$ \int \frac{cosx (1 + 4 cos2x)}{sinx + 4 sinx cos^2 x} dx $ is equal to

Q: $\large \int \frac{cosx (1 + 4 cos2x)}{sinx + 4 sinx cos^2 x} dx $ is equal to

(A) $\large \frac{1}{2} ln|sin^2 x + sin^2 2x| + c $

(B) $\large \frac{1}{2}ln |sinx| + \frac{1}{2} ln|sin x + 4sinx cos^2x| + c $

(D) $\large \frac{1}{2}ln |sinx| – ln|sin x + 4sinx cos^2x| + c $

Sol: $\large \int \frac{cosx (1 + 4 cos2x)}{sinx + 4 sinx cos^2 x} dx $

Put cosx = t

$\large = – \int \frac{t (1 + 4(2t^2 – 1))}{1-t^2 + 4 (1-t^2)t^2} dt $

$\large = \int \frac{t(8t^2 – 3)}{4t^4 – 3t^2 – 1} dt $

$\large = \frac{1}{2} ln |4t^4 – 3t^2 – 1| + c $

$\large = \frac{1}{2}ln |sinx| + \frac{1}{2} ln|sin x + 4sinx cos^2x| + c $