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Q21  Integrate the functions  \tan ^2 ( 2x-3 )

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Given function  \tan ^2 ( 2x-3 ),

Assume the 2x-3 =t

 \therefore 2dx =dt

\implies \int \tan^2(2x-3) dx = \frac{\int \tan^2(t)}{2}dt

=\frac{1}{2}\int (\sec^2t -1) dt                      \left [\because \tan^2t+1 = \sec^2 t \right ]

= \frac{1}{2}\left [ \tan t - t \right ] +C

Now, back substituting the value of t,

= \frac{1}{2}\left [ \tan(2x-3)-2x+3 \right ]+C

or \frac{1}{2} \tan(2x-3)-x+C , where C is any constant value.

Posted by

Divya Prakash Singh

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