Mathematics

# $\displaystyle \int x \sin \dfrac {x}{2}=?$

##### SOLUTION
Sol:
$\displaystyle \int x\sin \dfrac {x}{2}dx$
Let
$I=\displaystyle \int x\sin \dfrac {x}{2}dx$
We know that
$I=\displaystyle \int \Pi dx =I\displaystyle \int \Pi dx-\int \left (\dfrac {dI}{dx}\displaystyle \int \Pi dx\right)dx+C$
Now, using $LAT \varepsilon$
$I=\displaystyle \int { x\sin { \dfrac { x }{ 2 } dx } =x } \displaystyle \int { \sin { \dfrac { x }{ 2 } dx\quad - } } \left[ \displaystyle \int { \left( \dfrac { dx }{ dx } \displaystyle \int { \sin { \dfrac { x }{ 2 } } dx } \right) } dx \right] +C$
$=x\dfrac { \left[ -\cos { \dfrac { x }{ 2 } } \right] }{ \dfrac { 1 }{ 2 } } -\displaystyle \int { \dfrac { \left( -\cos { \dfrac { x }{ 2 } } \right) }{ \dfrac { 1 }{ 2 } } } dx+C$
$=-2x\cos \dfrac {x}{2}+\displaystyle \int 2\cos \dfrac {x}{2}dx+C$
$=2x\cos \dfrac {x}{2}+2\displaystyle \int \cos \dfrac {x}{2}dx+C$
$=-2cos \dfrac {x}{2}+2\left [\dfrac {\dfrac {\sin x}{2}}{\dfrac {1}{2}}\right ]+C$
$=-cos \dfrac {x}{2}+4\sin \dfrac {x}{2}+C$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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$\int_{}^{} {\frac{{ - 1}}{{\sqrt {1 - {x^2}} }}dx}$