Q117 / 160 · AP EAMCET 2022 (06 Jul Shift 2) · Mathematics
$$
\int_0^{\pi^2 / 4}(2 \sin \sqrt{x}+\sqrt{x} \cos \sqrt{x}) d x=
$$
- A$\frac{\pi}{2}$
- B$\pi$
- Correct answerC$\frac{\pi^2}{2}$
- D$\pi^2$
Solution
$\begin{aligned} & \text { Let } I=\int_0^{\pi^2 / 4} 2 \sin \sqrt{x}+\sqrt{x} \cos \sqrt{x} d x \\ & \begin{aligned} & I= 2 \int_0^{\pi^2 / 4} \sin \sqrt{x} d x+\int_0^{\pi^2 / 4} \sqrt{x} \cos \sqrt{x} d x \\ &\left.=2[x \sin \sqrt{x}]_0^{\pi^2 / 4}-2 \int_0^{\pi^2 / 4} \frac{d}{d x}(\sin \sqrt{x}) \int 1 d x\right] d x \\ &+\int_0^{\pi^2 / 4} \sqrt{x} \cos \sqrt{x} d x \\ &= 2\left[\frac{\pi^2}{4}\right]-2 \int_0^{\pi^2 / 4} \cos \sqrt{x} \times \frac{1}{2 \sqrt{x}} \times x d x \\ & I=+\int_0^{\pi^2 / 4} \sqrt{x} \cos \sqrt{x} d x \\ & 4=\frac{\pi^2}{2}\end{aligned}\end{aligned}$
Hence, the correct answer is option: 'C'.