Question

$\int \frac{4\sin x+5\cos x}{5\sin x+4\cos x}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \left(\frac{4\sin x+5\cos x}{5\sin x+4\cos x}\right)dx \\ \&\mathrm{let}4\sin x+5\cos x=A\left(5\sin x+4\cos x\right)+B\left(5\cos x-4\sin x\right)...\left(1\right) \\ \Rightarrow 4\sin x+5\cos x=\left(5A-4B\right)\sin x+\left(4A+5B\right)\cos x \\ \mathrm{By}\mathrm{equating}\mathrm{the}\mathrm{coefficients}\mathrm{of}\mathrm{like}\mathrm{terms}\mathrm{we}\mathrm{get}, \\ 5A-4B=4...\left(2\right) \\ 4A+5B=5...\left(3\right) \end{gathered}$$

By solving eq (2) and eq (3) we get,
$$\begin{gathered} A=\frac{40}{41},B=\frac{9}{41} \\ \mathrm{Thus},\mathrm{by}\mathrm{substituting}\mathrm{the}\mathrm{values}\mathrm{of}A\mathrm{and}B\mathrm{in}\mathrm{eq}\left(1\right),\mathrm{we}\mathrm{get} \\ I=\int \left[\frac{{\displaystyle \frac{40}{41}}\left(5\sin x+4\cos x\right)+{\displaystyle \frac{9}{41}}\left(5\cos x-4\sin x\right)}{\left(5\sin x+4\cos x\right)}\right]dx \\ =\frac{40}{41}\int dx+\frac{9}{41}\int \left(\frac{5\cos x-4\sin x}{5\sin x+4\cos x}\right)dx \\ \mathrm{Putting}5\sin x+4\cos x=t \\ \Rightarrow \left(5\cos x-4\sin x\right)dx=dt \\ \therefore I=\frac{40}{41}x+\frac{9}{41}\int \frac{1}{t}dt \\ =\frac{40}{41}x+\frac{9}{41}\ln \left|t\right|+C \\ =\frac{40}{41}x+\frac{9}{41}\ln \left|5\sin x+4\cos x\right|+C \end{gathered}$$