How do you integrate $\int x cos x$ $int xcosx$ by parts from $[0, \frac{π}{2}]$ $[0,pi/2]$ ?

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Answer 1 · 2016-11-21T09:01:30

$\int_{0}^{\frac{π}{2}} x cos x d x = \frac{π}{2} - 1$ $int_0^(pi/2)xcosxdx=pi/2-1$

Parts formula.

$\int u \frac{d v}{d x} d x = u v - \int v \frac{d u}{d x} d x$ $intu(dv)/(dx)dx=uv-intv(du)/(dx)dx$

$\int_{0}^{\frac{π}{2}} x cos x d x$ $int_0^(pi/2)xcosxdx$

$u = x \Rightarrow \frac{d u}{d x} = 1$ $u=x=>(du)/(dx)=1$

$\frac{d v}{d x} = cos x \Rightarrow v = sin x$ $(dv)/(dx)=cosx=>v=sinx$

$I = {[x sin x - \int (sin x) d x]}_{0}^{\frac{π}{2}}$ $I=[xsinx-int(sinx)dx]_0^(pi/2)$

$I = {[x sin x + cos x]}_{0}^{\frac{π}{2}}$ $I=[xsinx+cosx]_0^(pi/2)$

now substitute the limits in

$I = [(\frac{π}{2}) sin (\frac{π}{2}) + cos (\frac{π}{2})] - [0 \times sin 0 + cos 0]$ $I=[(pi/2)sin(pi/2)+cos(pi/2)]-[0xxsin0+cos0]$

$I=[(pi/2)cancel(sin(pi/2))^1+cancel(cos(pi/2))^0]-[cancel(0xxsin0)^0+cancel(cos0)^1]$

$int_0^(pi/2)xcosxdx=pi/2-1$

How do you integrate int xcosx∫xcosxint xcosx by parts from [0,pi/2][0,π2][0,pi/2]?