What is the integral of sec(aθ)tan(aθ) ?

The integral of sec(aθ)tan(aθ) is 1/a sec(aθ)+C

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integral of sec(aθ)tan(aθ)

\int sec (a θ) tan (a θ) d θ

\frac{1}{a} sec (a θ) + C

Solution steps

\int sec (a θ) tan (a θ) d θ

Apply u-substitution

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

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

Take the constant out:

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

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

Apply u-substitution

: \int 1 d v

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

Integral of a constant:

\int a d x = a x

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

Substitute back

= \frac{1}{a} \cdot 1 \cdot sec (a θ)

Simplify

= \frac{1}{a} sec (a θ)

Add a constant to the solution

= \frac{1}{a} sec (a θ) + C

y = 2, y = x^{2} - 2

\frac{\partial}{\partial y} (x^{\frac{1}{2}} y^{\frac{1}{2}})

12 x - x^{3}

x \to - 2 lim (3^{x + 1})

\int w^{2} ln (w) d w

What is the integral of sec(aθ)tan(aθ) ?
The integral of sec(aθ)tan(aθ) is 1/a sec(aθ)+C