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Question Number 157268 by gsk2684 last updated on 21/Oct/21

$$provethat$$$$\int \frac{x^2}{{(x\sin x+\cos x)}^2}dx=-\frac{x\sec x}{x\sin x+\cos x}+\tan x+c$$

Answered by Ar Brandon last updated on 18/Nov/21

$$I=\int \frac{x^2}{{(x\sin x+\cos x)}^2}dx=\int \frac{x\cos x}{{(x\sin x+\cos x)}^2}\cdot \frac{x}{\cos x}dx$$$$\{\begin{array}{l}\mathrm{u}\left(x\right)=\frac{x}{\cos x}\\ {\mathrm{v}}^{\prime }\left(x\right)=\frac{x\cos x}{{(x\sin x+\cos x)}^2}\end{array}\Rightarrow \{\begin{array}{l}{\mathrm{u}}^{\prime }\left(x\right)=\frac{\cos x+x\sin x}{{\cos }^{2}x}\\ \mathrm{v}\left(x\right)=-\frac{1}{x\sin x+\cos x}\end{array}$$$$I=-\frac{x}{(x\sin x+\cos x)\cos x}+\int \frac{dx}{{\cos }^{2}x}$$$$=-\frac{x\sec x}{x\sin x+\cos x}+\tan x+\mathrm{constant}$$

Commented by gsk2684 last updated on 21/Oct/21

$$thankyou$$

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