Question
Integrate the function in Exercise:$\tan^{-1}\sqrt{\frac{1-\text{x}}{1+\text{x}}}$
✓
Answer
$\text{I}=\tan^{-1}\sqrt{\frac{1-\text{x}}{1+\text{x}}}\text{dx}$
$\text{Let x}=\cos\theta\Rightarrow\text{dx}=-\sin\theta\text{d}\theta$
$\text{I}=\int\tan^{-1}\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}(-\sin\theta\text{d}\theta)$
$=-\int^{-1}\sqrt{\frac{2\sin^{2}\frac{\theta}{2}}{2\cos^{2}\frac{\theta}{2}}}\sin\theta\text{d}\theta$
$=-\int\tan^{-1}\tan\frac{\theta}{2}.\sin\theta\text{d}\theta$
$=-\frac{1}{2}\big[\theta.(-\cos\theta)-\int1.(-\cos\theta)\text{d}\theta\big]$
$=-\frac{1}{2}[-\theta\cos\theta+\sin\theta]$
$=+\frac{1}{2}\theta\cos\theta-\frac{1}{2}\sin\theta$
$=\frac{1}{2}\cos^{-1}\text{x}.\text{x}-\frac{1}{2}\sqrt{1-\text{x}^{2}}+\text{C}$
$=\frac{\text{x}}{2}\cos^{-1}\text{x}-\frac{1}{2}\sqrt{1-\text{x}^{2}}+\text{C}$
$=\frac{1}{2}\Big(\text{x}\cos^{-1}\text{x}-\sqrt{1-\text{x}^{2}}\Big)+\text{C}$
$\text{Let x}=\cos\theta\Rightarrow\text{dx}=-\sin\theta\text{d}\theta$
$\text{I}=\int\tan^{-1}\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}(-\sin\theta\text{d}\theta)$
$=-\int^{-1}\sqrt{\frac{2\sin^{2}\frac{\theta}{2}}{2\cos^{2}\frac{\theta}{2}}}\sin\theta\text{d}\theta$
$=-\int\tan^{-1}\tan\frac{\theta}{2}.\sin\theta\text{d}\theta$
$=-\frac{1}{2}\big[\theta.(-\cos\theta)-\int1.(-\cos\theta)\text{d}\theta\big]$
$=-\frac{1}{2}[-\theta\cos\theta+\sin\theta]$
$=+\frac{1}{2}\theta\cos\theta-\frac{1}{2}\sin\theta$
$=\frac{1}{2}\cos^{-1}\text{x}.\text{x}-\frac{1}{2}\sqrt{1-\text{x}^{2}}+\text{C}$
$=\frac{\text{x}}{2}\cos^{-1}\text{x}-\frac{1}{2}\sqrt{1-\text{x}^{2}}+\text{C}$
$=\frac{1}{2}\Big(\text{x}\cos^{-1}\text{x}-\sqrt{1-\text{x}^{2}}\Big)+\text{C}$
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