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Question Number 103860 by mathmax by abdo last updated on 17/Jul/20

find ∫ (dx/(cos^4 x))

finddxcos4x

Answered by Dwaipayan Shikari last updated on 17/Jul/20

∫(dx/(cos^4 x))=∫sec^4 xdx=∫sec^2 x(1+tan^2 x)dx=∫(1+t^2 )dt=t+(t^3 /3)+C  tanx+((tan^3 x)/3)+C           {put t=tanx

dxcos4x=sec4xdx=sec2x(1+tan2x)dx=(1+t2)dt=t+t33+Ctanx+tan3x3+C{putt=tanx

Answered by mathmax by abdo last updated on 17/Jul/20

I =∫  (dx/(cos^4 x))  we know  1+tan^2 x =(1/(cos^2 x)) ⇒(1/(cos^4 x)) =(1+tan^2 x)^2 ⇒  I =∫   (1+tan^2 x)^2  dx =_(tanx =u)    ∫ (1+u^2 )^2 (du/(1+u^2 )) =∫ (1+u^2 )du  =u +(u^3 /3)  +C =tanx +(1/3)tan^3 x +C .

I=dxcos4xweknow1+tan2x=1cos2x1cos4x=(1+tan2x)2I=(1+tan2x)2dx=tanx=u(1+u2)2du1+u2=(1+u2)du=u+u33+C=tanx+13tan3x+C.

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