Let I = ( x) d x Put x=t x = e ^ t d x = e ^ t dt I = t e ^ t dt = t e ^ t dt - [ d dt ( t ) e ^ t dt ] dt = t e ^ t - t e ^ t dt = e ^ t t - [ t e ^ t dt - ( d dt ( t ) e ^ t dt ) dt ] = e ^ t t - [ e ^ t t - (- t ) e ^ t dt ] = e ^ t t - e ^ t t - t e ^ t dt I = e ^ t ( t - t )- I + c _1 2 I = e ^ t ( t - t )+ c _1 I = e ^ t 2 ( t - t )+ c _1 2 I =x 2 [ ( x)- ( x)]+ c _r where c = c _1 2

( x) d x

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Question

$\int \sin (\log x) d x$

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Answer

Let $I =\int \sin (\log x) d x$
Put $\log x=t$
$\therefore x = e ^{ t }$
$\therefore d x = e ^{ t } dt$
$\therefore I =\int \sin t \cdot e ^{ t } dt$
$=\sin t \int e ^{ t } dt -\int\left[\frac{ d }{ dt }(\sin t ) \int e ^{ t } dt \right] dt$
$=\sin t \cdot e ^{ t }-\int \cos t \cdot e ^{ t } dt$
$= e ^{ t } \sin t -\left[\cos t \int e ^{ t } dt -\int\left(\frac{ d }{ dt }(\cos t ) \int e ^{ t } dt \right) dt \right]$
$= e ^{ t } \sin t -\left[ e ^{ t } \cos t -\int(-\sin t ) e ^{ t } dt \right]$
$= e ^{ t } \sin t - e ^{ t } \cos t -\int \sin t \cdot e ^{ t } dt$
$\therefore I = e ^{ t }(\sin t -\cos t )- I + c _1$
$\therefore 2 I = e ^{ t }(\sin t -\cos t )+ c _1$
$\therefore I =\frac{ e ^{ t }}{2}(\sin t -\cos t )+\frac{ c _1}{2}$
$\therefore I =\frac{x}{2}[\sin (\log x)-\cos (\log x)]+ c _r$
$\text { where } c =\frac{ c _1}{2}$

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