What is the integral of sec(5x-3)tan(5x-3) ?

The integral of sec(5x-3)tan(5x-3) is 1/5 sec(5x-3)+C

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integral of sec(5x-3)tan(5x-3)

\int sec (5 x - 3) tan (5 x - 3) d x

\frac{1}{5} sec (5 x - 3) + C

Solution steps

\int sec (5 x - 3) tan (5 x - 3) d x

Apply u-substitution

: \int sec (u) tan (u) \frac{1}{5} d u

= \int sec (u) tan (u) \frac{1}{5} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{5} \cdot \int sec (u) tan (u) d u

Apply u-substitution

: \int 1 d v

= \frac{1}{5} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= \frac{1}{5} \cdot 1 \cdot v

Substitute back

= \frac{1}{5} \cdot 1 \cdot sec (5 x - 3)

Simplify

= \frac{1}{5} sec (5 x - 3)

Add a constant to the solution

= \frac{1}{5} sec (5 x - 3) + C

f (x) = x^{3} (8 x^{2} + 2 x - 3)^{2}

\frac{d}{d x} (cos (- 4 x))

x \to \infty lim (\frac{x}{10})

What is the integral of sec(5x-3)tan(5x-3) ?
The integral of sec(5x-3)tan(5x-3) is 1/5 sec(5x-3)+C