Put x=t⇒x=t2⇒dx=2tdt∫0π2/4 cos⁡xdx=∫0π/2 (cos⁡t)⋅2tdt LL⇒ t=0,U.Lt=π24⇒t=π2 = 2t sint π20 -∫0π22sint dt =π-2

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$\int_{0}^{π^{2} / 4} \cos \sqrt{x} d x$

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$π - 2$

$π + 2$

$\frac{π}{2} - 1$

answer is A.

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Detailed Solution

$\begin{array}{l} Put \sqrt{x} = t \Rightarrow x = t^{2} \Rightarrow d x = 2 t d t \\ \int_{0}^{π^{2} / 4} \cos \sqrt{x} d x = \int_{0}^{π / 2} (\cos t) \cdot 2 t d t LL⇒ t = 0, U . L t = \sqrt{\frac{π^{2}}{4} \Rightarrow} t = \frac{π}{2} \end{array} = {(2 t \sin t)}^{\frac{π}{2}}_{0} - \int_{0}^{\frac{π}{2}} 2 \sin t d t = π - 2$