Question

Evaluate \[ \int\frac1{2\cot x-3\tan x}\,dx \]

• A. \(\frac15\log|2\cos^2x-3\sin^2x|+c\)
• B. \(-\frac1{10}\log|2-5\sin^2x|+c\)
• C. \(-\frac15\log|2-5\sin^2x|+c\)
• D. \(\frac1{10}\log|2\cos^2x-3\sin^2x|+c\)

Hint:

For integrals containing \(\tan x\) and \(\cot x\), convert to sine-cosine form before applying substitution.

Answer 1

The Correct Option is D

Concept: Convert cotangent and tangent into sine-cosine form.

Step 1: Rewrite denominator.
\[ I= \int \frac1{2\frac{\cos x}{\sin x}-3\frac{\sin x}{\cos x}}dx \] Take LCM. \[ = \int \frac{\sin x\cos x} {2\cos^2x-3\sin^2x} dx \]

Step 2: Substitute variable.
Let \[ t=\sin x \] Then \[ dt=\cos xdx \] So integral becomes \[ I= \int \frac{t}{2(1-t^2)-3t^2}dt \] \[ = \int \frac{t}{2-5t^2}dt \]

Step 3: Integrate.
Let \[ u=2-5t^2 \] Then \[ du=-10tdt \] Thus \[ I = -\frac1{10} \int\frac{du}{u} \] \[ = -\frac1{10}\log|u|+c \] \[ = -\frac1{10}\log|2-5\sin^2x|+c \] Since \[ 2-5\sin^2x=2\cos^2x-3\sin^2x \] Equivalent form: \[ \boxed{ \frac1{10}\log|2\cos^2x-3\sin^2x|+c } \]