$$\int_{-\frac{\pi}2}^{\frac{\pi}2} \frac{\sin^{2014}x}{\sin^{2014}x + \cos^{2014}x} \, dx$$

Question

ganeshie8 · Answer

even/odd stuff ha ?

anonymous · Answer

2014 is order of  derivative ?

hartnn · Answer

$\Large \int_a^b f(x) dx= \int_a^bf(a+b-x)dx $

use this!

klimenkov · Answer

@BSwan
2014 is the power.

hartnn · Answer

after you have proved that your function is even

anonymous · Answer

aha cool :)

hartnn · Answer

$\Large \int_{-a}^a f(x) dx = 2\int_0^a f(x)dx$
if f(x) is even function

hartnn · Answer

did u get what to do ?

klimenkov · Answer

$$2\int\limits_0^{\frac{\pi}2} \frac{\sin^{2014}x}{\sin^{2014}x + \cos^{2014}x} \, dx$$What's next?

hartnn · Answer

$\Large \int_a^b f(x) dx= \int_a^bf(a+b-x)dx$

replace x by pi/2 - x

hartnn · Answer

$I = 2\int\limits_0^{\frac{\pi}2} \frac{\sin^{2014}x}{\sin^{2014}x + \cos^{2014}x} \, dx ... ... (A)$

just giving a label, to be used later

klimenkov · Answer

$$2 \int\limits_0^{\frac{\pi}2} \frac{\cos^{2014}x}{\sin^{2014}x + \cos^{2014}x} \, dx$$

klimenkov · Answer

Oops...

hartnn · Answer

$I = 2 \int\limits_0^{\frac{\pi}2} \frac{\cos^{2014}x}{\sin^{2014}x + \cos^{2014}x} \, dx ... ... (B)$

Add (A) and (B)

klimenkov · Answer

$$I = -2 \int\limits_0^{\frac{\pi}2} \frac{\cos^{2014}x}{\sin^{2014}x + \cos^{2014}x} \, dx$$

hartnn · Answer

what ? why negative ?

hartnn · Answer

dx  remains as dx

hartnn · Answer

i am NOT doing any substitution

klimenkov · Answer

Yeah, everything is okay. My fault.
$$I + I = 2\int_0^{\frac\pi2}dx,$$$$I=\frac\pi2.$$Very nice, thank you.

hartnn · Answer

welcome ^_^