Question

can you solve this integral​

Answer 1

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▪ To Find :-

$I=\sf\int \dfrac{ \sin \sqrt\mathtt x}{ \sqrt{\mathtt x} } d\mathtt x$

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▪ Solution :-

$\large \bigstar \:  { \pmb{ \mathfrak{ Put \: \: }}} \: \: \bf\sqrt{x} = y : \\ \\ \sf : \longmapsto \bigg(\frac{1}{2 \sqrt{\mathtt x} } \bigg)d\mathtt x = dy \\ \\ \bf : \longmapsto \frac{1}{ x} \: dx = dy$

$\Large \bigstar \:   \underline{ \pmb{ \mathfrak{ \text{Now} , }}}$

$I = \displaystyle\sf\int \dfrac{ \sin \sqrt\mathtt x}{ \sqrt{\mathtt x} } .d\mathtt x =2 \int \sin y.dy \\ \\ \sf = - 2 \cos y + C$

$\large \bigstar \:   \underline{ \pmb{ \mathfrak{ Putting \: \: Back \: \: The \: \: \text Value \: \: of \: \: \bf y}}}$

$\:\: \:\:\:\large\purple{\bf I = - 2 cos \sqrt{x} + C}$

$\Large \red{\mathfrak{  \text{W}hich \:   \: is  \:  \: the  \:  \: required} }\\ \huge \red{\mathfrak{ \text{ A}nswer.}}$

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▪ Related Calculations :-

$\large \bigstar \:   \underline{ \pmb{ \mathfrak{ Derivative \: \: of \: \: \bf\sqrt x :-}}}$

$\sf \dfrac{d}{d \mathtt x} \sqrt{\mathtt x} \\ \\ = \sf\frac{d}{d\mathtt x} {\mathtt x}^{ 1/2} \\ \\ = \frac{1}{2} {\mathtt x}^{ - 1/2} \\ \\ = \sf \frac{1}{2} \times \frac{1}{ {\mathtt x}^{1/2} } \\ \\ = \bf \frac{1}{2 \sqrt{x} } \\ \\ : \longmapsto \bf\dfrac{d}{d \mathtt x} \sqrt{\mathtt x} = \frac{1}{2 \sqrt{x} }$

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Additional Information :-

$\large \bigstar \:   \underline{ \pmb{ \mathfrak{ Commonl \text y \: \: Used \: \: \text Integrals }}}$

$\maltese \: \: \: \displaystyle \bf\int \dfrac{1}{x} \: dx =log |x + | C \\ \\ \maltese \: \: \: \displaystyle \bf\int {a}^{x} \: dx = \frac{ {a}^{x} }{log \: a} + C \\ \\ \maltese \: \: \: \displaystyle \bf\int sinx \: dx = - cosx + C \\ \\ \maltese \: \: \: \displaystyle \bf\int cosx \: dx =sinx + C \\ \\ \maltese \: \: \: \displaystyle \bf\int sec {}^{2} x \: dx =tanc + C \\ \\ \maltese \: \: \: \displaystyle \bf\int {cosec}^{2} x \: dx = - cotx + C \\ \\ \maltese \: \: \: \displaystyle \bf\int secx.tanx \: dx =secx + C \\ \\ \maltese \: \: \: \displaystyle \bf\int cosecx.cotx \: dx = - cosecx + C$

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Answer 2

Here's The Answer

$Let \:\:\sqrt x = t\\\\ \implies\bigg(\frac{1}{2 \sqrt{ x} } \bigg)d x = dt \\ \\ \implies \frac{1}{\sqrt x} \: d x = 2dt$

$\displaystyle\int \dfrac{ \sin \sqrt x}{ \sqrt{x} } .d x =2 \int \sin t.dt \\ \\ = - 2 \cos t+ C$

Putting Value of y

$\implies \bf Integral=- 2 cos \sqrt{x} + C$

where C is the constant of Integration .