The value of int_0^{2pi} |sin^3 theta| , dthe | vedclass

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

Question

(C) The function $f(	heta) = |\sin^3 	heta|$ is periodic with period $\pi$. Thus,$\int_0^{2\pi} |\sin^3 	heta| \, d	heta = 2 \int_0^{\pi} |\sin^3

Vedclass · Accepted Answer

$8/3$

Vedclass · Answer

(C) The function $f(	heta) = |\sin^3 	heta|$ is periodic with period $\pi$. Thus,$\int_0^{2\pi} |\sin^3 	heta| \, d	heta = 2 \int_0^{\pi} |\sin^3 	heta| \, d	heta$. Since $\sin 	heta \ge 0$ for $	heta \in [0, \pi]$,we have $|\sin^3 	heta| = \sin^3 	heta$. Using the identity $\sin 3	heta = 3\sin 	heta - 4\sin^3 	heta$,we get $\sin^3 	heta = \frac{3\sin 	heta - \sin 3	heta}{4}$. Therefore,the integral becomes $2 \int_0^{\pi} \frac{3\sin 	heta - \sin 3	heta}{4} \, d	heta = \frac{1}{2} \int_0^{\pi} (3\sin 	heta - \sin 3	heta) \, d	heta$. Evaluating the integral: $\frac{1}{2} [-3\cos 	heta + \frac{1}{3}\cos 3	heta]_0^{\pi}$. $= \frac{1}{2} [(-3\cos \pi + \frac{1}{3}\cos 3\pi) - (-3\cos 0 + \frac{1}{3}\cos 0)]$. $= \frac{1}{2} [(3 - 1/3) - (-3 + 1/3)] = \frac{1}{2} [8/3 - (-8/3)] = \frac{1}{2} [16/3] = 8/3$.

The value of $\int_0^{2\pi} |\sin^3 \theta| \, d\theta$ is

Similar Questions

$\int_{\pi / 6}^{\pi / 3} \cos^{-4} x \, dx =$

Let $[\cdot]$ denote the greatest integer function. Then the value of $\int_0^3 \left( \frac{e^x + e^{-x}}{[x]!} \right) dx$ is :

If $(n - m)$ is odd and $|m| \ne |n|,$ then $\int_0^\pi {\cos mx \sin nx} \,dx$ is

If $I=\int_{1}^{2} \frac{dx}{\sqrt{2x^{3}-9x^{2}+12x+4}},$ then

The value of $\int_0^1 \frac{x^4 + 1}{x^2 + 1} \, dx$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module