Question

$\int \frac{1}{3+4\cot x}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{1}{3+4\cot x}dx \\ =\int \frac{1}{3+{\displaystyle \frac{4\cos x}{\sin x}}}dx \\ =\int \frac{\sin x}{3\sin x+4\cos x}dx \\ \mathrm{Let}\sin x=A\left(3\sin x+4\cos x\right)+B\left(3\cos x-4\sin x\right)...\left(1\right) \\ \Rightarrow \sin x=\left(3A-4B\right)\sin x+\left(4A+3B\right)\cos x \\ \mathrm{By}\mathrm{comparing}\mathrm{the}\mathrm{coefficients}\mathrm{of}\mathrm{both}\mathrm{sides}\mathrm{we}\mathrm{get}, \\ 3A-4B=1...\left(2\right) \\ 4A+3B=0...\left(3\right) \end{gathered}$$

Multiplying eq (2) by 3 and equation (3) by 4 , then by adding them we get

$$\begin{gathered} 9A-12B+16A+12B=3+0 \\ \Rightarrow 25A=3 \\ \Rightarrow A=\frac{3}{25} \\ \mathrm{Putting}\mathrm{value}\mathrm{of}A\mathrm{in}\mathrm{eq}\left(3\right)\mathrm{we}\mathrm{get}, \\ 4\times \frac{3}{25}+3B=0 \\ \Rightarrow 3B=-\frac{12}{25} \\ \Rightarrow B=-\frac{4}{25} \end{gathered}$$
$$\begin{gathered} \mathrm{Thus},\mathrm{by}\mathrm{substituting}\mathrm{the}\mathrm{value}\mathrm{of}A\mathrm{and}B\mathrm{in}\mathrm{eq}\left(1\right)\mathrm{we}\mathrm{get} \\ I=\int \left[\frac{{\displaystyle \frac{3}{25}}\left(3\sin x+4\cos x\right)-{\displaystyle \frac{4}{25}}\left(3\cos x-4\sin x\right)}{3\sin x+4\cos x}\right]dx \\ =\frac{3}{25}\int dx-\frac{4}{25}\int \left(\frac{3\cos x-4\sin x}{3\sin x+4\cos x}\right)dx \\ \mathrm{Putting}3\sin x+4\cos x=t \\ \Rightarrow \left(3\cos x-4\sin x\right)dx=dt \\ \therefore I=\frac{3}{25}\int dx-\frac{4}{25}\int \frac{dt}{t} \\ =\frac{3}{25}x-\frac{4}{25}\ln \left|t\right|+C \\ =\frac{3x}{25}-\frac{4}{25}\ln \left|3\sin x+4\cos x\right|+C \end{gathered}$$