Question

Evaluate \[ \int_0^{\pi^2/4}\left(2\sin\sqrt{x}+\sqrt{x}\cos\sqrt{x}\right)\,dx \]

• A. \(\frac{\pi}{2}\)
• B. \(\pi\)
• C. \(\frac{\pi^2}{2}\)
• D. \(\pi^2\)

Hint:

When an integral contains \(\sqrt{x}\), use the substitution \(t=\sqrt{x}\). This often converts the integral into a simple derivative form.

Answer 1

The Correct Option is C

Step 1: Use substitution.
Let \[ t=\sqrt{x} \] Then, \[ x=t^2 \] So, \[ dx=2t\,dt \] When \[ x=0,\quad t=0 \] When \[ x=\frac{\pi^2}{4},\quad t=\frac{\pi}{2} \]

Step 2: Transform the integral.
The given integral becomes \[ \int_0^{\pi/2}\left(2\sin t+t\cos t\right)2t\,dt \] \[ = \int_0^{\pi/2}\left(4t\sin t+2t^2\cos t\right)\,dt \]

Step 3: Identify the derivative form.
Observe that \[ \frac{d}{dt}\left(2t^2\sin t\right) = 4t\sin t+2t^2\cos t \] Therefore, \[ \int_0^{\pi/2}\left(4t\sin t+2t^2\cos t\right)\,dt = \left[2t^2\sin t\right]_0^{\pi/2} \]

Step 4: Evaluate the limits.
\[ \left[2t^2\sin t\right]_0^{\pi/2} = 2\left(\frac{\pi}{2}\right)^2\sin\frac{\pi}{2}-0 \] \[ = 2\cdot \frac{\pi^2}{4}\cdot 1 \] \[ = \frac{\pi^2}{2} \]

Step 5: Final conclusion.
Hence, \[ \boxed{\frac{\pi^2}{2}} \]