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

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Answer 1 · 2016-08-25T18:21:13

#f(x)=xsinx+cosx-2-1/sqrt2(pi/4+1)#.

#f(x)=intxcosxdx#

We have to integrate by using, the Rule of Integration by Parts :

#intuvdx=uintvdx-int((du)/dxintvdx)dx#

We take #u=x rArr (du)/dx=1#, and,

#v=cosx rArr intvdx=sinx#

Hence, #f(x)=xsinx-intsinxdx=xsinx+cosx+C#

To determine the const. of Integration #C#, we use,

#f(pi/4)=-2 rArr pi/4sin(pi/4)+cos(pi/4)+C=-2#

# rArr pi/4*1/sqrt2+1/sqrt2+C=-2#

#rArr C=-2-1/sqrt2(pi/4+1)#

Therefore, #f(x)=xsinx+cosx-2-1/sqrt2(pi/4+1)#.