F(x) = int (sin2xcos^2x-tanx)dxF(x)=∫(sin2xcos2x−tanx)dx
F(x) = int sin2xcos^2xdx-inttanxdx=I_1-I_2F(x)=∫sin2xcos2xdx−∫tanxdx=I1−I2
where I_1=int sin2xcos^2xdx and I_2=inttanxdxI1=∫sin2xcos2xdxandI2=∫tanxdx
I_1=int sin2xcos^2xdxI1=∫sin2xcos2xdx
let " "cos^2x=u=>-2sinxcosxdx=du=>sin2xdx=-du cos2x=u⇒−2sinxcosxdx=du⇒sin2xdx=−du
So I_1=-intudu=-u^2/2=-cos^4x/2I1=−∫udu=−u22=−cos4x2
Again I_2=inttanxdx=logabs(secx)I2=∫tanxdx=log|secx|
Therefore
F(x)=I_1-I_2=-cos^4x/2-logabs(secx)+cF(x)=I1−I2=−cos4x2−log|secx|+c
where c = integration constant
again it is given that F(pi/4)=1F(π4)=1
F(pi/4)=-cos^4(pi/4)/2-logabs(sec(pi/4))+cF(π4)=−cos4(π4)2−log∣∣sec(π4)∣∣+c
=>1=-(1/sqrt2)^4/2-logabs(sqrt2)+c⇒1=−(1√2)42−log∣∣√2∣∣+c
=>1=-1/8-1/2logabs(2)+c⇒1=−18−12log|2|+c
=>c=1+1/8+1/2logabs(2)=9/8+1/2logabs(2)⇒c=1+18+12log|2|=98+12log|2|
Hence
F(x)==-cos^4x/2-logabs(secx)+9/8+1/2logabs(2)F(x)==−cos4x2−log|secx|+98+12log|2|
