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Q22  Integrate the functions  \sec ^2 ( 7- 4x )

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Given function  \sec ^2 ( 7- 4x ),

Assume the 7-4x=t

 \therefore -4dx =dt

\implies \int \sec^2(7-4x)dx = \frac{-1}{4}\int \sec^2t dt

=-\frac{1}{4}(\tan t) +C                      

Now, back substituted the value of t.

=-\frac{1}{4}\tan(7-4x)+C, where C is any constant value.

Posted by

Divya Prakash Singh

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