Question:
\(\int \frac{1}{2\cot x-3\tan x}dx =\)
\(\int \frac{1}{2\cot x-3\tan x}dx =\)
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Convert trig integrals into \(\sin^2 x\) or \(\cos^2 x\) substitutions.
Updated On: Jun 22, 2026
- \(-\frac{1}{10}\log|2-5\sin^2 x|+c\)
- \(\frac{1}{10}\log|3+2\cos^2 x|+c\)
- \(-\frac{1}{10}\log|2\cot x+3\tan x|+c\)
- \(\frac{1}{10}\log|2\cot x-3\tan x|+c\) \bigskip
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The Correct Option is A
Solution and Explanation
Concept:
Convert to sin-cos form.
Step 1: Simplify.
\[ 2\cot x-3\tan x=\frac{2\cos x}{\sin x}-\frac{3\sin x}{\cos x} \] \[ =\frac{2\cos^2 x-3\sin^2 x}{\sin x\cos x} \]
Step 2: Integral form.
\[ \int \frac{\sin x\cos x}{2\cos^2 x-3\sin^2 x}dx \] Let \(t=\sin^2 x\)
Step 3: Result.
\[ -\frac{1}{10}\log|2-5\sin^2 x|+c \] \[ \boxed{(A)} \]
Step 1: Simplify.
\[ 2\cot x-3\tan x=\frac{2\cos x}{\sin x}-\frac{3\sin x}{\cos x} \] \[ =\frac{2\cos^2 x-3\sin^2 x}{\sin x\cos x} \]
Step 2: Integral form.
\[ \int \frac{\sin x\cos x}{2\cos^2 x-3\sin^2 x}dx \] Let \(t=\sin^2 x\)
Step 3: Result.
\[ -\frac{1}{10}\log|2-5\sin^2 x|+c \] \[ \boxed{(A)} \]
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