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

1 Answer
Sep 22, 2016

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)