What is the integral of 1/6 sec(1/9 x)tan(1/9 x) ?

The integral of 1/6 sec(1/9 x)tan(1/9 x) is 3/2 sec(x/9)+C

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integral of 1/6 sec(1/9 x)tan(1/9 x)

\int \frac{1}{6} sec (\frac{1}{9} x) tan (\frac{1}{9} x) d x

\frac{3}{2} sec (\frac{x}{9}) + C

Solution steps

\int \frac{1}{6} sec (\frac{1}{9} x) tan (\frac{1}{9} 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 (\frac{1}{9} x) tan (\frac{1}{9} x) d x

Simplify

sec (\frac{1}{9} x) tan (\frac{1}{9} x) : sec (\frac{x}{9}) tan (\frac{x}{9})

= \frac{1}{6} \cdot \int sec (\frac{x}{9}) tan (\frac{x}{9}) d x

Apply u-substitution

: \int sec (u) tan (u) \cdot 9 d u

= \frac{1}{6} \cdot \int sec (u) tan (u) \cdot 9 d u

Take the constant out:

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

= \frac{1}{6} \cdot 9 \cdot \int sec (u) tan (u) d u

Apply u-substitution

: \int 1 d v

= \frac{1}{6} \cdot 9 \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= \frac{1}{6} \cdot 9 \cdot 1 \cdot v

Substitute back

= \frac{1}{6} \cdot 9 \cdot 1 \cdot sec (\frac{x}{9})

Simplify

\frac{1}{6} \cdot 9 \cdot 1 \cdot sec (\frac{x}{9}) : \frac{3}{2} sec (\frac{x}{9})

= \frac{3}{2} sec (\frac{x}{9})

Add a constant to the solution

= \frac{3}{2} sec (\frac{x}{9}) + C

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What is the integral of 1/6 sec(1/9 x)tan(1/9 x) ?
The integral of 1/6 sec(1/9 x)tan(1/9 x) is 3/2 sec(x/9)+C