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

Answer 1

Luke Alvarez

#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)#

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Answer 2

Charles Anctil

The integral ( F(x) = \int (\sin(2x)\cos^2(x) - \tan(x)) , dx ) evaluates to ( F(x) = \frac{1}{3}\cos^3(x) - \ln|\cos(x)| + C ). If ( F\left(\frac{\pi}{3}\right) = 1 ), then ( \frac{1}{3}\cos^3\left(\frac{\pi}{3}\right) - \ln|\cos\left(\frac{\pi}{3}\right)| + C = 1 ). Solving for ( C ), we get ( C = \frac{2}{3} - \ln\left(\frac{\sqrt{3}}{2}\right) ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.