What is the integral from 0 to x of tan(t) ?

The integral from 0 to x of tan(t) is -ln|cos(x)|

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integral from 0 to x of tan(t)

\int_{0}^{x} tan (t) d t

- ln ∣ cos (x) ∣

Solution steps

\int_{0}^{x} tan (t) d t

Rewrite using trig identities

= \int_{0}^{x} \frac{sin ( t )}{cos ( t )} d t

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - [ln ∣ u ∣]_{1}^{c o s (x)}

Compute the boundaries:

ln ∣ cos (x) ∣

= - ln ∣ cos (x) ∣

\int_{0}^{3} 8 x d x

\int_{0}^{\frac{π}{4}} cos^{3} (x) d x

\int_{0}^{2} 3^{x} (1 - 2^{x}) d x

\int_{- 6}^{6} ∣ x ∣ d x

\int_{\frac{π}{6}}^{\frac{π}{2}} \frac{sin ( x )}{x} d x

What is the integral from 0 to x of tan(t) ?
The integral from 0 to x of tan(t) is -ln|cos(x)|