First slide
Methods of integration
Question

 Evaluate 13+sin2xdx 

Moderate
Solution

I=13+sin2xdx=13sin2x+cos2x+2sinxcosxdx=sec2x3tan2x+2tanx+3dx[Dividing Nr and Dr by cos2 x]                Putting tan x = t and sec2 x dx = dt, we get I=dt3t2+2t+3=13dtt2+23t+1=13dtt+132+2232I=131223tan1t+13223+C=122tan13t+122+C=122tan13tan x+122+C

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