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

The integral of (-1/5 sec(1-5x)) is 1/25 ln|tan(1-5x)+sec(1-5x)|+C

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

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

\frac{1}{25} ln ∣ tan (1 - 5 x) + sec (1 - 5 x) ∣ + C

Solution steps

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

Take the constant out:

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

= - \frac{1}{5} \cdot \int sec (1 - 5 x) d x

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= - \frac{1}{5} (- \frac{1}{5} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = 1 - 5 x

= - \frac{1}{5} (- \frac{1}{5} ln ∣ tan (1 - 5 x) + sec (1 - 5 x) ∣)

Simplify

- \frac{1}{5} (- \frac{1}{5} ln ∣ tan (1 - 5 x) + sec (1 - 5 x) ∣) : \frac{1}{25} ln ∣ tan (1 - 5 x) + sec (1 - 5 x) ∣

= \frac{1}{25} ln ∣ tan (1 - 5 x) + sec (1 - 5 x) ∣

Add a constant to the solution

= \frac{1}{25} ln ∣ tan (1 - 5 x) + sec (1 - 5 x) ∣ + C

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What is the integral of (-1/5 sec(1-5x)) ?
The integral of (-1/5 sec(1-5x)) is 1/25 ln|tan(1-5x)+sec(1-5x)|+C