Integrate $\int_{0}^{\pi}{\frac{x\cos{x}}{1+\sin^{2}{x}}dx}$

Question

Integrate $$\displaystyle \int_{0}^{\pi}{\frac{x\cos{x}}{1+\sin^{2}{x}}dx}$$

Answer 1

With some calculations, we obtain

$$ \int_{0}^{\pi} \frac{x \cos x}{1+\sin^2 x} \, dx = 4 \chi_{2}(1-\sqrt{2})$$

where

$$ \chi_{2}(z) = \sum_{n=0}^{\infty} \frac{z^{2n+1}}{(2n+1)^2} $$

is the Legendre chi function of order 2. By exploiting some identities involving dilogarithm, we find that

$$ \chi_{2}(1-\sqrt{2}) = \frac{1}{4} \log^2 (1+\sqrt{2}) - \frac{3}{8}\zeta(2). $$

This gives the answer

$$\int_{0}^{\pi} \frac{x \cos x}{1+\sin^2 x} \, dx = \log^2(1+\sqrt{2}) - \frac{\pi^2}{4}. $$

Some detailed calculations, though written in Korean, can be found in my blog posting.

Answer 2

Here's a more elementary way:$$J=\int_0^\pi\frac{x \cos(x) dx}{1+\sin^2(x)}= \int_0^\frac{\pi}{2} \frac{x \cos(x)dx}{1+\sin^2(x)}+\int_0^\frac{\pi}{2} \frac{-x \sin(x)dx}{1+\cos^2(x)}-\frac{\pi}{2} \int_\frac{\pi}{2}^0 \frac{- \sin (x) dx}{1+\cos^2(x)}$$ $$J=I_1-I_2-\frac{\pi^2}{8}$$

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$$I_1=\int_0^\frac{\pi}{2} \frac{ x\cos(x) dx}{1+\sin^2(x)}= \sum_{k \ge 0}(-1)^k \int_0^\frac{\pi}{2} x \cos(x) \sin^{2k}(x)dx$$ $$= \sum_{k \ge 0}(-1)^k \left( \frac{\pi}{2}\frac{1}{2k+1}- \frac{1}{2k+1}\int_0^\frac{\pi}{2} \sin^{2k+1}(x) dx \right)$$ , using the beta function, $$= \sum_{k \ge 0}(-1)^k \left( \frac{\pi}{2}\frac{1}{2k+1}- \frac{1}{2k+1}\frac{\Gamma(k+1)\sqrt{\pi}}{\Gamma(k+1+\frac{1}{2})} \right)$$ , and the Legendre duplication formula, $$= \sum_{k \ge 0}(-1)^k \left( \frac{\pi}{2}\frac{1}{2k+1}- \frac{1}{2k+1}\frac{\Gamma(k+1)^22^{2k+1}}{\Gamma(2k+2)} \right)$$ $$= \sum_{k \ge 0}(-1)^k \left( \frac{\pi}{2}\frac{1}{2k+1}- \frac{1}{(2k+1)^2}\frac{2^{2k+1}}{\binom{2k}{k}} \right)$$ $$= \frac{\pi}{2}\arctan(1)-\sum_{k \ge 0}(-1)^k \frac{1}{(2k+1)^2}\frac{2^{2k+1}}{\binom{2k}{k}} $$

$$\stackrel{\text{Mathematica}}{=} \frac{\pi^2}{8}-\frac{\pi^2}{8}+ \frac{1}{2}\sinh^{-1}(1)^2 $$

$$I_1= \frac{1}{2}\sinh^{-1}(1)^2 $$

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$$I_2=\int_0^\frac{\pi}{2} \frac{ x\sin(x) dx}{1+\cos^2(x)}= \sum_{k \ge 0}(-1)^k \int_0^\frac{\pi}{2} x \sin(x) \cos^{2k}(x)dx$$ $$= \sum_{k \ge 0}(-1)^k \frac{1}{2k+1}\int_0^\frac{\pi}{2} \cos^{2k+1}(x)dx$$ $$= \sum_{k \ge 0}(-1)^k \frac{1}{2k+1}\int_0^\frac{\pi}{2} \sin^{2k+1}(x)dx$$ $$I_2=\frac{\pi^2}{8}- \frac{1}{2} \sinh^{-1}(1)^2$$

