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Find the value of $\displaystyle \int_0^{\frac{\pi}{4}}\left({\frac{x}{x\sin x+\cos x}}\right)^2dx$

My approach is as follow $(x\sin x+\cos x)^{-1}=t$

$$-\frac{x\cos x}{(x\sin x+\cos x)^2}dx=dt$$

After this steps I am not able to approach it

janmarqz
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Samar Imam Zaidi
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    https://math.stackexchange.com/questions/366509/evaluating-this-integral-small-int-frac-x2-dx-x-sin-x-cos-x2 – lab bhattacharjee Apr 01 '20 at 13:43
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Let $$I(x)=\int \frac{x^2 dx}{(x\sin +\cos x)^2}= \int x \sec x \frac{(x \cos x)}{(x \sin x+\cos )^2} dx$$. Let us integrate by parts, and taking $(x\sin x+\cos x)=t$. We get $$I(x)=-x \sec x (x\sin x+\cos x)^{-1} +\int \frac{(\sec x +x\sec x \tan x)}{(x\sin x+\cos x)} dx=$$ $$I(x)=-x \sec x (x\sin x+\cos x)^{-1}+ \int \sec^2 x ~dx$$ $$I(x)=-x \sec x (x\sin x+\cos x)^{-1}+ \tan x+C$$ So the required definite integral in $(0, \pi/4)$ is $$I=\frac{4-\pi}{4+\pi}$$

Z Ahmed
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