int_0^{pi / 2} sin ^5left(frac{x}{2}right) cd | vedclass

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

Question

(C) Let $I = \int_0^{\pi / 2} \sin ^5 \left(\frac{x}{2}\right) \sin x \, dx$. Using the identity $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$,we get

Vedclass · Accepted Answer

$\frac{1}{14 \sqrt{2}}$

Vedclass · Answer

(C) Let $I = \int_0^{\pi / 2} \sin ^5 \left(\frac{x}{2}\right) \sin x \, dx$. Using the identity $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$,we get: $I = \int_0^{\pi / 2} \sin ^5 \left(\frac{x}{2}\right) \cdot 2 \sin \frac{x}{2} \cos \frac{x}{2} \, dx$ $I = 2 \int_0^{\pi / 2} \sin ^6 \left(\frac{x}{2}\right) \cos \frac{x}{2} \, dx$ Let $t = \sin \frac{x}{2}$. Then $dt = \frac{1}{2} \cos \frac{x}{2} \, dx$,which implies $2 \, dt = \cos \frac{x}{2} \, dx$. When $x = 0$,$t = \sin(0) = 0$. When $x = \pi / 2$,$t = \sin(\pi / 4) = \frac{1}{\sqrt{2}}$. Substituting these into the integral: $I = 2 \int_0^{1/\sqrt{2}} t^6 (2 \, dt) = 4 \int_0^{1/\sqrt{2}} t^6 \, dt$ $I = 4 \left[ \frac{t^7}{7} \right]_0^{1/\sqrt{2}} = \frac{4}{7} \left( \frac{1}{\sqrt{2}} \right)^7$ Since $(\sqrt{2})^7 = 2^3 \cdot \sqrt{2} = 8 \sqrt{2}$,we have: $I = \frac{4}{7 \cdot 8 \sqrt{2}} = \frac{1}{7 \cdot 2 \sqrt{2}} = \frac{1}{14 \sqrt{2}}$.

$t \ (\text{in second})$	$0$	$2$	$4$	$6$	$8$	$10$
$v \ (\text{in m/s})$	$0$	$12$	$16$	$20$	$35$	$60$

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

Similar Questions

$\int_{1/e}^e |\log x| \, dx = $

Let the domain of the function $f(x) = \log_2 \log_4 \log_6(3 + 4x - x^2)$ be $(a, b)$. If $\int_0^{b-a} [x^2] dx = p - \sqrt{q} - \sqrt{r}$,where $p, q, r \in \mathbb{N}$,$\gcd(p, q, r) = 1$,and $[\cdot]$ is the greatest integer function,then $p + q + r$ is equal to

If $\int_{\log _{e} 2}^{x} (e^{t}-1)^{-1} dt = \log _{e} \frac{3}{2}$, then the value of $x$ is

$\int_0^{\pi / 4} [\sqrt{\tan x} + \sqrt{\cot x}] \, dx$ is equal to

The velocity of a particle which starts from rest is given by the following table. The total distance travelled (in metre) by the particle in $10 \ s$, using the Trapezoidal rule, is:
$t \ (\text{in second})$ $0$ $2$ $4$ $6$ $8$ $10$
$v \ (\text{in m/s})$ $0$ $12$ $16$ $20$ $35$ $60$

Vedclass Test Series

Exam Paper Generator

Online Exam Module