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

The integral of 1/(9sec(x)) is 1/9 sin(x)+C

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integral of 1/(9sec(x))

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

\frac{1}{9} sin (x) + C

Solution steps

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

Take the constant out:

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

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

Rewrite using trig identities

= \frac{1}{9} \cdot \int cos (x) d x

Use the common integral:

\int cos (x) d x = sin (x)

= \frac{1}{9} sin (x)

Add a constant to the solution

= \frac{1}{9} sin (x) + C

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What is the integral of 1/(9sec(x)) ?
The integral of 1/(9sec(x)) is 1/9 sin(x)+C