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Question: Answered & Verified by Expert
$$
\int_0^{\pi^2 / 4}(2 \sin \sqrt{x}+\sqrt{x} \cos \sqrt{x}) d x=
$$
MathematicsDefinite IntegrationAP EAMCETAP EAMCET 2022 (06 Jul Shift 2)
Options:
  • A $\frac{\pi}{2}$
  • B $\pi$
  • C $\frac{\pi^2}{2}$
  • D $\pi^2$
Solution:
2604 Upvotes Verified Answer
The correct answer is: $\frac{\pi^2}{2}$
$\begin{aligned} & \text { Let } I=\int_0^{\pi^2 / 4} 2 \sin \sqrt{x}+\sqrt{x} \cos \sqrt{x} d x \\ & \begin{aligned} & I= 2 \int_0^{\pi^2 / 4} \sin \sqrt{x} d x+\int_0^{\pi^2 / 4} \sqrt{x} \cos \sqrt{x} d x \\ &\left.=2[x \sin \sqrt{x}]_0^{\pi^2 / 4}-2 \int_0^{\pi^2 / 4} \frac{d}{d x}(\sin \sqrt{x}) \int 1 d x\right] d x \\ &+\int_0^{\pi^2 / 4} \sqrt{x} \cos \sqrt{x} d x \\ &= 2\left[\frac{\pi^2}{4}\right]-2 \int_0^{\pi^2 / 4} \cos \sqrt{x} \times \frac{1}{2 \sqrt{x}} \times x d x \\ & I=+\int_0^{\pi^2 / 4} \sqrt{x} \cos \sqrt{x} d x \\ & 4=\frac{\pi^2}{2}\end{aligned}\end{aligned}$

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