int_0^pi x sin^3 x , dx = | vedclass

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

Question

(B) Let $I = \int_0^\pi x \sin^3 x \, dx$ ..... $(i)$Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we have:$I = \int_0^\pi (\pi - x

Vedclass · Accepted Answer

$\frac{2\pi}{3}$

Vedclass · Answer

(B) Let $I = \int_0^\pi x \sin^3 x \, dx$ ..... $(i)$Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we have:$I = \int_0^\pi (\pi - x) \sin^3(\pi - x) \, dx = \int_0^\pi (\pi - x) \sin^3 x \, dx$ ..... $(ii)$Adding $(i)$ and $(ii)$:$2I = \int_0^\pi (x + \pi - x) \sin^3 x \, dx = \pi \int_0^\pi \sin^3 x \, dx$Using the identity $\sin 3x = 3 \sin x - 4 \sin^3 x$,we get $\sin^3 x = \frac{3 \sin x - \sin 3x}{4}$.$2I = \pi \int_0^\pi \frac{3 \sin x - \sin 3x}{4} \, dx = \frac{\pi}{4} \left[ -3 \cos x + \frac{\cos 3x}{3} \right]_0^\pi$$2I = \frac{\pi}{4} \left[ (-3(-1) + \frac{-1}{3}) - (-3(1) + \frac{1}{3}) \right] = \frac{\pi}{4} \left[ (3 - \frac{1}{3}) - (-3 + \frac{1}{3}) \right]$$2I = \frac{\pi}{4} \left[ \frac{8}{3} + \frac{8}{3} \right] = \frac{\pi}{4} \left( \frac{16}{3} \right) = \frac{4\pi}{3}$Therefore,$I = \frac{2\pi}{3}$.

$\int_0^\pi x \sin^3 x \, dx = $

Similar Questions

If $I_{1} = \int_{0}^{1} (1 - x^{50})^{100} dx$ and $I_{2} = \int_{0}^{1} (1 - x^{50})^{101} dx$ such that $I_{2} = \alpha I_{1}$,then $\alpha$ is equal to:

$\int_{-1}^1 x|x| \, dx =$

If $\int_0^{2024 \pi} \frac{2023^{\sin ^2 x}}{2023^{\sin ^2 x}+2023^{\cos ^2 x}} d x=k$,then $\left(\frac{2 k}{\pi}+1\right)=$

The value of the integral $\int_{4}^{10} \frac{[x^2]}{[(x-14)^2] + [x^2]} dx$,where $[x]$ denotes the greatest integer function,is

If $\int_{0}^{1} \tan ^{-1} x \, dx = p$,then the value of $\int_{0}^{1} \tan ^{-1}\left(\frac{1-x}{1+x}\right) \, dx$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module