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

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(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

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

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

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