What is F(x) = int sin2xcos^2x-tan^3x dxF(x)=∫sin2xcos2x−tan3xdx if F(pi/3) = 1 F(π3)=1?
1 Answer
Explanation:
F(x)=int(sin2xcos^2x-tan^3x)dxF(x)=∫(sin2xcos2x−tan3x)dx
Note that
=2intsinxcos^3xdx-inttanx(sec^2x-1)dx=2∫sinxcos3xdx−∫tanx(sec2x−1)dx
=2intcos^3xsinxdx-inttanxsec^2xdx+inttanxdx=2∫cos3xsinxdx−∫tanxsec2xdx+∫tanxdx
For the first integral, let
=-2intu^3du-inttanxsec^2xdx+inttanxdx=−2∫u3du−∫tanxsec2xdx+∫tanxdx
=-2(u^4/4)-inttanxsec^2xdx+inttanxdx=−2(u44)−∫tanxsec2xdx+∫tanxdx
=-cos^4x/2-inttanxsec^2xdx+inttanxdx=−cos4x2−∫tanxsec2xdx+∫tanxdx
Now, let
=-cos^4x/2-intvdv+inttanxdx=−cos4x2−∫vdv+∫tanxdx
=-cos^4x/2-v^2/2+inttanxdx=−cos4x2−v22+∫tanxdx
=-cos^4x/2-tan^2x/2+inttanxdx=−cos4x2−tan2x2+∫tanxdx
=-cos^4x/2-tan^2x/2+intsinx/cosxdx=−cos4x2−tan2x2+∫sinxcosxdx
Again, let
=-cos^4x/2-tan^2x/2-int(dw)/w=−cos4x2−tan2x2−∫dww
=-cos^4x/2-tan^2x/2-lnabsw=−cos4x2−tan2x2−ln|w|
F(x)=-cos^4x/2-tan^2x/2-lnabscosx+CF(x)=−cos4x2−tan2x2−ln|cosx|+C
Apply the original condition
1=-cos^4(pi/3)/2-tan^2(pi/3)/2-lnabscos(pi/3)+C1=−cos4(π3)2−tan2(π3)2−ln∣∣cos(π3)∣∣+C
1=-(1/2)^4/2-(sqrt3)^2/2-ln(1/2)+C1=−(12)42−(√3)22−ln(12)+C
Note that
1=-(1/16)/2-3/2+ln(2)+C1=−1162−32+ln(2)+C
32/32=-1/32-48/32+ln(2)+C3232=−132−4832+ln(2)+C
C=81/32-ln(2)C=8132−ln(2)
Thus:
F(x)=-cos^4x/2-tan^2x/2-lnabscosx+81/32-ln(2)F(x)=−cos4x2−tan2x2−ln|cosx|+8132−ln(2)
