What is the integral of tan(5v)sec(5v) ?

The integral of tan(5v)sec(5v) is 1/5 sec(5v)+C

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integral of tan(5v)sec(5v)

\int tan (5 v) sec (5 v) d v

\frac{1}{5} sec (5 v) + C

Solution steps

\int tan (5 v) sec (5 v) d v

Apply u-substitution

: \int \frac{1}{5} d u

= \int \frac{1}{5} d u

Integral of a constant:

\int a d x = a x

= \frac{1}{5} u

Substitute back

u = sec (5 v)

= \frac{1}{5} sec (5 v)

Add a constant to the solution

= \frac{1}{5} sec (5 v) + C

\frac{d y}{d t} = 20 - \frac{y}{230 + t}

\frac{1}{( x + 3 ) ^{2}}

\int tanh (x) d x

\frac{\partial}{\partial z} (x y z^{2})

\frac{d}{d x} (sin (3 x) + cos (3 x))

What is the integral of tan(5v)sec(5v) ?
The integral of tan(5v)sec(5v) is 1/5 sec(5v)+C