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

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

Question

(C) Let $I = \int_{-\pi/2}^{\pi/2} \sin^4 x \cos^6 x \, dx$. Since $f(x) = \sin^4 x \cos^6 x$ is an even function,$f(-x) = f(x)$,we have: $I = 2 \int_

Vedclass · Accepted Answer

$\frac{3\pi}{256}$

Vedclass · Answer

(C) Let $I = \int_{-\pi/2}^{\pi/2} \sin^4 x \cos^6 x \, dx$. Since $f(x) = \sin^4 x \cos^6 x$ is an even function,$f(-x) = f(x)$,we have: $I = 2 \int_0^{\pi/2} \sin^4 x \cos^6 x \, dx$. Using the Wallis formula $\int_0^{\pi/2} \sin^m x \cos^n x \, dx = \frac{[(m-1)(m-3)...(1)][(n-1)(n-3)...(1)]}{(m+n)(m+n-2)...(2)} 	imes \frac{\pi}{2}$ (if both $m, n$ are even): Here $m=4$ and $n=6$. $I = 2 	imes \left[ \frac{(3 	imes 1) 	imes (5 	imes 3 	imes 1)}{(10 	imes 8 	imes 6 	imes 4 	imes 2)} 	imes \frac{\pi}{2} ight]$. $I = 2 	imes \left[ \frac{3 	imes 15}{3840} 	imes \frac{\pi}{2} ight] = \frac{45\pi}{3840} = \frac{3\pi}{256}$.

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

Similar Questions

The derivative of $F(x) = \int_{x^2}^{x^3} \frac{1}{\log t} \, dt$,$(x > 0)$ is

The value of $\mathop {\lim }\limits_{x \to 0} \frac{{\int_0^x {\cos {t^2}dt} }}{x}$ is

If $f(x) = \int_{9x^2}^{x^4} 5^{\sqrt{t}} dt$,then $\lim_{h \to 0} \frac{f(3 + h) - f(3 - h)}{h}$ is equal to

Let $H(x) = \int_{x^2}^{x^3} (x + 1) \sin(t^3) dt$. Then $\lim_{x \to 1} \frac{H(x)}{x - 1}$ is equal to:

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module