Evaluate the definite integral: int_0^{pi^2 / | vedclass

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

Question

(C) Let $I = \int_0^{\pi^2 / 4} (2 \sin \sqrt{x} + \sqrt{x} \cos \sqrt{x}) \, dx$. Substitute $t = \sqrt{x}$,so $x = t^2$ and $dx = 2t \, dt$. When $x

Vedclass · Accepted Answer

$\frac{\pi^2}{2}$

Vedclass · Answer

(C) Let $I = \int_0^{\pi^2 / 4} (2 \sin \sqrt{x} + \sqrt{x} \cos \sqrt{x}) \, dx$. Substitute $t = \sqrt{x}$,so $x = t^2$ and $dx = 2t \, dt$. When $x = 0$,$t = 0$. When $x = \frac{\pi^2}{4}$,$t = \frac{\pi}{2}$. Substituting these into the integral: $I = \int_0^{\pi/2} (2 \sin t + t \cos t) (2t) \, dt = \int_0^{\pi/2} (4t \sin t + 2t^2 \cos t) \, dt$. Using integration by parts on the second term $\int 2t^2 \cos t \, dt$: Let $u = 2t^2$ and $dv = \cos t \, dt$,then $du = 4t \, dt$ and $v = \sin t$. $\int 2t^2 \cos t \, dt = 2t^2 \sin t - \int 4t \sin t \, dt$. Substituting this back into the expression for $I$: $I = \int_0^{\pi/2} 4t \sin t \, dt + [2t^2 \sin t]_0^{\pi/2} - \int_0^{\pi/2} 4t \sin t \, dt$. $I = [2t^2 \sin t]_0^{\pi/2} = 2(\frac{\pi}{2})^2 \sin(\frac{\pi}{2}) - 0 = 2(\frac{\pi^2}{4})(1) = \frac{\pi^2}{2}$.

Evaluate the definite integral: $\int_0^{\pi^2 / 4} (2 \sin \sqrt{x} + \sqrt{x} \cos \sqrt{x}) \, dx$

Similar Questions

$\int_0^{16} \frac{\sqrt{x}}{1+\sqrt{x}} d x=$

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

$\int_{\pi / 6}^{\pi / 3} \tan ^{3} x \cdot \sin ^{2} 3 x\left(2 \sec ^{2} x \cdot \sin ^{2} 3 x+3 \tan x \cdot \sin 6 x\right) d x$ is equal to

Prove that $\int_{0}^{1} \sin^{-1} x \, dx = \frac{\pi}{2} - 1$.

$\int_0^{\pi /4} (\cos x - \sin x) dx + \int_{\pi /4}^{5\pi /4} (\sin x - \cos x) dx + \int_{2\pi }^{\pi /4} (\cos x - \sin x) dx$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module