What is the integral of 1/6 sec(x)tan(x) ?

The integral of 1/6 sec(x)tan(x) is 1/6 sec(x)+C

integral of 1/6 sec(x)tan(x)

\int \frac{1}{6} sec (x) tan (x) d x

\frac{1}{6} sec (x) + C

Solution steps

\int \frac{1}{6} sec (x) tan (x) d x

Take the constant out:

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

= \frac{1}{6} \cdot \int sec (x) tan (x) d x

Apply u-substitution

= \frac{1}{6} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= \frac{1}{6} \cdot 1 \cdot u

Substitute back

u = sec (x)

= \frac{1}{6} \cdot 1 \cdot sec (x)

Simplify

= \frac{1}{6} sec (x)

Add a constant to the solution

= \frac{1}{6} sec (x) + C

\frac{d}{d x} ((a - \frac{b}{x})^{2})

n = 1 \sum \infty \frac{2}{n \cdot ( n + 1 )}

d er i v a t i v e g (x) = (4 x^{3} + 6 x^{2})^{5}

\frac{d}{d θ} (2 sin^{2} (θ))

\frac{d}{d x} ((a x - b) (c x^{2}))

What is the integral of 1/6 sec(x)tan(x) ?
The integral of 1/6 sec(x)tan(x) is 1/6 sec(x)+C