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

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Answer 1 · 2016-10-09T14:08:16

$F(x) = int (sin2xcos^2x-tanx)dx$

$F(x) = int sin2xcos^2xdx-inttanxdx=I_1-I_2$

where $I_1=int sin2xcos^2xdx and I_2=inttanxdx$

$I_1=int sin2xcos^2xdx$
let $" "cos^2x=u=>-2sinxcosxdx=du=>sin2xdx=-du$

So $I_1=-intudu=-u^2/2=-cos^4x/2$

Again $I_2=inttanxdx=logabs(secx)$

Therefore

$F(x)=I_1-I_2=-cos^4x/2-logabs(secx)+c$

where c = integration constant

again it is given that $F(pi/4)=1$

$F(pi/4)=-cos^4(pi/4)/2-logabs(sec(pi/4))+c$

$=>1=-(1/sqrt2)^4/2-logabs(sqrt2)+c$

$=>1=-1/8-1/2logabs(2)+c$

$=>c=1+1/8+1/2logabs(2)=9/8+1/2logabs(2)$

Hence

$F(x)==-cos^4x/2-logabs(secx)+9/8+1/2logabs(2)$

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