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

Evaluate the integral: $\int_0^{\pi /2} \frac{1 + 2\cos x}{(2 + \cos x)^2} dx$

Let $\int_0^1 f(x) \, dx = 1$,$\int_0^1 x f(x) \, dx = a$,and $\int_0^1 x^2 f(x) \, dx = a^2$. Then the value of $\int_0^1 (x - a)^2 f(x) \, dx$ is:

If $[x]$ is the greatest integer not exceeding $x$,then $\int_{-0.5}^{1.5} x^2[x] d x=$

$\int_{0}^{1} \left(\frac{x^{2}-2}{x^{2}+1}\right) dx =$

$ \int_{-5}^{5} |x+2| \, dx $ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module