What is the integral from 0 to pi/6 of 8sec^3(x) ?

The integral from 0 to pi/6 of 8sec^3(x) is 8(1/3+1/4 ln(3))

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 8sec^3(x)

\int_{0}^{\frac{π}{6}} 8 sec^{3} (x) d x

8 (\frac{1}{3} + \frac{1}{4} ln (3))

Decimal

4.86389 \dots

Solution steps

\int_{0}^{\frac{π}{6}} 8 sec^{3} (x) d x

Take the constant out:

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

= 8 \cdot \int_{0}^{\frac{π}{6}} sec^{3} (x) d x

Apply Integral Reduction

= 8 ([\frac{sec ^{2} ( x ) sin ( x )}{2}]_{0}^{\frac{π}{6}} + \frac{1}{2} \cdot \int_{0}^{\frac{π}{6}} sec (x) d x)

\int_{0}^{\frac{π}{6}} sec (x) d x = \frac{1}{2} ln (3)

= 8 ([\frac{sec ^{2} ( x ) sin ( x )}{2}]_{0}^{\frac{π}{6}} + \frac{1}{2} \cdot \frac{1}{2} ln (3))

Simplify

8 ([\frac{sec ^{2} ( x ) sin ( x )}{2}]_{0}^{\frac{π}{6}} + \frac{1}{2} \cdot \frac{1}{2} ln (3)) : 8 ([\frac{1}{2} sec (x) tan (x)]_{0}^{\frac{π}{6}} + \frac{1}{4} ln (3))

= 8 ([\frac{1}{2} sec (x) tan (x)]_{0}^{\frac{π}{6}} + \frac{1}{4} ln (3))

Compute the boundaries:

\frac{1}{3}

= 8 (\frac{1}{3} + \frac{1}{4} ln (3))

\frac{2}{x}, 8 x, \frac{x}{8}

x \to 5 lim (\frac{20 - 4 x}{x - 5})

\int_{0}^{\infty} x^{- 2} d x

x = 2 y^{2}, x + y = 1

\int_{0}^{a} sin^{2} (\frac{nπ x}{a}) d x

What is the integral from 0 to pi/6 of 8sec^3(x) ?
The integral from 0 to pi/6 of 8sec^3(x) is 8(1/3+1/4 ln(3))