What is the integral of sec(32w)tan(32w) ?

The integral of sec(32w)tan(32w) is 1/32 sec(32w)+C

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integral of sec(32w)tan(32w)

\int sec (32 w) tan (32 w) d w

\frac{1}{32} sec (32 w) + C

Solution steps

\int sec (32 w) tan (32 w) d w

Apply u-substitution

: \int \frac{1}{32} d u

= \int \frac{1}{32} d u

Integral of a constant:

\int a d x = a x

= \frac{1}{32} u

Substitute back

u = sec (32 w)

= \frac{1}{32} sec (32 w)

Add a constant to the solution

= \frac{1}{32} sec (32 w) + C

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

\int \frac{1}{( 2 x ) ^{2}} d x

\int \frac{x ^{2} - 1}{x ^{2}} d x

\int - \frac{1}{3 y} d y

\int e^{\frac{- 1}{2} x} d x

What is the integral of sec(32w)tan(32w) ?
The integral of sec(32w)tan(32w) is 1/32 sec(32w)+C