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Question Number 141872 by mohammad17 last updated on 24/May/21

Commented by mohammad17 last updated on 24/May/21

$$helpmesirplease$$

Commented by mohammad17 last updated on 24/May/21

$$????$$

Answered by Dwaipayan Shikari last updated on 24/May/21

$$\frac{y^{1/4}}{{sin}^{-1}x}=u$$$$\frac{xcosx}{4}log\left(sinx\right)-log\left({sin}^{-1}x\right)=log\left(u\right)$$$$\Rightarrow \frac{{xcos}^{2}x}{4sinx}+log\left(sinx\right)(-\frac{x}{4}sinx+\frac{cosx}{4})-\frac{1}{{sin}^{-1}x\sqrt{1-x^2}}=\frac{1}{u}.\frac{du}{dx}$$$$\frac{du}{dx}=\frac{\sqrt[4]{y}}{{sin}^{-1}x}(\frac{{xcos}^{2}x}{4sinx}+log\left(sinx\right)(-\frac{x}{4}sinx+\frac{cosx}{4})-\frac{1}{{sin}^{-1}x\sqrt{1-x^2}})$$

Commented by mohammad17 last updated on 24/May/21

$$thankyousir$$

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