Question

Evaluate : $\int \frac{2\cos x+3\sin x}{4\cos x+4\sin x}dx$

Answer 1

$\int \frac{2\cos x+3\sin x}{4\cos x+4\sin x}$

Let $2\cos x+3\sin x=a(4\cos x+4\sin x)+b\frac{d}{dx}(4\cos x-4\sin x)$

$4a+4b=2$

$4a-4b=3$

$8a=5,8b=-1$

$a=\frac{5}{8},b=\frac{-1}{8}$

So $\int \frac{\frac{5}{8}(4\cos x+4\sin x)+\left(\frac{-1}{8}\right)(4\cos x-4\sin x)}{4\cos x+4\sin x}$

$=\frac{5}{8}\int 1dx+(-\frac{1}{8})\int \frac{d(4\cos x+4\sin x)}{4\cos x+4\sin x}$

$=\frac{5}{8}x-\frac{1}{8}\log (4\cos x+4\sin x)+c$