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

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Answer 1 · 2016-11-21T09:01:30

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

Parts formula.

$intu(dv)/(dx)dx=uv-intv(du)/(dx)dx$

$int_0^(pi/2)xcosxdx$

$u=x=>(du)/(dx)=1$

$(dv)/(dx)=cosx=>v=sinx$

$I=[xsinx-int(sinx)dx]_0^(pi/2)$

$I=[xsinx+cosx]_0^(pi/2)$

now substitute the limits in

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

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

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

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