What is the integral from 1 to x of tan(t) ?

The integral from 1 to x of tan(t) is -ln|cos(x)|+ln(cos(1))

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integral from 1 to x of tan(t)

\int_{1}^{x} tan (t) d t

- ln ∣ cos (x) ∣ + ln (cos (1))

Solution steps

\int_{1}^{x} tan (t) d t

Rewrite using trig identities

= \int_{1}^{x} \frac{sin ( t )}{cos ( t )} d t

Apply u-substitution

: \int_{c o s (1)}^{c o s (x)} - \frac{1}{u} d u

= \int_{c o s (1)}^{c o s (x)} - \frac{1}{u} d u

Take the constant out:

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

= - \int_{c o s (1)}^{c o s (x)} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - [ln ∣ u ∣]_{c o s (1)}^{c o s (x)}

Compute the boundaries:

ln ∣ cos (x) ∣ - ln (cos (1))

= - (ln ∣ cos (x) ∣ - ln (cos (1)))

Simplify

= - ln ∣ cos (x) ∣ + ln (cos (1))

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What is the integral from 1 to x of tan(t) ?
The integral from 1 to x of tan(t) is -ln|cos(x)|+ln(cos(1))