int_{0}^{pi} sqrt{frac{1 + cos 2x}{2}} , dx i | vedclass

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

Question

(B) We know that $1 + \cos 2x = 2 \cos^2 x$. Substituting this into the integral,we get: $I = \int_{0}^{\pi} \sqrt{\frac{2 \cos^2 x}{2}} \, dx = \int_

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) We know that $1 + \cos 2x = 2 \cos^2 x$. Substituting this into the integral,we get: $I = \int_{0}^{\pi} \sqrt{\frac{2 \cos^2 x}{2}} \, dx = \int_{0}^{\pi} |\cos x| \, dx$. Since $\cos x \ge 0$ for $x \in [0, \pi/2]$ and $\cos x \le 0$ for $x \in [\pi/2, \pi]$,we split the integral: $I = \int_{0}^{\pi/2} \cos x \, dx - \int_{\pi/2}^{\pi} \cos x \, dx$. Evaluating the integrals: $I = [\sin x]_{0}^{\pi/2} - [\sin x]_{\pi/2}^{\pi}$. $I = (\sin(\pi/2) - \sin(0)) - (\sin(\pi) - \sin(\pi/2))$. $I = (1 - 0) - (0 - 1) = 1 + 1 = 2$.

$\int_{0}^{\pi} \sqrt{\frac{1 + \cos 2x}{2}} \, dx$ is equal to

Similar Questions

$[x]$ represents the greatest integer function. If $\int_{\sqrt{3}}^{\sqrt{18}}[x] \, dx = a + b\sqrt{2} + c\sqrt{3}$,then $a + b + c =$

The value of $\int_{-2}^{3} |1 - x^2| dx$ is

Dividing the interval $[0, 6]$ into $6$ equal parts and by using the trapezoidal rule,the value of $\int_0^6 x^3 \, dx$ is approximately:

If $a \in Z^{+}$,$[x]$ is the greatest integer not more than $x$,and $\int_0^a 2^{[x]} dx = 127$,then $a =$

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module