I = ∫ 8x/4cos^2+sin^2x dx x ∈ to (0, π) is equal to

Question

I = \(\int_\limits0^\pi \frac{8x}{4\ \cos^2x\ +\ \sin^2x} dx\) is equal to

Answer 1

Answer is: \((2 \pi^2)\) 

\(I = \int_\limits0^\pi \frac{8xdx}{4\ \cos^2x\ +\ \sin^2x} \)   ....(i) 

\(I = \int_\limits0^\pi \frac{8(\pi-x)dx}{4\ \cos^2x\ +\ \sin^2x} \)   .....(ii) 

Adding (i) and (ii) 

\(2I = \int_\limits0^\pi \frac{8\pi dx}{4\ \cos^2x\ +\ \sin^2x} \)