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Question Number 60376 by rahul 19 last updated on 20/May/19

Commented by rahul 19 last updated on 20/May/19

$$dx{}^{\ast }$$

Commented by Mr X pcx last updated on 20/May/19

$$changementx={tan}^2\theta give$$$$I=\int \frac{1}{tan\theta (1+{tan}^2\theta )}arcos\left(cos\left(2\theta \right)\right)2tan\theta (1+{tan}^2\theta )d\theta$$$$=4\theta =4arctan\left(\sqrt{x}\right)+C.$$

Answered by tanmay last updated on 20/May/19

$$x={tan}^2\theta$$$$dx=2tan\theta \times {sec}^2\theta d\theta$$$$\int \frac{1}{tan\theta {sec}^2\theta }\times 2\theta \times 2tan\theta {sec}^2\theta d\theta$$$$=4\times \frac{{\theta }^2}{2}+c$$$$=2{\left({tan}^{-1}\sqrt{x}\right)}^2+c$$$$crosscheck...$$$$y=2{\left({tan}^{-1}\sqrt{x}\right)}^2+c$$$$\frac{dy}{dx}=4\left({tan}^{-1}\sqrt{x}\right)\times \frac{1}{1+x}\times \frac{1}{2\sqrt{x}}$$$$=2{tan}^{-1}\sqrt{x}\times \frac{1}{\sqrt{x}\times (1+x)}$$$$={cos}^{-1}\left(\frac{1-x}{1+x}\right)\times \frac{1}{\sqrt{x}\times (1+x)}$$$$$$$$\alpha ={tan}^{-1}\sqrt{x}\nwarrow$$$$tan\alpha =\sqrt{x}$$$$cos2\alpha =\frac{1-{tan}^2\alpha }{1+{tan}^2\alpha }=\frac{1-x}{1+x}\Uparrow$$$$2\alpha ={cos}^{-1}\left(\frac{1-x}{1+x}\right)\Uparrow$$$$2{tan}^{-1}\sqrt{x}=cos\left(\frac{1-x}{1+x}\right)\to \to \to \uparrow$$$$$$

Commented by rahul 19 last updated on 20/May/19

$$thankyousir!$$$$noneoftheoptioniscorrect.$$

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