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Thus

$$J=\sinh^{-1}(1)^2- \frac{\pi^2}{4}.$$

Answer 3

Via integration by parts, we convert the integral into $$ \begin{aligned} \int_0^\pi\frac{x \cos x}{1+\sin ^2 x} d x&=\int_0^\pi x d\left(\tan ^{-1}\left(\sin x\right)\right) \\ & =\left[x \tan ^{-1}(\sin x)\right]_0^\pi-\int_0^\pi \tan ^{-1}(\sin x) d x \\ & =-2 \int_0^{\frac{\pi}{2}} \tan ^{-1}(\sin x) d x \end{aligned} $$ Let’s consider the parametrised integral $$I(a)= \int_0^{\frac{\pi}{2}} \tan ^{-1}(a\sin x) d x $$ Differentiating $I(a)$ w.r.t. $a$ yields $$ \begin{aligned} I^{\prime}(a) & =\int_0^{\frac{\pi}{2}} \frac{\sin x}{1+a^2 \sin ^2 x} d x \\ & =\int_0^1 \frac{d c}{1+a^2\left(1-c^2\right)}, \quad \textrm{ where }c=\cos x \\ & =\int_0^1 \frac{d c}{1+a^2-a^2 c^2} \\ & =\frac{1}{2 a \sqrt{1+a^2}}\left[\ln \frac{\sqrt{1+a^2}+a c}{\sqrt{1+a^2}-a c}\right]_0^1 \\ & =\frac{1}{2 a \sqrt{1+a^2}} \ln \left(\frac{\sqrt{1+a^2}+a}{\sqrt{1+a^2}-a}\right) \end{aligned} $$ $$ \begin{aligned} I(1) & =I(1)-I(0) \\ & =\int_0^1 \frac{1}{2a \sqrt{1+a^2}} \ln \left(\frac{\sqrt{1+a^2}+a}{\sqrt{1+a^2}-a}\right) d a \\ & =\int_0^{\sinh ^{-1}(1)} \frac{\theta}{\sinh \theta} d \theta\\&= \underbrace{ \int_0^{\infty} \frac{\theta}{\sinh \theta} d \theta}_{J} - \underbrace{ \int_{\sinh ^{-1}(1)}^{\infty} \frac{\theta}{\sinh \theta} d \theta}_{K} \end{aligned} $$

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$$ J \stackrel{\theta=-\ln u}{=} 2 \int_0^{1} \frac{\ln u}{u^2-1} d u=\frac{\pi^2}{4} $$

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For the last integral $K$, letting $v=\sinh \theta$ gives

$$ \begin{aligned} & K=\int_1^{\infty} \frac{\operatorname{arcsinh} v}{v \sqrt{1+v^2}} d v \\ & =\int_0^1 \frac{\operatorname{arcsinh}(v)}{\sqrt{1+v^2}} d v \\ & =\int_0^1 \operatorname{arccsch} v d(\operatorname{arcsinh} v) \\ & =[\operatorname{arccsch}v \operatorname{arcsinh} v]_0^1 -\int_0^1 \frac{\operatorname{arcsinh} v}{v \sqrt{1+v^2}} d v\\&= \operatorname{arcsinh}^2(1)-I(1) \quad (\textrm{ via IBP and } v=\sinh \theta) \end{aligned} $$ $$\therefore 2I(1)=\frac{\pi^2}{4} - \operatorname{arcsinh}^2(1) $$ Now we can conclude $$ \boxed{I=-2I(1)=\log^2 (\sqrt 2+1)-\frac{\pi^2}{4}} $$