GATE 2021 BT – Question 59

Correct answer: 2

Explanation

**Step 1: substitute.** Let $u=\sqrt x$, so $x=u^2$ and $dx=2u\,du$. The limits change:
- $x=0\Rightarrow u=0$,
- $x=\pi^2/4\Rightarrow u=\pi/2$.

$$\int_0^{\pi^2/4}\sin\sqrt x\,dx=2\int_0^{\pi/2}u\sin u\,du .$$

**Step 2: integrate by parts.** With $f=u$ and $dg=\sin u\,du$:
$$\int u\sin u\,du=-u\cos u+\int\cos u\,du=-u\cos u+\sin u .$$

**Step 3: evaluate.**
$$2\Big[-u\cos u+\sin u\Big]_0^{\pi/2}=2\Big[(0+1)-(0+0)\Big]=\mathbf{2}.$$