Question

What is `f(x) = int xsinx^2 + tan^2x -cos^2x dx` if `f(0)=-3`?

Answer 1

`f(x)=-1/2cosx^2+tanx-3/2x-1/2sinxcosx-5/2`

Explanation:

`f(x)=intxsinx^2+tan^2x-cos^2xdx`

The first term is of the form `g'(x)g(x)`, so integrates to `-1/2cosx^2`.

The second term we transform by use of a standard trig identity:
`tan^2x=sec^2x-1`. We know that `d/dx(tanx)=sec^2x`, so the second term integrates to `tanx-x`.

The third term we transform by use of another standard trig identity:
`cos^2x=1/2(1+cos2x)`
So
`int-cos^2xdx=-1/2int1+cos2xdx=-1/2(x+1/2sin2x)`
Recall that `sin2x=2sinxcosx`, and so the third term integrates to
`-1/2(x+sinxcosx)`.

Put the three integrated terms together:
`f(x)=-1/2cosx^2+tanx-3/2x-1/2sinxcosx+C`

Solve for C using the given condition, `f(0)=-3`:
`-3=-1/2+0-3/2*0-1/2*0*1+C`
`-3=-1/2+C`
`C=-5/2`

So
`f(x)=-1/2cosx^2+tanx-3/2x-1/2sinxcosx-5/2`