Question

${\int }_{\frac{{\pi }^2}{16}}^{\frac{{\pi }^2}{4}}\frac{sin\sqrt{x}}{\sqrt{x}}dx=$

• A. $\sqrt{2}$
• B. $\frac{1}{\sqrt{2}}$
• C. $2\sqrt{2}$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is A $\sqrt{2}$

Put $\sqrt{x}=t\Rightarrow \frac{1}{2\sqrt{x}}dx=dt$
$x=\frac{{\pi }^2}{16}\Rightarrow t=\frac{\pi }{4},x=\frac{{\pi }^2}{4}\Rightarrow t=\frac{\pi }{2}$
${\int }_{\frac{{\pi }^2}{16}}^{\frac{{\pi }^2}{4}}\frac{sin\sqrt{x}}{\sqrt{x}}dx={\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}[sint.2dt=-2cost]{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}=-2[0-\frac{1}{\sqrt{2}}]=\sqrt{2}$