What is #f(x) = int xsinx dx# if #f(pi/4)=-2 #?

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Answer 1 · 2016-06-29T09:24:33

#= - x cos x + sin x + 1/sqrt 2 (pi/4 -1) - 2#

#f(x) = int dx \ (x \ sinx)#

just IBP its is easiest way
Wikipedia

Here
#u = x , u' =1 #

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

So we have

#- x cos x - int dx \ (- cos x * 1)#

#= - x cos x + int dx \ ( cos x)#

#= - x cos x + sin x + C#

now using the IV

#-2 = - pi/4 1/sqrt 2 + 1/sqrt 2 + C#

# C = 1/sqrt 2 (pi/4 -1) -2 #