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Q17  Integrate the rational functions \frac{\cos x }{(1- \sin x )( 2- \sin x )} 

[Hint : Put \sin x = t]

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Given function \frac{\cos x }{(1- \sin x )( 2- \sin x )}

Applying the given hint: putting \sin x =t

We get, \cos x dx =dt

\therefore \int \frac{\cos x }{(1- \sin x )( 2- \sin x )}dx = \int \frac{dt}{(1-t)(2-t)}

Partial fraction of above equation,

\frac{1}{(1-t)(2-t)} =\frac{A}{(1-t)}+\frac{B}{(2-t)}

1 = A(2-t)+B(1-t)                                      ................(1)

Now, substituting t = 2\ and\ 1 in equation (1), we get

A=1\ and\ B=-1

\therefore \frac{1}{(1-t)(2-t)} = \frac{1}{(1-t)} - \frac{1}{(2-t)}

\implies \int \frac{\cos x }{(1-\sin x)(2-\sin x )}dx = \int \left \{ \frac{1}{1-t}-\frac{1}{(2-t)} \right \}dt

= -\log|1-t| +\log|2-t| +C

= \log\left | \frac{2-t}{1-t} \right |+C

Back substituting the value of t in the above equation, we get

= \log\left | \frac{2-\sin x}{1- \sin x} \right |+C

 

Posted by

Divya Prakash Singh

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