int_0^{pi /2} sin^4 x cos^6 x , dx equals | vedclass

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

Question

(B) Using the Wallis formula for the integral $I = \int_0^{\pi /2} \sin^m x \cos^n x \, dx = \frac{\Gamma(\frac{m+1}{2}) \Gamma(\frac{n+1}{2})}{2 \Gam

Vedclass · Accepted Answer

$\frac{3\pi}{512}$

Vedclass · Answer

(B) Using the Wallis formula for the integral $I = \int_0^{\pi /2} \sin^m x \cos^n x \, dx = \frac{\Gamma(\frac{m+1}{2}) \Gamma(\frac{n+1}{2})}{2 \Gamma(\frac{m+n+2}{2})}$.Here,$m = 4$ and $n = 6$.$I = \frac{\Gamma(\frac{4+1}{2}) \Gamma(\frac{6+1}{2})}{2 \Gamma(\frac{4+6+2}{2})} = \frac{\Gamma(5/2) \Gamma(7/2)}{2 \Gamma(6)}$.Using $\Gamma(n+1) = n\Gamma(n)$ and $\Gamma(1/2) = \sqrt{\pi}$:$\Gamma(5/2) = \frac{3}{2} \cdot \frac{1}{2} \cdot \sqrt{\pi} = \frac{3\sqrt{\pi}}{4}$.$\Gamma(7/2) = \frac{5}{2} \cdot \frac{3}{2} \cdot \frac{1}{2} \cdot \sqrt{\pi} = \frac{15\sqrt{\pi}}{8}$.$\Gamma(6) = 5! = 120$.$I = \frac{(\frac{3\sqrt{\pi}}{4}) (\frac{15\sqrt{\pi}}{8})}{2 \cdot 120} = \frac{45\pi}{32 \cdot 240} = \frac{45\pi}{7680} = \frac{3\pi}{512}$.

$\int_0^{\pi /2} \sin^4 x \cos^6 x \, dx$ equals

Similar Questions

The value of $x$ that maximises the value of the integral $\int\limits_x^{x + 3} {t(5 - t)\,dt}$ is

The value of the integral $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^4 x \left( 1 + \log \left( \frac{2 + \sin x}{2 - \sin x} \right) \right) dx$ is

$\int_{-3\pi}^{3\pi} \sin^2 \theta \sin^2 2\theta \, d\theta$ is equal to -

The points of extremum of $\int_0^{x^2} \frac{t^2-5t+4}{2+e^t} dt$ are

Let $f(x) = \left| \begin{array}{ccc} \sec x & \cos x & \sec^2 x + \cot x \csc x \\ \cos^2 x & \cos^2 x & \csc^2 x \\ 1 & \cos^2 x & \cos^2 x \end{array} \right|$,then $\int_0^{\pi /2} f(x) dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module