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Question

How to integrate $$\int_0^{\pi/2} \frac{x(1+\sin^2 x)\cos x}{(3+\sin^2 x)(1+3\sin^2 x)}\,dx$$

My attempt

\begin{align*}I &= \int_0^{\pi/2} \frac{x(1+\sin^2 x)\cos x}{(3+\sin^2 x)(1+3\sin^2 x)}\,dx \\&= \frac{1}{4}\int_0^{\pi/2} \frac{x\cos x}{1+3\sin^2 x}\,dx+\frac{1}{4}\int_0^{\pi/2} \frac{x\cos x}{3+\sin^2 x}\,dx\\&= \frac{1}{4\sqrt{3}}\int_0^{\pi/2} x\,d \left(\arctan \left(\sqrt{3}\sin x\right) + \arctan \left(\frac{1}{\sqrt{3}}\sin x\right)\right)\\&= \frac{\pi^2}{16\sqrt{3}} - \frac{1}{4\sqrt{3}}\int_0^{\pi/2} \arctan \left(\sqrt{3}\sin x\right) + \arctan \left(\frac{1}{\sqrt{3}}\sin x\right)\,dx\end{align*}

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Martin.s
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  • Seems to be a known one, found 3 post referencing it but all incomplete regarding obtention of the final value in term of log. https://artofproblemsolving.com/community/c7h1288609p6815856 (search P7), https://math.stackexchange.com/q/1099180/399263 using $\chi_2$ function, and https://math.stackexchange.com/q/1893393/399263 an attempt by Feynmann trick. – zwim Feb 19 '24 at 13:16

1 Answers1

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Consider $I(a)=\int_0^{\pi/2}\arctan (a \sin x)dx$ $$I’(a)= \int_0^{\pi/2}\frac{\sin x}{ 1+a^2 \sin^2x}dx =\frac{\text{arcsinh}\ a}{a\sqrt{1+a^2}}$$ Then \begin{align} K=& \int_0^{\pi/2} \arctan \left(\sqrt{3}\sin x\right) + \arctan \left(\frac{\sin x}{\sqrt{3}}\right)\,dx\\ =& \int_0^{\sqrt3}I’(a)da + \int_0^{1/\sqrt3}I’(a)\overset{a \to 1/a}{da}\\ =& \int_0^\infty \frac{\text{arccsch}\ a}{\sqrt{1+a^2}}da - \int_0^{\sqrt3} \frac{\text{arccsch}\ a}{\sqrt{1+a^2}}- \frac{\text{arcsinh}\ a}{a\sqrt{1+a^2}}\ da\\ =&\ \frac{\pi^2}4 - \text{arcsinh}\ a\ \text{arccsch}\ a\bigg|_0^{\sqrt3} =\frac{\pi^2}4-\text{arcsinh}\sqrt3\ \text{arccsch}\sqrt3 \end{align} and \begin{align*}& \int_0^{\pi/2} \frac{x(1+\sin^2 x)\cos x}{(3+\sin^2 x)(1+3\sin^2 x)}\,dx\\ =& \ \frac{\pi^2}{16\sqrt{3}} - \frac{1}{4\sqrt{3}}K = \frac{1}{4\sqrt{3}} \text{arcsinh}\sqrt3\ \text{arccsch}\sqrt3 \end{align*}

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