How do you find the integral of $sec (3 x) sec (3 x)$ $sec(3x)sec(3x)$ ?

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Answer 1 · 2018-07-23T10:35:29

$\frac{1}{3} tan (3 x) + C$ $1/3\tan(3x)+C$

Given that

$\int \sec(3x)\sec(3x)\ dx$

$=\int \sec^2(3x)\ dx$

$=1/3\int \sec^2(3x)\3 dx$

$=1/3\int \sec^2(3x)\d(3x)$

$=1/3\tan(3x)+C$

How do you find the integral of sec(3x)sec(3x)sec(3x)sec(3x)sec(3x)sec(3x)?