int_{-frac{pi}{4}}^{frac{pi}{4}} sin^2 x , dx | vedclass

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

Question

(B) Since $f(x) = \sin^2 x$ is an even function,we can write: $\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sin^2 x \, dx = 2 \int_{0}^{\frac{\pi}{4}} \sin^

Vedclass · Accepted Answer

$\frac{\pi}{4} - \frac{1}{2}$

Vedclass · Answer

(B) Since $f(x) = \sin^2 x$ is an even function,we can write: $\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sin^2 x \, dx = 2 \int_{0}^{\frac{\pi}{4}} \sin^2 x \, dx$ Using the identity $\sin^2 x = \frac{1 - \cos(2x)}{2}$: $2 \int_{0}^{\frac{\pi}{4}} \frac{1 - \cos(2x)}{2} \, dx = \int_{0}^{\frac{\pi}{4}} (1 - \cos(2x)) \, dx$ Integrating term by term: $[x - \frac{\sin(2x)}{2}]_{0}^{\frac{\pi}{4}}$ Substituting the limits: $(\frac{\pi}{4} - \frac{\sin(\frac{\pi}{2})}{2}) - (0 - \frac{\sin(0)}{2})$ $= \frac{\pi}{4} - \frac{1}{2} - 0 = \frac{\pi}{4} - \frac{1}{2}$ Thus,the correct option is $B$.

$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sin^2 x \, dx = $ . . . . . . .

Similar Questions

$\int_0^\infty {\frac{{x\,dx}}{{(1 + x)(1 + {x^2})}}} = $

The value of $\int_0^1 (1 + e^{-x^2}) \,dx$ is:

$\int_{-2}^3 |1-x^2| dx =$

$\int_0^k(\sqrt{k}-\sqrt{t})^2 \, dt =$

Evaluate the definite integral $\int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module