Question

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

Answer 1

Put $\begin{array}{c}2\sin x-3\sin x=\left(\frac{Ad\left(4\cos x+5\sin x\right)}{dx}\right)+\frac{B\left(\cos x+5\sin x\right)}{dx}\\ \Rightarrow 2\cos x-3\sin x=A\left(-4\sin x+5\cos x\right)+B\left(4\cos x+5\sin x\right)\\ \Rightarrow 2\cos x-3\sin x=\sin x\left(-4A+5B\right)+\cos x\left(5A+4B\right)\end{array}$

Comparing the co-efficient at sinx and cosx on both sides we have

$\begin{array}{c}-4A+5B=-3------\left(i\right)\\ 5A+4B=2-------\left(ii\right)\end{array}$

On solving (i) and (ii) we have

$-20A+25B=-15$

$\underset{–}{20A+16B=8}$

$41B=-7$

$B=\frac{-7}{41}$

Put $B=\frac{-7}{41}$ in equation (i)

we get,

$\begin{array}{c}-4A+5B=-3\\ -4A+5\left(\frac{-7}{41}\right)=-3\\ -4A-\frac{35}{41}=-3\\ -4A=-3+\frac{35}{41}\\ -4A=\frac{-123+35}{41}\\ -4A=\frac{-88}{41}\\ A=\frac{22}{41}\end{array}$

Now,

$\begin{array}{c}I=\int \frac{2\cos x-3\sin }{4\cos x+5\sin x}dx\\ I=\int \frac{A\left(-4\sin x+5\cos x\right)+B\left(4\cos x+5\sin x\right)}{4\cos x+5\sin x}dx\\ I=\int \frac{\frac{22}{41}\left(-4\sin x+5\cos x\right)}{\left(4\cos x+5\sin x\right)}+\frac{\frac{-7}{41}\left(4\cos x+5\sin x\right)}{\left(4\cos x+5\sin x\right)}dx\end{array}$

Hence we get,

$I=\frac{22}{41}\log \left|4\cos x+5\sin x\right|\frac{-7}{41}x+c$