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Question

Findsec x1+cosec xdx.


Solution

Let I=sec x1+cosec xdx=sin xcos x(1+sin x)dx=sin x cos x(1sin x)(1+sin x)2dx
Put sin x=tcos xdx=dt    I=t(1t)(1+t)2dt
Consider t(1t)(1+t)2=A1+t+B(1+t)2+C1t    t=A(1t2)+B(1t)+C(1+t)2
On comparing the coefficients of like terms on both sides, we get: A=14,B=12,C=14.
So, I=(14×11+t12×1(1+t)2×14×11t)dt=14log|1+t|+12+2t14log|1t|+c
I=14log|1+sin x1sin x|+12+2 sin x+c.

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