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Determine exactly $\int_{0}^{\pi/2}\dfrac{x\cos x}{1+\sin^2 x} dx$.

I think this integral might involve the natural logarithm, but I don't think there's a clean method to simplify it. A lot of integration techniques don't seem to work (e.g. integration by parts, u-substitution, etc.). I don't think this integral can be solved using normal calculus.

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    This integral is well beyond the usual calculus level, and the answer is $$\int_{0}^{\frac{\pi}{2}}\frac{x\cos x}{1+\sin^2 x},\mathrm{d}x=\frac{1}{2}\log^2(1+\sqrt{2}).$$ This is closely related to Legendre $\chi$-function. See (1), (2), (3) for various methods. – Sangchul Lee Oct 21 '19 at 05:33
  • @SangchuiLee Those links only seem to integrate to $\pi$. If they can go to $\pi/2$, it'd be great if you left an answer that did it. – J.G. Oct 21 '19 at 06:24

2 Answers2

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Integrating by parts $$\int\dfrac{x\cos x}{1+\sin^2 x} dx=x \tan ^{-1}(\sin (x))-\int\tan ^{-1}(\sin (x))\,dx$$ that is to say that $$\int_{0}^{\frac{\pi}{2}}\dfrac{x\cos x}{1+\sin^2 x} dx=\frac{\pi ^2}{8}-\int_{0}^{\frac{\pi}{2}}\tan ^{-1}(\sin (x))\,dx$$ Now using $$\tan ^{-1}(\sin (x))=\sum_{n=0}^\infty (-1)^n\frac{ \sin ^{2 n+1}(x)}{2 n+1}$$ $$\int_{0}^{\frac{\pi}{2}}\tan ^{-1}(\sin (x))\,dx=\sum_{n=0}^\infty \frac{(-1)^n }{2 n+1}\int_{0}^{\frac{\pi}{2}}\sin ^{2 n+1}(x)\,dx=\sum_{n=0}^\infty(-1)^n\frac{\sqrt{\pi } \,\Gamma (n+1)}{2 (2 n+1)\, \Gamma \left(n+\frac{3}{2}\right)}$$ and the summation is $$\frac{\pi ^2}{8}-\frac{1}{2} \sinh ^{-1}(1)^2$$ So, for the integral, the result is just $$\int_{0}^{\frac{\pi}{2}}\dfrac{x\cos x}{1+\sin^2 x} dx=\frac{1}{2} \sinh ^{-1}(1)^2=\frac{1}{2}\log^2(1+\sqrt{2})$$ already given by Sangchul Lee.

Claude Leibovici
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$$I=\int_{0}^{\pi/2} x \frac{\cos x}{1+\sin^2 x} dx = x \tan^{-1} x)|_{0}^{\pi/2}-\int_{0}^{\pi/2} \tan^{-1}\sin x ~ dx.$$ Please see @Claude Lebovici's answer below.

Z Ahmed
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