Question

Evaluate $\int \frac{2\cos x+3\sin x}{4\cos x+5\sin x}dx$.

Answer 1

This is based on a special substitution.

Let $2\cos x+3\sin x=A(4\cos x+5\sin x)+B\frac{d}{dx}(4\cos x+5\sin x)$

On equating coefficients, we have $4A+5B=2$, $5A-4B=3$

On solving these,

$A=\frac{23}{41},B=-\frac{2}{41}$

$\therefore \int \frac{2\cos x+3\sin x}{4\cos x+5\sin x}dx=\int \frac{A(4\cos x+5\sin x)+B\frac{d}{dx}(4\cos x+5\sin x)}{4\cos x+5\sin x}dx$

$=\int Adx+\int B\frac{-4\sin x+5\cos x}{4\cos x+5\sin x}dx$

$=\frac{23}{41}\int dx-\frac{2}{41}\int \frac{\frac{d}{dx}(4\cos x+5\sin x)}{4\cos x+5\sin x}dx$

$=\frac{23}{41}x-\frac{2}{41}\ln |4\cos x+5\sin x|+c$