What is the integral of sec(9x)tan(9x) ?

The integral of sec(9x)tan(9x) is 1/9 sec(9x)+C

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integral of sec(9x)tan(9x)

\int sec (9 x) tan (9 x) d x

\frac{1}{9} sec (9 x) + C

Solution steps

\int sec (9 x) tan (9 x) d x

Apply u-substitution

: \int \frac{1}{9} d u

= \int \frac{1}{9} d u

Integral of a constant:

\int a d x = a x

= \frac{1}{9} u

Substitute back

u = sec (9 x)

= \frac{1}{9} sec (9 x)

Add a constant to the solution

= \frac{1}{9} sec (9 x) + C

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What is the integral of sec(9x)tan(9x) ?
The integral of sec(9x)tan(9x) is 1/9 sec(9x)+C