∫sin2xsin4x+cos4xdx=?
2tan-1(tan2x)+c
tan-1(xtan2x)+c
tan-1(tan2x)+c
None of these
Explanation For The Correct Option:
Evaluating the integral:
∫sin2xsin4x+cos4xdx
⇒∫sin2xsin2x2+cos2x2⇒∫sin2xsin2x+cos2x2-2sin2xcos2xdx∵a+b2=a2+b2+2ab⇒a2+b2=a+b2-2ab⇒∫sin2x1-2sin2xcos2xdx∵sin2x+cos2x=1⇒∫sin2x1-sin2x22dx∵sin2x=2sinxcosx⇒2∫sin2x2-sin2x2dx⇒2∫sin2x1+cos22xdx∵sin2θ=1-cos2θ
Substituting t=cos(2x) then also putting value of dx after differentiating t w.r.t. x,
dt=-2sin2xdx
=-∫dt1+t2=-tan-1t+C[Cisintegratingconstant]=-tan-1cos(2x)+csubstitutingbackt=cos(2x)
Therefore, ∫sin2xsin4x+cos4xdx=tan-1cos2x+c
Hence, option (D) is the correct answer.