What is the integral of tan(x+1) ?

The integral of tan(x+1) is -ln|cos(x+1)|+C

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integral of tan(x+1)

\int tan (x + 1) d x

- ln ∣ cos (x + 1) ∣ + C

Solution steps

\int tan (x + 1) d x

Rewrite using trig identities

= \int \frac{sin ( x + 1 )}{cos ( x + 1 )} d x

Apply u-substitution

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

= \int - \frac{1}{u} d u

Take the constant out:

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

= - \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - ln ∣ u ∣

Substitute back

u = cos (x + 1)

= - ln ∣ cos (x + 1) ∣

Add a constant to the solution

= - ln ∣ cos (x + 1) ∣ + C

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