What is the integral of tan(1+pit-t) ?

The integral of tan(1+pit-t) is -1/(pi-1)ln|cos(1+pit-t)|+C

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integral of tan(1+pit-t)

\int tan (1 + π t - t) d t

- \frac{1}{π - 1} ln ∣ cos (1 + π t - t) ∣ + C

Solution steps

\int tan (1 + π t - t) d t

Apply u-substitution

: \int tan (u) \frac{1}{π - 1} d u

= \int tan (u) \frac{1}{π - 1} d u

Take the constant out:

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

= \frac{1}{π - 1} \cdot \int tan (u) d u

Rewrite using trig identities

= \frac{1}{π - 1} \cdot \int \frac{sin ( u )}{cos ( u )} d u

Apply u-substitution

: \int - \frac{1}{v} d v

= \frac{1}{π - 1} \cdot \int - \frac{1}{v} d v

Take the constant out:

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

= \frac{1}{π - 1} (- \int \frac{1}{v} d v)

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{π - 1} (- ln ∣ v ∣)

Substitute back

= \frac{1}{π - 1} (- ln ∣ cos (1 + π t - t) ∣)

Simplify

= - \frac{1}{π - 1} ln ∣ cos (1 + π t - t) ∣

Add a constant to the solution

= - \frac{1}{π - 1} ln ∣ cos (1 + π t - t) ∣ + C

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What is the integral of tan(1+pit-t) ?
The integral of tan(1+pit-t) is -1/(pi-1)ln|cos(1+pit-t)|+C