Question

The value of \[ \int \frac{dx}{\sin x\,(2+3\cos x)} \] is

• A. \[ \frac12\log|1+\sin x| +\frac1{10}\log|\sin x-1| -\frac35\log|2+3\sin x| +C \]
• B. \[ \frac12\log|1+\cos x| +\frac1{10}\log|\cos x-1| -\frac35\log|2+3\cos x| +C \]
• C. \[ \frac12\log|1+\cos x| +\frac1{10}\log|\cos x-1| -\frac35\log|2+3\cos x| +C \]
• D. \[ \frac12\log|1+\sin x| +\frac1{10}\log|\cos x-1| -\frac35\log|2+\cos x| +C \]

Hint:

For integrals involving \[ \sin x,\quad \cos x \] and rational expressions in \(\cos x\), try \[ t=\cos x \] so that \[ dt=-\sin x\,dx. \] The integral often reduces to partial fractions.

Answer 1

The Correct Option is C

Concept: Put \[ t=\cos x \] so that \[ dt=-\sin x\,dx. \] This converts the trigonometric integral into a rational function integral.

Step 1: Substitute \(t=\cos x\). \[ dt=-\sin x\,dx \] Hence \[\begin{aligned} I &= \int \frac{dx}{\sin x(2+3\cos x)} \\ &= -\int \frac{dt} {(1-t^2)(2+3t)} \end{aligned}\]

Step 2: Use partial fractions. \[ -\frac1{(1-t^2)(2+3t)} = \frac{1}{2(1+t)} -\frac1{10(1-t)} -\frac{9}{10(2+3t)} \] Therefore, \[\begin{aligned} I &= \int \left[ \frac{1}{2(1+t)} -\frac1{10(1-t)} -\frac{9}{10(2+3t)} \right]dt \end{aligned}\]

Step 3: Integrate term by term. \[\begin{aligned} I &= \frac12\log|1+t| +\frac1{10}\log|t-1| -\frac35\log|2+3t| +C \end{aligned}\]

Step 4: Replace \(t=\cos x\). \[\begin{aligned} I &= \frac12\log|1+\cos x| +\frac1{10}\log|\cos x-1| -\frac35\log|2+3\cos x| +C \end{aligned}\] \[\begin{aligned} \boxed{ \frac12\log|1+\cos x| +\frac1{10}\log|\cos x-1| -\frac35\log|2+3\cos x| +C } \end{aligned}\] Hence, option \(\mathbf{(C)}\) is correct.