What is F(x) = int sin2xcos^2x-tan^3x dx if F(pi/3) = 1 ?
1 Answer
Explanation:
F(x)=int(sin2xcos^2x-tan^3x)dx
Note that
=2intsinxcos^3xdx-inttanx(sec^2x-1)dx
=2intcos^3xsinxdx-inttanxsec^2xdx+inttanxdx
For the first integral, let
=-2intu^3du-inttanxsec^2xdx+inttanxdx
=-2(u^4/4)-inttanxsec^2xdx+inttanxdx
=-cos^4x/2-inttanxsec^2xdx+inttanxdx
Now, let
=-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
=-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
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
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)