# I cant understand what will I do to slove it I tried it many times but I did not get rigjt answer

Nishant Vora IIT Patna
6 years ago
u can proceed like this
Himanshu
12 Points
6 years ago
Nishant sir I think something is wrong You did not late correctly You should check tanx/2=t is right but its differentiation side is not correct
mycroft holmes
272 Points
6 years ago
We first assume that a>b, to avoid denominator going to 0 in the domain.

Write the integral as $I_1+I_2+I_3+I_4$

where $I_1 = \int_0^{\frac{\pi}{2}} \frac {\sin^2 x}{a-b \cos x} \ dx$
I2, I3, I4 similarly defined in the other quadrants.

Now, and with the transformation  we obtain, I4 = Iand I3 = I2.

Hence the given integral can be written as $2(I_1 + I_2) =2 \left( \int_{0}^{\frac{\pi}{2}} \frac {\sin^2 x}{a-b \cos x} \ dx + \int_{\frac{\pi}{2}}^{ \pi} \frac {\sin^2 x}{a-b \cos x} \ dx\right )$

Now, we transforming the second integral with the substitution, we get

the integral as $\int_{0}^{\frac{\pi}{2}} \frac {\sin^2 x}{a-b \cos x} \ dx + \int_{0}^{\frac{\pi}{2}} \frac {\sin^2 x}{a+b \cos x} \ dx$
$= \int_{0}^{\frac{\pi}{2}} \frac {\sin^2 x}{a^2-b^2 \cos^2 x} \ dx$

$= \frac{1}{b^2}\int_{0}^{\frac{\pi}{2}} \frac {b^2 - b^2\cos^2 x}{a^2-b^2 \cos^2 x} \ dx$

$= \frac{b^2-a^2}{b^2}\int_{0}^{\frac{\pi}{2}} \frac {1}{a^2-b^2 \cos^2 x} \ dx + \frac{1}{b^2}\int_{0}^{\frac{\pi}{2}} \ dx$
and putting tan x = t, we get

$= \frac{\pi}{2b^2} - \frac{\sqrt{a^2-b^2}}{b^3}\frac{\pi}{2}$

$= \frac{\pi}{2} \left( \frac{b - \sqrt{a^2-b^2}}{b^3} \right)$