The value of int_{0}^{20pi} (sin^4 x + cos^4 | vedclass

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

Question

(C) We know that $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - \frac{1}{2} \sin^2(2x)$.Using the identity $\sin^2(2x) = \f

Vedclass · Accepted Answer

$15\pi$

Vedclass · Answer

(C) We know that $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - \frac{1}{2} \sin^2(2x)$.Using the identity $\sin^2(2x) = \frac{1 - \cos(4x)}{2}$, we get $\sin^4 x + \cos^4 x = 1 - \frac{1}{2} \left( \frac{1 - \cos(4x)}{2} \right) = 1 - \frac{1}{4} + \frac{1}{4} \cos(4x) = \frac{3}{4} + \frac{1}{4} \cos(4x)$.Now, integrating from $0$ to $20\pi$:$\int_0^{20\pi} (\frac{3}{4} + \frac{1}{4} \cos 4x) dx = [\frac{3}{4}x + \frac{1}{16} \sin 4x]_0^{20\pi}$.Substituting the limits: $(\frac{3}{4} \cdot 20\pi + \frac{1}{16} \sin(80\pi)) - (0 + 0) = 15\pi + 0 = 15\pi$.

The value of $\int_{0}^{20\pi} (\sin^4 x + \cos^4 x) dx$ is equal to:

Similar Questions

Evaluate the following definite integral as the limit of a sum:
$\int_{1}^{4}(x^{2}-x) dx$

$\int_0^{\pi /2} \frac{\cos x}{(1 + \sin x)(2 + \sin x)} \,dx = $

$\int_0^1 x(1-x)^n dx = $ . . . . . . .

If $[x]$ is the greatest integer function not greater than $x$,then $\int_0^8 [x] dx$ is equal to

Evaluate the integral: $\int_0^\pi \left(\cos^2 \left(\frac{3\pi}{8} - \frac{x}{4}\right) - \cos^2 \left(\frac{11\pi}{8} + \frac{x}{4}\right)\right) dx$

Vedclass Test Series

Exam Paper Generator

Online Exam Module