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Question

15-4 cos x dx

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Solution

Let I= 15-4 cos xdxPutting cos x=1-tan2 x21+tan2 x2 I= 15-4 1-tan2 x21+tan2 x2dx = 1+tan2 x25 1+tan2 x2-4+4 tan2 x2dx = sec2 x2 9 tan2 x2+1dxLet tan x2=t12 sec2 x2dx=dtsec2 x2dx=2dtI=2dt9t2+1 =29dtt2+19 =29 dtt2+132 =29×3 tan-1 t13+C =23 tan-1 3t+C =23 tan-1 3 tan x2+C

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