What is the integral of sec(ax)tan(ax) ?

The integral of sec(ax)tan(ax) is 1/a sec(ax)+C

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integral of sec(ax)tan(ax)

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

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

Solution steps

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

Apply u-substitution

: \int \frac{sec ( u ) tan ( u )}{a} d u

= \int \frac{sec ( u ) tan ( u )}{a} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{a} \cdot \int sec (u) tan (u) d u

Apply u-substitution

: \int 1 d v

= \frac{1}{a} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= \frac{1}{a} \cdot 1 \cdot v

Substitute back

= \frac{1}{a} \cdot 1 \cdot sec (a x)

Simplify

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

Add a constant to the solution

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

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What is the integral of sec(ax)tan(ax) ?
The integral of sec(ax)tan(ax) is 1/a sec(ax)+C