int x operatorname{Tan}^{-1} sqrt{frac{1+x^2} | vedclass

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Question

(B) Let $I = \int x \operatorname{Tan}^{-1} \sqrt{\frac{1+x^2}{1-x^2}} \, dx$. Put $x^2 = \cos 	heta$, so $2x \, dx = -\sin 	heta \, d	heta$, which

Vedclass · Accepted Answer

$\frac{x^2}{4}\left(\pi-\operatorname{Cos}^{-1} x^2\right)+\frac{1}{4} \sqrt{1-x^4}+c$

Vedclass · Answer

(B) Let $I = \int x \operatorname{Tan}^{-1} \sqrt{\frac{1+x^2}{1-x^2}} \, dx$. Put $x^2 = \cos 	heta$, so $2x \, dx = -\sin 	heta \, d	heta$, which means $x \, dx = -\frac{1}{2} \sin 	heta \, d	heta$. The integral becomes $I = \int \operatorname{Tan}^{-1} \sqrt{\frac{1+\cos 	heta}{1-\cos 	heta}} \left(-\frac{1}{2} \sin 	hetaight) d	heta$. Using $\sqrt{\frac{1+\cos 	heta}{1-\cos 	heta}} = \sqrt{\frac{2\cos^2(	heta/2)}{2\sin^2(	heta/2)}} = \cot(	heta/2) = 	an(\frac{\pi}{2} - \frac{	heta}{2})$. So, $I = -\frac{1}{2} \int (\frac{\pi}{2} - \frac{	heta}{2}) \sin 	heta \, d	heta = -\frac{\pi}{4} \int \sin 	heta \, d	heta + \frac{1}{4} \int 	heta \sin 	heta \, d	heta$. $I = \frac{\pi}{4} \cos 	heta + \frac{1}{4} [-	heta \cos 	heta + \int \cos 	heta \, d	heta] = \frac{\pi}{4} \cos 	heta - \frac{1}{4} 	heta \cos 	heta + \frac{1}{4} \sin 	heta + c$. Since $\cos 	heta = x^2$, $	heta = \operatorname{Cos}^{-1} x^2$, and $\sin 	heta = \sqrt{1-x^4}$. $I = \frac{\pi}{4} x^2 - \frac{1}{4} x^2 \operatorname{Cos}^{-1} x^2 + \frac{1}{4} \sqrt{1-x^4} + c = \frac{x^2}{4} (\pi - \operatorname{Cos}^{-1} x^2) + \frac{1}{4} \sqrt{1-x^4} + c$.

$\int x \operatorname{Tan}^{-1} \sqrt{\frac{1+x^2}{1-x^2}} \, dx =$

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