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Question Number 103860 by mathmax by abdo last updated on 17/Jul/20
find∫dxcos4x
Answered by Dwaipayan Shikari last updated on 17/Jul/20
∫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=∫dxcos4xweknow1+tan2x=1cos2x⇒1cos4x=(1+tan2x)2⇒I=∫(1+tan2x)2dx=tanx=u∫(1+u2)2du1+u2=∫(1+u2)du=u+u33+C=tanx+13tan3x+C.
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