What is the integral of 1/(sec(3x-1)) ?

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

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integral of 1/(sec(3x-1))

\int \frac{1}{sec ( 3 x - 1 )} d x

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

Solution steps

\int \frac{1}{sec ( 3 x - 1 )} d x

Use the basic trigonometric identity:

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

= \int cos (3 x - 1) d x

Apply u-substitution

: \int cos (u) \frac{1}{3} d u

= \int cos (u) \frac{1}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int cos (u) d u

Use the common integral:

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

= \frac{1}{3} sin (u)

Substitute back

u = 3 x - 1

= \frac{1}{3} sin (3 x - 1)

Add a constant to the solution

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

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