int_0^pi sin^5left( frac{x}{2} right) , dx eq | vedclass

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

Question

(A) Let $I = \int_0^\pi \sin^5\left( \frac{x}{2} \right) \, dx$. Substitute $t = \frac{x}{2}$,then $dx = 2 \, dt$. When $x = 0$,$t = 0$. When $x = \pi

Vedclass · Accepted Answer

$\frac{16}{15}$

Vedclass · Answer

(A) Let $I = \int_0^\pi \sin^5\left( \frac{x}{2} \right) \, dx$. Substitute $t = \frac{x}{2}$,then $dx = 2 \, dt$. When $x = 0$,$t = 0$. When $x = \pi$,$t = \frac{\pi}{2}$. So,$I = \int_0^{\pi/2} \sin^5(t) \cdot 2 \, dt = 2 \int_0^{\pi/2} \sin^5(t) \, dt$. Using Wallis' Formula,$\int_0^{\pi/2} \sin^n(x) \, dx = \frac{(n-1)(n-3)\dots(1)}{n(n-2)\dots(2)} \cdot \frac{\pi}{2}$ (if $n$ is even) or $\frac{(n-1)(n-3)\dots(2)}{n(n-2)\dots(1)}$ (if $n$ is odd). For $n = 5$,$\int_0^{\pi/2} \sin^5(t) \, dt = \frac{4 \cdot 2}{5 \cdot 3 \cdot 1} = \frac{8}{15}$. Therefore,$I = 2 \cdot \frac{8}{15} = \frac{16}{15}$.

$\int_0^\pi \sin^5\left( \frac{x}{2} \right) \, dx$ equals

Similar Questions

$\int_0^{\pi / 2} \sin ^m x \cos ^4 x \, dx = \frac{7 \pi}{2048} \Rightarrow m = ?$

$\int_0^{\frac{\pi}{2}} \sin^6 x \cos^4 x \, dx =$

The value of $\int_{0}^{1} x^{2}(1-x^{2})^{3/2} dx$ is

Let $F(x) = \int_x^{x^2+\frac{\pi}{6}} 2 \cos^2 t \, dt$ for all $x \in \mathbb{R}$ and $f: [0, \frac{1}{2}] \rightarrow [0, \infty)$ be a continuous function. For $a \in [0, \frac{1}{2}]$,if $F'(a) + 2$ is the area of the region bounded by $x=0, y=0, y=f(x)$ and $x=a$,then $f(0)$ is

$\int_{-\pi/2}^{\pi/2} \sin^4 x \cos^6 x \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module