Math, asked by IamRocketBoy, 19 hours ago

PLEASE HELP ME ASAP

TOPIC - INTEGRATION

♡♡Find The Value Of Given Integral ♡♡

 \large \int \dfrac{sin \sqrt{x} }{ \sqrt{x} } dx \\

Answers

Answered by SparklingBoy
362
  • Let The Given Integral be = I

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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}ns\text{w}er.}}

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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 Knowledge :-

\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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Answered by BrainlyTurtle
232

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 .

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