Question

What is `F(x) = int sin2xcos^2x-tan^3x dx` if `F(pi/3) = 1`?

Answer 1

`F(x)=-cos^4x/2-tan^2x/2-lnabscosx+81/32-ln(2)`

Explanation:

`F(x)=int(sin2xcos^2x-tan^3x)dx`

Note that `sin2x=2sinxcosx`. Also, rewrite `tan^3x` as `tanxtan^2x=tanx(sec^2x-1)`.

`=2intsinxcos^3xdx-inttanx(sec^2x-1)dx`

`=2intcos^3xsinxdx-inttanxsec^2xdx+inttanxdx`

For the first integral, let `u=cosx` so `du=-sinxdx`:

`=-2intu^3du-inttanxsec^2xdx+inttanxdx`

`=-2(u^4/4)-inttanxsec^2xdx+inttanxdx`

`=-cos^4x/2-inttanxsec^2xdx+inttanxdx`

Now, let `v=tanx` so that `dv=sec^2xdx`:

`=-cos^4x/2-intvdv+inttanxdx`

`=-cos^4x/2-v^2/2+inttanxdx`

`=-cos^4x/2-tan^2x/2+inttanxdx`

`=-cos^4x/2-tan^2x/2+intsinx/cosxdx`

Again, let `w=cosx` so `dw=-sinxdx`:

`=-cos^4x/2-tan^2x/2-int(dw)/w`

`=-cos^4x/2-tan^2x/2-lnabsw`

`F(x)=-cos^4x/2-tan^2x/2-lnabscosx+C`

Apply the original condition `F(pi/3)=1`:

`1=-cos^4(pi/3)/2-tan^2(pi/3)/2-lnabscos(pi/3)+C`

`1=-(1/2)^4/2-(sqrt3)^2/2-ln(1/2)+C`

Note that `ln(1/2)=ln(2^-1)=-ln(2)`:

`1=-(1/16)/2-3/2+ln(2)+C`

`32/32=-1/32-48/32+ln(2)+C`

`C=81/32-ln(2)`

Thus:

`F(x)=-cos^4x/2-tan^2x/2-lnabscosx+81/32-ln(2)`