I have define integral:
$$\int_0^\pi\frac{x\sin x\,dx}{1+\cos^2x}=\frac{\pi}{2}\int_0^\pi\frac{\sin x\,dx}{1+\cos^2x}$$
my question is how do we get that $\pi/2$
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Yeah that right,thanks for editing – user5927353 Aug 03 '16 at 13:51
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You have not checked any of answers to your questions, even they provide full details – Gathdi Aug 11 '16 at 16:05
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Hint. One may use $$ \int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx $$ with $$ a=0,\quad b=\pi, \quad f(x)=\frac{x \sin x}{1+\cos^2x}. $$
Olivier Oloa
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@user5927353 One may observe that $f(\pi-x)=\frac{(\pi-x) \sin x}{1+\cos^2x}$, since $\sin(\pi-x)=\sin x$ and $\cos^2(\pi-x)=\cos^2 x$. – Olivier Oloa Aug 03 '16 at 14:00
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Write: $$I=\int_0^\pi\frac{x\sin x\,dx}{1+\cos^2x},$$ $$I=\int_0^\pi (x-\frac{\pi}{2})\frac{\sin x\,dx}{1+\cos^2x}+\frac{\pi}{2}\int_0^\pi\frac{\sin x\,dx}{1+\cos^2x},$$ $$I=I_1+\frac{\pi}{2}\int_0^\pi\frac{\sin x\,dx}{1+\cos^2x}.$$ Now $$I_1\stackrel{\text{substitute: }x\to (x-\frac{\pi}{2})}=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{x\cos{x}}{1+\sin^2{x}} dx=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} f(x)=0,$$ where in the last step we used the fact that $f(x)$ is odd.
Finally, $$I=\frac{\pi}{2}\int_0^\pi\frac{\sin x\,dx}{1+\cos^2x}.$$
alans
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