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Q33  Integrate the functions  \frac{1}{1- \tan x }

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Given function  \frac{1}{1- \tan x }

Assume that I = \int \frac{1}{1- \tan x } dx

 Now solving the assumed integral;

I = \int \frac{1}{1-\frac{\sin x}{\cos x }} dx

= \int \frac{\cos x }{\cos x - \sin x } dx

= \frac{1}{2}\int \frac{2\cos x }{\cos x - \sin x } dx

= \frac{1}{2}\int \frac{(\cos x -\sin x ) +(\cos x +\sin x ) }{(\cos x - \sin x) } dx

=\frac{1}{2}\int 1 dx + \frac{1}{2} \int \frac{\cos x +\sin x }{\cos x -\sin x } dx

=\frac{1}{2}(x) + \frac{1}{2} \int \frac{\cos x +\sin x }{\cos x -\sin x } dx

Now, to solve further we will assume \cos x - \sin x =t

Or, (-\sin x-\cos x )dx =dt

\therefore I = \frac{x}{2}+ \frac{1}{2}\int \frac{-(dt)}{t}

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

Now, back substituting the value of t,

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

Posted by

Divya Prakash Singh

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