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$$\int_{0}^{\pi}\frac{x^2}{(1+x \sin x)^2}\,\mathrm dx$$ I got $$2I=\int_{0}^{\pi}\frac{x^2}{(1+x \sin x)^2}\,\mathrm dx + \int_{0}^{\pi}\frac{x^2}{(1 - x\cos x + \sin x)^2}\,\mathrm dx $$ After this step I got stuck...

According to me I tried $\cos x=t$ put I failed.

Is this the way to solve or have some different method? Any help will be appreciated. Thanks and if you need more details I will help

vitamin d
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Diwakar yadav
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    Integrals of the form $\int R(x,\sin x),\mathrm{d}x$ (as well as $\int R(x,\cos x),\mathrm{d}x$ and $\int R(x,e^x),\mathrm{d}x$) for rational functions $R(x,y)$ are notorious for their non-integrability in elementary terms (in general sense). Do you have any particular reason to believe that this integral has an elementary closed-form? – Sangchul Lee Jun 17 '21 at 18:14
  • How did you get $2I$? – Laxmi Narayan Bhandari Jun 18 '21 at 05:37

0 Answers0