What is #int_(pi/8)^((11pi)/12) x^2-cos^2x+xsinxdx#?

1 Answer
Jun 29, 2016

#approx 10#

Explanation:

#int_(pi/8)^((11pi)/12)dx \ (x^2-cos^2x+xsinx)#

using identity # cos 2 z = 2 cos^2 z - 1 \implies cos ^2 z = (cos 2z + 1)/(2)#

#= int_(pi/8)^((11pi)/12)dx \ (x^2-(cos 2x + 1)/(2)+ color{red}{xsinx})#

first 2 terms are trivial, we do the term in red using IBP, ie

#int dx \ (u v') = uv - int dx \ (u' v)#

Here

#u = x, u' = 1#
#v' = sin x, v = -cos x#

so we have
#- x cos x - int dx \ (-cos x)#
#color{red}{= - x cos x + intdx \ (cos x)}#

so the integral becomes

#= [-x cos x] \_(pi/8)^((11pi)/12) + int_(pi/8)^((11pi)/12)dx \ (x^2-(cos 2x + 1)/(2)+ cos x) #

#= [-x cos x + x^3/3-(sin 2x)/4 - x/2+ sin x ] \_(pi/8)^((11pi)/12)#

crunch it in calculator ....

#approx 10#