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प्रश्न
Find `int (2cos x)/((1-sinx)(1+sin^2 x)) dx`
उत्तर
Let `sin x = t => cos x dx = dt`
`int (2dt)/((1-t)(1+t^2))`
Using partial fraction
`2/((1-t)(1+t^2)) = A/((1-t)) + (Bt + C)/((1+t^2))`
On solving A = 1, B =1, C = 1
`int (2dt)/((1-t)(1+t^2)) = int (dt)/((1-t)) + int ((1+t))/((1+t^2)) dt`
`= int (dt)/((1-t)) + int (dt)/(1+t^2) + int (tdt)/((1+t^2))`
`= -In (1-t) + tan^(-1) t + 1/2 In (1+t^2)`
`= In sqrt(1+t^2)/(1-t) + tan^(-1) t + C`
Replacing the value of t
`= In sqrt(1+sin^2x)/(1-sinx) + tan^(-1)(sin x) + C`
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