What is the integral from 0 to pi/(12) of 6tan(3x) ?

The integral from 0 to pi/(12) of 6tan(3x) is ln(2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to pi/(12) of 6tan(3x)

\int_{0}^{\frac{π}{12}} 6 tan (3 x) d x

ln (2)

Decimal Notation

0.69314 \dots

Solution steps

\int_{0}^{\frac{π}{12}} 6 tan (3 x) d x

Take the constant out:

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

= 6 \cdot \int_{0}^{\frac{π}{12}} tan (3 x) d x

Rewrite using trig identities

= 6 \cdot \int_{0}^{\frac{π}{12}} \frac{sin ( 3 x )}{cos ( 3 x )} d x

Apply u-substitution

: \int_{1}^{\frac{2}{2}} - \frac{1}{3 u} d u

= 6 \cdot \int_{1}^{\frac{2}{2}} - \frac{1}{3 u} d u

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

= 6 (- \int_{\frac{2}{2}}^{1} - \frac{1}{3 u} d u)

Take the constant out:

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

= 6 (- (- \frac{1}{3} \cdot \int_{\frac{2}{2}}^{1} \frac{1}{u} d u))

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 6 (- (- \frac{1}{3} [ln ∣ u ∣]_{\frac{2}{2}}^{1}))

Simplify

6 (- (- \frac{1}{3} [ln ∣ u ∣]_{\frac{2}{2}}^{1})) : 2 [ln ∣ u ∣]_{\frac{1}{2}}^{1}

= 2 [ln ∣ u ∣]_{\frac{1}{2}}^{1}

Compute the boundaries:

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

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

Simplify

= ln (2)

\frac{\partial}{\partial x} (a ln (x^{2} + y^{2}))

x \to 2 lim (2 x^{3} - 3 x^{2} + 4 x - 9)

y^{^{''}} + 2 y^{^{'}} + y = x e^{- x}

x \to 0 lim (\frac{3}{x ( x + 3 ) ^{2}})

\int_{1}^{e} x ln (x) d x

What is the integral from 0 to pi/(12) of 6tan(3x) ?
The integral from 0 to pi/(12) of 6tan(3x) is ln(2)