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I can`t understand what will I do to slove it I tried it many times but I did not get rigjt answer

Himanshu , 8 Years ago
Grade 12th pass
anser 3 Answers
Nishant Vora

Last Activity: 8 Years ago

u can proceed like this
223-311_Capture.PNG

Himanshu

Last Activity: 8 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

Last Activity: 8 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 = I1 and 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)
 
 

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