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)#