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Integral Calculus

definite integarl of (1+2cosx)/(2+cosx)2dx from 0 to π/2 is?

Profile image of pihu
16 Years agoGrade
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1 Answer

Profile image of Jitender Singh
12 Years ago
Ans:
I = \int_{0}^{\pi /2}\frac{(1+2cosx)}{(2+cosx)^{2}}dx
t = tan\frac{x}{2}
dt = \frac{1}{2}sec^{2}\frac{x}{2}dx
sinx = \frac{2t}{1+t^{2}}, cosx = \frac{1-t^{2}}{1+t^{2}}
I = \int_{0}^{1} \frac{-2(t^{2}-3)}{t^{4}+6t^{2}+9}dt
I = -2\int_{0}^{1} ((\frac{1}{t^{2}+3})-(\frac{6}{(t^{2}+3)^{2}}))dt
I = (\frac{2t}{t^{2}+3})_{0}^{1} = \frac{1}{2}
Thanks & Regards
Jitender Singh
IIT Delhi
askIITians Faculty

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