What is f(x) = int xsin2x+cosx dx if f(pi/2)=-4 ?

1 Answer
Apr 24, 2018

see below

Explanation:

I=int(xsin2x+cosx)dx

we have two integrals, the first one needs to be done by parts, the second is a standard integral. We will omit the constant of integration until the end.

f(x)=I=int(xsin2x)dx+intcosxdx=I_1+I_2

I_2=intcosxdx=sinx--(1)

I_1=intxsin2xdx

IBP

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

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

(dv)/(dx)=sin2x=>v=-1/2cos2x

I_1=-1/2xcos2x+int1/2cos2xdx

I_1=-1/2xcos2x+1/4sin2x

:.I=I-1+I_2=f(x)=-1/2xcos2x+1/4sin2x+sinx+c

nowf(pi/2)=-4

-4=-1/2pi/2cos2(pi/2)+1/4sin(2(pi/2))+sin(pi/2)+c

-4=-(pi/4)cancel(cos(pi))^(-1)+cancel(1/4sinpi)^0+cancel(sin(pi/2))^1+c

=>-4=pi/4+1+c

c=-5-pi/4

f(x)=-1/2xcos2x+1/4sin2x+sinx-5-pi/4