$\int_{π^{2} / 16}^{π^{2} / 4} \frac{sin \sqrt{x}}{\sqrt{x}} d x =$

\sqrt{2}

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1 / \sqrt{2}

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\sqrt{2}

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π / 2

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Solution

The correct option is A $\sqrt{2}$
Let $x^{0.5} = t$
$\frac{1}{2 x^{0.5}} d x = d t$
Putting the values back in the given expression,
$\int 2 s i n t d t$
Integrating the expression, indefinitely
= $- 2 c o s t + c$
Putting back the value of t = $x^{0.5}$
= $- 2 c o s (x^{0.5}) + c$
Applying the limits,
$(2 c o s (\frac{π}{4})) - (2 c o s (\frac{π}{2}))$
= $2^{0.5}$

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Similar questions

\int_{\frac{π^{2}}{16}}^{\frac{π^{2}}{4}} \frac{s i n \sqrt{x}}{\sqrt{x}} d x =

\int_{- π / 2}^{π / 2} cos x

(\frac{1 + x}{1 - x}) d x

If, $f (x) = c o s [π^{2}] x + c o s [- π^{2}] x,$ where [x] stands for greatest integer function, then

π

s i n \frac{A}{2} + s i n \frac{B}{2} + s i n \frac{C}{2} = 1 + 4 s i n [\frac{π - A}{4}] s i n [\frac{π - B}{4}] s i n [\frac{π - C}{4}]

f (x) = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{1 - sin x}{(π - 2 x)^{2}} \cdot \frac{log sin x}{log (1 + π^{2} - 4 π x + 4 x^{2})}, & x \neq \frac{π}{2} k, & x = \frac{π}{2} \end{matrix}

x = \frac{π}{2}

k

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