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Find the integral $$I=\int_{-\pi}^{+\pi}\dfrac{x\sin{x}\arctan{e^x}}{1+\cos^2{x}}dx$$

My try: let $$I=\int_{-\pi}^{0}\dfrac{x\sin{x}\arctan{e^x}}{1+\cos^2{x}}dx+\int_{0}^{+\pi}\dfrac{x\sin{x}\arctan{e^x}}{1+\cos^2{x}}dx=I_{1}+I_{2}$$

for $I_{2}$,we let $x=-t$

StubbornAtom
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  • I am getting $I\approx 3.87578$ with a numerical quadrature method and am not sure how to obtain an exact solution yet. – Riemann1337 Oct 26 '13 at 09:20

2 Answers2

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Putting $x=-t,$

$$I_1=\int_{-\pi}^0\frac{x\sin x\arctan(e^x)}{1+\cos^2x}dx$$

$$=\int_{\pi}^0\frac{(-t)\sin(-t)\arctan(e^{-t})}{1+\cos^2(-t)}(-dt)$$

$$=-\int_{\pi}^0\frac{t\sin t\arctan(e^{-t})}{1+\cos^2t}dt$$

$$=\int_0^{\pi}\frac{t\sin t\arctan(e^{-t})}{1+\cos^2t}dt$$

Now, $\arctan \frac1x=\text{arccot} x=\frac\pi2-\arctan x$

$$\implies I_1=\int_0^{\pi}\frac{t\sin t\left(\frac\pi2-\arctan(e^t)\right)}{1+\cos^2t}dt $$

$$\implies I=I_1+I_2=\frac\pi2 \int_0^{\pi}\frac{t\sin t}{1+\cos^2t}dt $$

$$\text{Applying }\int_a^bf(x)dx=\int_a^bf(a+b-x)dx\text{ on }I \text{ with }a=0,b=\pi$$

$$ I=\frac\pi2\int_0^{\pi}\frac{t\sin t}{1+\cos^2t}dt=\frac\pi2\int_0^{\pi}\frac{(\pi-t)\sin(\pi- t)}{1+\cos^2(\pi-t)}dt=\frac\pi2\int_0^{\pi}\frac{(\pi-t)\sin t}{1+\cos^2t}dt$$

$$\implies I=\frac{\pi}2\left(\int_0^{\pi}\frac{\pi\sin t}{1+\cos^2t}dt-I\right)$$ Can you take it from here?

lab bhattacharjee
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$$I=\int_{-\pi}^{\pi}\frac{x\sin x\arctan e^x}{1+\cos^2x}dx$$ $$=\int_{-\pi}^{\pi}\frac{x\sin x\arctan e^{-x}}{1+\cos^2x}dx$$ $$=\frac{1}{2}\int_{-\pi}^{\pi}\frac{x\sin x(\arctan e^x+\arctan e^{-x})}{1+\cos^2x}dx$$ $$=\frac{\pi}{4}\int_{-\pi}^{\pi}\frac{x\sin x}{1+\cos^2x}dx$$ $$=\frac{\pi}{2}\int_{0}^{\pi}\frac{x\sin x}{1+\cos^2x}dx$$ $$=\frac{\pi^2}{2^2}\int_{0}^{\pi}\frac{\sin x}{1+\cos^2x}dx$$ $$=\frac{\pi^2}{2^2}(-\arctan(\cos x))|_0^\pi=\frac{\pi^3}{8}$$

JamesJ
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