The value of the integral `int (cos^3x+cos^5 x)/(sin^2 x+sin^4 x) dx` is (A) `sin x-6tan^(-1) (sin x)+C` (B) `sin x-2 (sin x)^(-1)+C` (C) `sin x-2 (sin x)^(-1)-6tan^(-1) (sin x)+C` (D) `sin x-2 (sin x)^(-1)+5tan^(-1) (sin x)+C`

Explanation

Let `f(sinx,cosx)=(cos^(3)x+cos^(5)x)/(sin^(2)x+sin^(4)x)`
`impliesf(sinx,cosx)=-(cos^(3)x+cos^(5)x)/(sin^(2)x+sin^(4)x)=-f(sinx,cosx)`
`:.` Let us substitute `sinx=t`
so that `cosxdx=dt`
Thus `int(cos^(3)x+cos^(5)x)/(sin^(2)x+sin^(4)x)dx=int((cos^(2)x+cos^(4)x))/(sin^(2)x+sin^(4)x)cosxdx`
`=int((1-sin^(2)x)+(1-sin^(2)x)^(2))/(sin^(2)x+sin^(4)x)dx`
`=int(1-t^(2)+1+t^(4)-2t^(2))/(t^(2)+t^(4))dt`
`=int(2-3t^(2)+t^(4))/(t^(2)+t^(4))dt`
`=int((t^(2)-2)(t^(2)-1))/(t^(2)(1+t^(2)))dt`
`=int(1+(2)/(t^(2))-(6)/(1+t^(2)))dt`
`=t-(2)/(t)-6tan^(-1)t+C`
`=sinx-2cosecx-6tan^(-1)(sinx)+C`