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

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Answer 1 · 2016-09-22T20:45:46

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

#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)#