What is the integral of 1/(sec(5x+1)) ?

The integral of 1/(sec(5x+1)) is 1/5 sin(5x+1)+C

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integral of 1/(sec(5x+1))

\int \frac{1}{sec ( 5 x + 1 )} d x

\frac{1}{5} sin (5 x + 1) + C

Solution steps

\int \frac{1}{sec ( 5 x + 1 )} d x

Use the basic trigonometric identity:

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

= \int cos (5 x + 1) d x

Apply u-substitution

: \int cos (u) \frac{1}{5} d u

= \int cos (u) \frac{1}{5} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int cos (u) d u

Use the common integral:

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

= \frac{1}{5} sin (u)

Substitute back

u = 5 x + 1

= \frac{1}{5} sin (5 x + 1)

Add a constant to the solution

= \frac{1}{5} sin (5 x + 1) + C

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What is the integral of 1/(sec(5x+1)) ?
The integral of 1/(sec(5x+1)) is 1/5 sin(5x+1)+C