This is my question, $$\int_{-\pi}^{\pi} {2x(1+\sin x)\over1+\cos^2x} \,\, \mathrm{d}x$$

(1) $\pi^2\over4$

(2) $\pi^2$

(3) zero

(4) $\pi\over2$

I first broke the function into parts: $\int_{-\pi}^{\pi} {2x\over1+\cos^2x}$+ $\int_{-\pi}^{\pi} {2x\sin x\over1+\cos^2x}$. Here $\int_{-\pi}^{\pi} {2x\over1+\cos^2x}$being an odd function becomes zero and$\int_{-\pi}^{\pi} {2x\sin x\over1+\cos^2x}$being an even function can be written as 4$\int_{0}^{\pi} {x\sin x\over1+\cos^2x}dx$ and now I used the property $\int_{0}^{a} {f(a-x)}dx$. Then,I'm stuck.

Please help me through this.
Please explain any steps that you give.
Thanks in advance.


I'm not an expert on definite integrals such as this, but it looks like you're on the right track. I don't think you quite finished what your property was, but I think I see the direction it was headed in. So from

$$4\int_0^\pi\frac{x\sin xdx}{1+\cos^2x}$$


$$u=\pi-x,du=-dx$$ $$4\int_0^\pi\frac{x\sin xdx}{1+\cos^2x}=-4\int_\pi^0\frac{(\pi-u)\sin(\pi-u)du}{1+\cos^2(\pi-u)}=$$ $$4\int_0^\pi\frac{(\pi-u)\sin udu}{1+(-\cos u)^2}$$

Finally, since the variable is just a placeholder, let's just replace $u$ with $x$ again to get

$$I=4\int_0^\pi\frac{x\sin xdx}{1+\cos^2x}=4\int_0^\pi\frac{(\pi-x)\sin xdx}{1+\cos^2x}$$

Summing these gives $$2I=4\int_0^\pi\frac{\pi\sin xdx}{1+\cos^2x}$$ $$u=\cos x,du=-\sin xdx$$ $$I=-2\pi\int_1^{-1}\frac{du}{1+u^2}=2\pi(\tan^{-1}u]_{-1}^1)=2\pi(\frac{\pi}4+\frac\pi4)=\pi^2$$

So assuming I've done everything right, I arrive at answer 2.

  • $\begingroup$ @Mike..Yeah i was certainly stuck at that.And,yes you're right.Thanks for the help. $\endgroup$ – Yash Lekhwani May 11 '14 at 11:33

Using symmetry we get $$J:=\int_{-\pi}^\pi{2x(1+\sin x)\over 1+\cos^2 x}\ dx=4\int_0^\pi x\>{\sin x\over 1+\cos^2 x}\ dx\ .$$ Partial integration then gives $$J=4\left(-x\arctan(\cos x)\biggr|_0^\pi +\int_0^\pi\arctan(\cos x)\ dx\right)\ .$$ As $x\mapsto \cos x$ is odd with respect to $x={\pi\over2}$ the last integral vanishes, so that we are left with $$J=-4\pi\arctan(\cos\pi)=-4\pi\bigl(-{\pi\over4}\bigr)=\pi^2\ .$$


Using $\displaystyle\int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx\ \ \ \ (1)$

If $\displaystyle f(x)=\frac{2x(1+\sin x)}{1+\cos^2x}, f\left(-\pi+\pi-x\right)=f(-x)=\frac{-2x(1-\sin x)}{1+\cos^2(-x)}=\frac{2x\sin x-2x}{1+\cos^2x}$

$$I=\int_{-\pi}^{\pi}\frac{2x(1+\sin x)}{1+\cos^2x}\ dx=\int_{-\pi}^{\pi}\frac{2x\sin x-2x}{1+\cos^2x}\ dx$$

$$\implies I+I=4\int_{-\pi}^{\pi}\frac{x\sin x}{1+\cos^2x}\ dx$$

Again if $\displaystyle g(x)=\frac{x\sin x}{1+\cos^2x}, g(-x)=\frac{(-x)\sin(-x)}{1+\cos^2(-x)}=\frac{x\sin x}{1+\cos^2x}$ i.e., $g(x)$ is even

$$\text{ As for even }g(x),\int_{-a}^ag(x)dx=2\int_0^ag(x)dx,$$

$$\int_{-\pi}^{\pi}\frac{x\sin x}{1+\cos^2x}\ dx=2\int_0^{\pi}\frac{x\sin x}{1+\cos^2x}\ dx$$

Again, using $(1),$

$$J=\int_0^{\pi}\frac{x\sin x}{1+\cos^2x}\ dx=\int_0^{\pi}\frac{(\pi-x)\sin(\pi-x)}{1+\cos^2(\pi-x)}\ dx=\int_0^{\pi}\frac{(\pi-x)\sin x}{1+\cos^2x}\ dx$$

$$\implies J+J=\pi\int_0^{\pi}\frac{\sin x}{1+\cos^2x}\ dx$$

Set $\displaystyle \cos x=u$

  • $\begingroup$ @YashLekhwani, How about this? $\endgroup$ – lab bhattacharjee May 11 '14 at 13:30
  • $\begingroup$ yeah I did nearly this. P.S.you'll are very good at explaining stuff:) $\endgroup$ – Yash Lekhwani May 11 '14 at 13:33

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