I have taken $a=e$ but using partial integration, getting no luck.Please help.I avoided purposefully 6 pages of needless scribble as that may deviate the answerer.
Asked
Active
Viewed 64 times
0
-
The value of this integral should be $$\frac{\pi}{2}$$ – Dr. Sonnhard Graubner May 12 '18 at 15:37
1 Answers
4
We know that $\displaystyle\int_{-a}^af(x)\,dx = \int_0^af(x)\,dx+\int_0^af(-x)dx$
so,$I = \displaystyle\int_{-\pi}^\pi\dfrac{\sin^2(x)}{1+a^x}dx $
$I = \displaystyle\int_0^\pi\dfrac{\sin^2(x)}{1+a^x}+\dfrac{\sin^2(x)}{1+a^{-x}}\,dx$
$I = \displaystyle\int_0^\pi\dfrac{\sin^2(x)}{1+a^x}+\dfrac{a^x\cdot\sin^2(x)}{1+a^{x}}\,dx$
$I = \displaystyle\int_0^\pi\dfrac{\sin^2(x)(1+a^x)}{1+a^x}$
$I =\displaystyle\int_0^\pi\sin^2(x)\,dx$
I assume you can continue from here. ask if you need help
HINT: let $\sin^2(x) = \dfrac{1-\cos(2x)}{2}$
The Integrator
- 3,200
-
I forgot it but I thought of dividing the limits although sorry I have been too dumb. I am just stupid! – Saradamani May 12 '18 at 15:45
-
@Saradamani It alright, we all make mistakes ( Me more than most lol). Just be careful not do it again in the future. – The Integrator May 12 '18 at 15:46