Question

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

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \left(\frac{2\sin x+3\cos x}{3\sin x+4\cos x}\right)dx \\ \&\mathrm{let}2\sin x+3\cos x=A\left(3\sin x+4\cos x\right)+B\left(3\cos x-4\sin x\right)...\left(1\right) \\ \Rightarrow 2\sin x+3\cos x=\left(3A-4B\right)\sin x+\left(4A+3B\right)\cos x \end{gathered}$$

By comparing the coefficients of like terms we get,

$$\begin{gathered} 3A-4B=2...\left(2\right) \\ 4A-3B=3...\left(3\right) \end{gathered}$$

Multiplying eq (2) by 3 and eq (3) by 4 and then adding,

$$\begin{gathered} 9A-12B+16A+12B=6+12 \\ \Rightarrow 25A=18 \\ \Rightarrow A=\frac{18}{25} \\ \mathrm{Putting}\mathrm{value}\mathrm{of}A=\frac{18}{25}\mathrm{in}\mathrm{eq}\left(2\right)\mathrm{we}\mathrm{get}, \\ 3\times \frac{18}{25}-4B=2 \\ \Rightarrow \frac{54}{25}-2=4B \\ \Rightarrow \frac{4}{25\times 4}=B \\ \Rightarrow B=\frac{1}{25} \end{gathered}$$
Thus, substituting the values of A,B and C in eq (1) we get ,

$$\begin{gathered} I=\int \left[\frac{{\displaystyle \frac{18}{25}}\left(3\sin x+4\cos x\right)+{\displaystyle \frac{1}{25}}\left(3\cos x-4\sin x\right)}{\left(3\sin x+4\cos x\right)}\right]dx \\ =\frac{18}{25}\int dx+\frac{1}{25}\int \left(\frac{3\cos x-4\sin x}{3\sin x+4\cos x}\right)dx \\ \mathrm{Putting}3\sin x+4\cos x=t \\ \Rightarrow \left(3\cos x-4\sin x\right)dx=dt \\ \therefore I=\frac{18}{25}\int dx+\frac{1}{25}\int \frac{1}{t}dt \\ =\frac{18x}{25}+\frac{1}{25}\ln \left|t\right|+C \\ =\frac{18x}{25}+\frac{1}{25}\ln \left|3\sin x+4\cos x\right|+C \end{gathered}$$