Question

$\int \frac{8\cot x+1}{3\cot x+2}dx$

Answer 1

$$\begin{gathered} LetI=\int \left(\frac{8\cot x+1}{3\cot x+2}\right)dx \\ =\int \left(\frac{8{\displaystyle \frac{\cos x}{\sin x}}+1}{{\displaystyle \frac{3\cos x}{\sin x}}+2}\right)dx \\ =\int \left(\frac{8\cos x+\sin x}{3\cos x+2\sin x}\right)dx \\ \mathrm{Now},\mathrm{let}8\cos x+\sin x=A\left(3\cos x+2\sin x\right)+B\left(-3\sin x+2\cos x\right)...\left(1\right) \\ \Rightarrow 8\cos x+\sin x=3A\cos x+2A\sin x-3B\sin x+2B\cos x \\ \Rightarrow 8\cos x+\sin x=\left(3A+2B\right)\cos x+\left(2A-3B\right)\sin x \\ \mathrm{Equating}\mathrm{the}\mathrm{coefficients}\mathrm{of}\mathrm{like}\mathrm{terms}\mathrm{we}\mathrm{get}, \\ 2\mathrm{A}-3\mathrm{B}=1...\left(2\right) \\ 3\mathrm{A}+2\mathrm{B}=8...\left(3\right) \end{gathered}$$

Solving eq (2) and eq (3) we get,
A = 2, B = 1
Thus, by substituting the values of A and B in eq (1) we get ,


$$\begin{gathered} I=\int \left[\frac{2\left(3\cos x+2\sin x\right)+1\left(-3\sin x+2\cos x\right)}{\left(3\cos x+2\sin x\right)}\right]dx \\ =2\int \left(\frac{3\cos x+2\sin x}{3\cos x+2\sin x}\right)dx+\int \left(\frac{-3\sin x+2\cos x}{3\cos x+2\sin x}\right)dx \\ =2\int dx+\int \left(\frac{-3\sin x+2\cos x}{3\cos x+2\sin x}\right)dx \\ \mathrm{Putting}3\cos x+2\sin x=t \\ \Rightarrow \left(-3\sin x+2\cos x\right)dx=dt \\ \therefore I=2\int dx+\int \frac{1}{t}dt \\ =2x+\ln \left|t\right|+C \\ =2x+\ln \left|3\cos x+2\sin x\right|+C \end{gathered}$$