What is the integral of sec(17w)tan(17w) ?

The integral of sec(17w)tan(17w) is 1/17 sec(17w)+C

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integral of sec(17w)tan(17w)

\int sec (17 w) tan (17 w) d w

\frac{1}{17} sec (17 w) + C

Solution steps

\int sec (17 w) tan (17 w) d w

Apply u-substitution

: \int \frac{1}{17} d u

= \int \frac{1}{17} d u

Integral of a constant:

\int a d x = a x

= \frac{1}{17} u

Substitute back

u = sec (17 w)

= \frac{1}{17} sec (17 w)

Add a constant to the solution

= \frac{1}{17} sec (17 w) + C

\frac{d}{d x} (ln (2 x - 1) + 2)

\frac{sin ( q ) + cos ( q )}{cos ( q )}

y^{^{''}} - 10 y^{^{'}} + 21 y = 0

\frac{4}{s ( s ^{2} + 2 s + 4 )}

x \to - 8 lim (3 x + 1)

What is the integral of sec(17w)tan(17w) ?
The integral of sec(17w)tan(17w) is 1/17 sec(17w)+C