Integrate the function $\cot x \log \sin x$.
(N/A) Let $I = \int \cot x \log \sin x \, dx$.
Substitute $t = \log \sin x$.
Differentiating both sides with respect to $x$,we get $\frac{dt}{dx} = \frac{1}{\sin x} \cdot \cos x = \cot x$.
Therefore,$dt = \cot x \, dx$.
Substituting these into the integral,we get $I = \int t \, dt$.
Integrating with respect to $t$,we get $I = \frac{t^2}{2} + C$.
Substituting back $t = \log \sin x$,we get $I = \frac{1}{2}(\log \sin x)^2 + C$,where $C$ is an arbitrary constant.
Substitute $t = \log \sin x$.
Differentiating both sides with respect to $x$,we get $\frac{dt}{dx} = \frac{1}{\sin x} \cdot \cos x = \cot x$.
Therefore,$dt = \cot x \, dx$.
Substituting these into the integral,we get $I = \int t \, dt$.
Integrating with respect to $t$,we get $I = \frac{t^2}{2} + C$.
Substituting back $t = \log \sin x$,we get $I = \frac{1}{2}(\log \sin x)^2 + C$,where $C$ is an arbitrary constant.
Explore More
Similar Questions
$\int \frac{e^x(1+x)}{\sin ^2(x \cdot e^x)} dx = $ . . . . . . $+ C$.
$\int e^{\sqrt{x}} \, dx = $ . . . . . . $+ c ; x > 0$
$\int \frac{\sin (\tan ^{-1} x)}{1+x^2} d x=$ . . . . . . $+C$.
Integrate the function: $e^{3 \log x}(x^{4}+1)^{-1}$
Medium
View Solution$\int \frac{dx}{e^{-2x}(e^{2x} + 1)^2} = $
Easy
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo