Question

Evaluate!!

Definite integrals.....


Please Help!!!!

Answer 1

Final Answer : $\pi/2$

Logic used :
1) Property of Definite Integrals :
$\int _{ - b} ^{b} f(x)dx = \int _{ 0} ^{b} (f(x) + f( - x))dx$
where f(x) is continuous from x =- b to x = b.

2) $\cos(2x) = 1 - 2 \sin^2(x)$

Steps:
1) Here,
$f(x) = \frac{ { \sin }^{2}(x) }{1 + {a}^{x} } \: and \: b = \pi$
Then,
$f(x) + f( - x) = { \sin }^{2} (x)$

2) Using Logic 1 :

$\int _{ - \pi} ^{\pi} \frac{ { \sin }^{2} (x)}{1 + {a}^{x} } dx = \int _{ 0} ^{\pi} (f(x) + f( - x))dx \\ \\ = > \int _{0} ^{\pi} { \sin}^{2} (x) \\ \\ = > \int _{0} ^{\pi} \frac{1 - \cos(2x) }{2} dx \\ \\ = > \int _{0} ^{\pi} \frac{1}{2} dx - \int _{0} ^{\pi} \frac{ \cos(2x) }{2} \\ \\ = > \frac{\pi}{2} - 0 \\ = > \frac{\pi}{2}$