What is the integral from 0 to pi/6 of tan(2θ) ?

The integral from 0 to pi/6 of tan(2θ) is 1/2 ln(2)

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integral from 0 to pi/6 of tan(2θ)

\int_{0}^{\frac{π}{6}} tan (2 θ) d θ

\frac{1}{2} ln (2)

Decimal Notation

0.34657 \dots

Solution steps

\int_{0}^{\frac{π}{6}} tan (2 θ) d θ

Rewrite using trig identities

= \int_{0}^{\frac{π}{6}} \frac{sin ( 2 θ )}{cos ( 2 θ )} d θ

Apply u-substitution

: \int_{1}^{\frac{1}{2}} - \frac{1}{2 u} d u

= \int_{1}^{\frac{1}{2}} - \frac{1}{2 u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{\frac{1}{2}}^{1} - \frac{1}{2 u} d u

Take the constant out:

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

= - (- \frac{1}{2} \cdot \int_{\frac{1}{2}}^{1} \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - (- \frac{1}{2} [ln ∣ u ∣]_{\frac{1}{2}}^{1})

Simplify

= \frac{1}{2} [ln ∣ u ∣]_{\frac{1}{2}}^{1}

Compute the boundaries:

ln (2)

= \frac{1}{2} ln (2)

\int_{1}^{2} V d V

\int_{x^{2}}^{1} 4 x d x

\int_{- \infty}^{0} e^{- 5 x} d x

\int_{- 1}^{1} 3 x^{2} x^{2} + 1 d x

\int_{- 1}^{1} 6^{x} d x

What is the integral from 0 to pi/6 of tan(2θ) ?
The integral from 0 to pi/6 of tan(2θ) is 1/2 ln(2)