The value of the integral int_{-pi/4}^{pi/4} | vedclass

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

Question

(NONE) Let $I = \int_{-\pi/4}^{\pi/4} \sin^4 x \, dx$. Since $f(x) = \sin^4 x$ is an even function,$I = 2 \int_0^{\pi/4} \sin^4 x \, dx$.
Using the i

Vedclass · Accepted Answer

$-1/2$

Vedclass · Answer

(NONE) Let $I = \int_{-\pi/4}^{\pi/4} \sin^4 x \, dx$. Since $f(x) = \sin^4 x$ is an even function,$I = 2 \int_0^{\pi/4} \sin^4 x \, dx$.
Using the identity $\sin^2 x = \frac{1 - \cos 2x}{2}$,we have $\sin^4 x = (\frac{1 - \cos 2x}{2})^2 = \frac{1}{4}(1 - 2\cos 2x + \cos^2 2x)$.
Further,$\cos^2 2x = \frac{1 + \cos 4x}{2}$,so $\sin^4 x = \frac{1}{4}(1 - 2\cos 2x + \frac{1 + \cos 4x}{2}) = \frac{1}{4}(\frac{3}{2} - 2\cos 2x + \frac{1}{2}\cos 4x) = \frac{3}{8} - \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x$.
Integrating this,$I = 2 \int_0^{\pi/4} (\frac{3}{8} - \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x) \, dx = 2 [\frac{3}{8}x - \frac{1}{4}\sin 2x + \frac{1}{32}\sin 4x]_0^{\pi/4}$.
Evaluating at the limits: $I = 2 [(\frac{3}{8} \cdot \frac{\pi}{4} - \frac{1}{4}\sin \frac{\pi}{2} + \frac{1}{32}\sin \pi) - (0)] = 2 [\frac{3\pi}{32} - \frac{1}{4}] = \frac{3\pi}{16} - \frac{1}{2}$.

The value of the integral $\int_{-\pi/4}^{\pi/4} \sin^4 x \, dx$ is

Similar Questions

If $p(x)$ is a cubic polynomial with $p(1)=3, p(0)=2$ and $p(-1)=4$,then $\int_{-1}^1 p(x) dx$ is

The greatest value of the function $F(x) = \int_1^x {|t| \, dt}$ on the interval $\left[ -\frac{1}{2}, \frac{1}{2} \right]$ is:

The integral $\int_{0}^{2} ||x-1|-x| dx$ is equal to (in $.5$)

The value of $12 \int \limits_0^3 \left| x^2 - 3x + 2 \right| dx$ is $.............$

The positive value of $x$ satisfying the equation $\int_x^1(1-t) dt = \frac{1}{2}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module