Question

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

Answer 1

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`

now`f(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`