Mathematics

# Solve : $\displaystyle2 \underset{0}{\overset{\pi / 2}{\int}} x \, sin \, x \, dx$

##### SOLUTION
$I=2\displaystyle\int^{\dfrac{\pi}{2}}_0x\sin xdx$
Integration by parts
$\displaystyle\int^b_af(x)g(x)dx=f(x)\displaystyle\int g(x)dx\displaystyle\int^b_a-\displaystyle\int^b_af'(x)g(x)dx$
$I=2\left[x[-\cos x]^{\dfrac{\pi}{2}}_0-\displaystyle\int^{\dfrac{\pi}{2}}_0(1)(-\cos x)dx\right]$
$=2\left[-\left(\dfrac{\pi}{2}\cos \dfrac{\pi}{2}-0\right)+\displaystyle\int^{\dfrac{\pi}{2}}_0\cos xdx\right]$
$=2\left[\left[(0-0)+\sin x\right]^{\dfrac{\pi}{2}}_0\right]$
$=2\left[\sin\dfrac{\pi}{2}-\sin (0)\right]\Rightarrow 2\left[1-0\right]\Rightarrow 2$
$\therefore 2\displaystyle\int^{\dfrac{\pi}{2}}_0x\sin xdx=2$.

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

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