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

The value of the integral $I = \int_{\frac{\pi}{24}}^{\frac{5\pi}{24}} \frac{dx}{1+\sqrt[3]{\tan 2x}}$ is:

$\int_{ - \pi }^{\pi } {\frac{{2x(1 + \sin x)}}{{1 + {{\cos }^2}x}}dx} $ is

$\int_0^\pi \frac{x \tan x}{\sec x + \tan x} \,dx = $

$\int_{-\pi / 2}^{\pi / 2} \sin |x| \, dx$ is equal to

The option$(s)$ with the values of $a$ and $L$ that satisfy the following equation is(are) $\frac{\int_0^{4 \pi} e^t(\sin^6 at + \cos^4 at) dt}{\int_0^{\pi} e^t(\sin^6 at + \cos^4 at) dt} = L$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module