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

The integral from 0 to pi/6 of tan^3(2x) is -1/2 ln(2)+3/4

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/6 of tan^3(2x)

\int_{0}^{\frac{π}{6}} tan^{3} (2 x) d x

- \frac{1}{2} ln (2) + \frac{3}{4}

Decimal Notation

0.40342 \dots

Solution steps

\int_{0}^{\frac{π}{6}} tan^{3} (2 x) d x

Rewrite using trig identities

= \int_{0}^{\frac{π}{6}} (- 1 + sec^{2} (2 x)) tan (2 x) d x

Apply u-substitution

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

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

Expand

\frac{- 1 + u ^{2}}{2 u} : - \frac{1}{2 u} + \frac{u}{2}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int_{1}^{2} \frac{u}{2} d u = \frac{3}{4}

= - \frac{1}{2} ln (2) + \frac{3}{4}

\int_{0}^{x^{2}} \frac{1}{1 + t} d t

\int_{0}^{a} \frac{1}{a ^{2} + x ^{2}} d x

\int_{0}^{1} 5 x + 4 d x

\int_{0}^{x} 9 + x^{2} d x

\int_{0}^{5} (2 x + 1) d x

What is the integral from 0 to pi/6 of tan^3(2x) ?
The integral from 0 to pi/6 of tan^3(2x) is -1/2 ln(2)+3/4