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

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

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integral of sec(32x)tan(32x)

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

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

Solution steps

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

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 x)

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

Add a constant to the solution

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

\frac{1}{x + 2}

\int \frac{7 t ^{3} + t ^{2}}{t} d t

2^{8 - x^{2}}

\int \frac{1}{4 x ^{2} + 2} d x

n = 0 \sum \infty \frac{( n ^{n} )}{n !}

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