int_0^{pi / 2} sin ^8 x cos ^2 x , dx is equa | vedclass

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

Question

(D) We use the Wallis formula for the definite integral $\int_0^{\pi / 2} \sin^m x \cos^n x \, dx = \frac{\Gamma(\frac{m+1}{2}) \Gamma(\frac{n+1}{2})}

Vedclass · Accepted Answer

$\frac{7 \pi}{512}$

Vedclass · Answer

(D) We use the Wallis formula for the definite integral $\int_0^{\pi / 2} \sin^m x \cos^n x \, dx = \frac{\Gamma(\frac{m+1}{2}) \Gamma(\frac{n+1}{2})}{2 \Gamma(\frac{m+n+2}{2})}$.Here, $m = 8$ and $n = 2$.Substituting these values, we get $\int_0^{\pi / 2} \sin^8 x \cos^2 x \, dx = \frac{\Gamma(\frac{9}{2}) \Gamma(\frac{3}{2})}{2 \Gamma(\frac{12}{2})} = \frac{\Gamma(\frac{9}{2}) \Gamma(\frac{3}{2})}{2 \Gamma(6)}$.Using $\Gamma(n+1) = n \Gamma(n)$, we have $\Gamma(\frac{9}{2}) = \frac{7}{2} \cdot \frac{5}{2} \cdot \frac{3}{2} \cdot \frac{1}{2} \sqrt{\pi}$ and $\Gamma(\frac{3}{2}) = \frac{1}{2} \sqrt{\pi}$.Also, $\Gamma(6) = 5! = 120$.Thus, the integral is $\frac{(\frac{7 \cdot 5 \cdot 3 \cdot 1}{16} \sqrt{\pi}) \cdot (\frac{1}{2} \sqrt{\pi})}{2 \cdot 120} = \frac{\frac{105}{32} \pi}{240} = \frac{105 \pi}{7680} = \frac{7 \pi}{512}$.

$\int_0^{\pi / 2} \sin ^8 x \cos ^2 x \, dx$ is equal to

Similar Questions

Let $f : (-1, 1) \to R$ be a continuous function. If $\int\limits_0^{\sin x} {f(t)dt} = \frac{\sqrt{3}}{2}x$,then $f\left(\frac{\sqrt{3}}{2}\right)$ is equal to

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

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

$\int_0^2 x^{\frac{5}{2}} \sqrt{2-x} \, dx =$

$\int_0^\pi x \sin^7 x \cos^6 x \, dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module