IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

joyfrancis

I = ₀

^pi/2 (1+2cosx)/(2+cosx)

Multiply divide by (2-cosx)

=> I = ₀^pi/2 (2 + 3cosx -2cos²x) / 3+sin²x

=> I =₀^pi/22/(3+sin²x) + ₀^pi/23cosx/(3+sin²x) - ₀^pi/22cos²x/(3+sin²x)

Let ₀^pi/22/(3+sin²x) = I₁ ; ₀^pi/23cosx/(3+sin²x) =I₂ ; ₀^pi/22cos²x/(3+sin²x)=I₃

For I₁ and I₃...divide by cos²x and put tanx = t

For I₂....put sinx = t