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Question

1sin4 x+cos4 x dx

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Solution

We have,I=dxsin4x+cos4x
Dividing numerator and denominator by cos4x
I=sec4x dxtan4x+1=sec2x sec2x dxtan4x+1=1+tan2x sec2x dxtan4x+1Putting tan x=tsec2x dx=dtI=1+t2 dtt4+1=1t2+1 dtt2+1t2=1+1t2t-1t2+2dtPutting t-1t=p1+1t2dt=dpI=1p2+22dp=12 tan-1p2+C=12 tan-1t-1t2+C=12 tan-1 t2-12 t+C=12 tan-1 tan2x-12 tan x+C=12 tan-1-2×1-tan2x2 tan x+C=12 tan-1-2tan 2x+C=12 tan-1-2 cot 2x+C

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