta\text{d}\theta-\int\Big\{\Big(\frac{\text{d}}{\text{d}\theta}\theta\Big)\int\sin\theta\text{d}\theta\Big\}\text{d}\theta\Big]$
$=-\frac{1}{2}\big[\theta(-\cos\theta)-\int1.(-\cos\theta)\text{d}\theta\big]$
$=-\frac{1}{2}\big[-\theta\cos\theta+\sin\theta\big]+\text{C}$
$=-\frac{1}2{}\Big[-\theta.\cos\theta+\sqrt{1-\cos^2\theta}\Big]+\text{C}$
$=-\frac{1}{2}\Big[-\cos^{-1}\text{x.x}+\sqrt{1-\text{x}^2}\Big]+\text{C}$ $\big[\because\theta=\cos^{-1}\text{x}\big]$
$=\frac{\text{x}\cos^{-1}\text{x}}{2}-\frac{\sqrt{1-\text{x}^2}}{2}+\text{C}$

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