Evaluate the integral: int_0^pi frac{x}{sin x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $I = \int_0^\pi \frac{x}{\sin x}(3 \cos^2 x + 2 \sin x + \sin^3 x - 3) dx$. Since $\cos^2 x = 1 - \sin^2 x$,the expression becomes $3(1 - \sin

Vedclass · Accepted Answer

$\frac{\pi(5 \pi-12)}{4}$

Vedclass · Answer

(A) Let $I = \int_0^\pi \frac{x}{\sin x}(3 \cos^2 x + 2 \sin x + \sin^3 x - 3) dx$. Since $\cos^2 x = 1 - \sin^2 x$,the expression becomes $3(1 - \sin^2 x) + 2 \sin x + \sin^3 x - 3 = -3 \sin^2 x + 2 \sin x + \sin^3 x$. Dividing by $\sin x$,we get $-3 \sin x + 2 + \sin^2 x$. Thus,$I = \int_0^\pi x(\sin^2 x - 3 \sin x + 2) dx \dots (1)$. Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$: $I = \int_0^\pi (\pi - x)(\sin^2 x - 3 \sin x + 2) dx \dots (2)$. Adding $(1)$ and $(2)$: $2I = \int_0^\pi \pi(\sin^2 x - 3 \sin x + 2) dx$. $2I = \pi \int_0^\pi (\frac{1 - \cos 2x}{2} - 3 \sin x + 2) dx = \pi \int_0^\pi (\frac{5}{2} - \frac{1}{2} \cos 2x - 3 \sin x) dx$. $2I = \pi [\frac{5}{2}x - \frac{1}{4} \sin 2x + 3 \cos x]_0^\pi$. $2I = \pi [(\frac{5\pi}{2} - 0 + 3(-1)) - (0 - 0 + 3(1))] = \pi [\frac{5\pi}{2} - 3 - 3] = \pi [\frac{5\pi}{2} - 6] = \frac{\pi(5\pi - 12)}{2}$. Therefore,$I = \frac{\pi(5\pi - 12)}{4}$.

Evaluate the integral: $\int_0^\pi \frac{x}{\sin x}(3 \cos^2 x + 2 \sin x + \sin^3 x - 3) dx$

Similar Questions

$\int_{0}^{1} \sin \left( 2 \tan^{-1} \sqrt{\frac{1+x}{1-x}} \right) \, dx = $

The value of the integral $\int_{-\pi/4}^{\pi/4} \left( \frac{32 \cos^4 x}{1 + e^{\sin x}} \right) dx$ is:

$\int_0^\pi \frac{x \sin x}{1+\cos ^2 x} d x=$

If $\int\limits_0^1 \frac{\ln x}{\sqrt{1 - x^2}} dx = k \int\limits_0^\pi \ln(1 + \cos x) dx$,then the value of $k$ is:

$\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{d x}{1+\cos x}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module