What is the integral of sec(13w)tan(13w) ?

The integral of sec(13w)tan(13w) is 1/13 sec(13w)+C

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integral of sec(13w)tan(13w)

\int sec (13 w) tan (13 w) d w

\frac{1}{13} sec (13 w) + C

Solution steps

\int sec (13 w) tan (13 w) d w

Apply u-substitution

: \int \frac{1}{13} d u

= \int \frac{1}{13} d u

Integral of a constant:

\int a d x = a x

= \frac{1}{13} u

Substitute back

u = sec (13 w)

= \frac{1}{13} sec (13 w)

Add a constant to the solution

= \frac{1}{13} sec (13 w) + C

\int_{0}^{\frac{π}{4}} tan^{3} (x) d x

(2 y^{2} x + 3 x^{2}) d x + 2 x^{2} y d y = 0

\frac{d}{d x} (\frac{x ^{3}}{x + 2})

x^{9} y^{^{'}} - 9 x^{8} y = 5 y^{2}

\frac{d}{d t} (t - 2 (t - 1) u_{(1)} (t) + (t - 2) u_{(2)} (t) (t))

What is the integral of sec(13w)tan(13w) ?
The integral of sec(13w)tan(13w) is 1/13 sec(13w)+C