What is the integral from 0 to 1/2 of 5arccos(x) ?

The integral from 0 to 1/2 of 5arccos(x) is 5(pi/6-(sqrt(3))/2+1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 1/2 of 5arccos(x)

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

5 (\frac{π}{6} - \frac{3}{2} + 1)

Decimal Notation

3.28786 \dots

Solution steps

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

Take the constant out:

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

= 5 \cdot \int_{0}^{\frac{1}{2}} arccos (x) d x

Apply Integration By Parts

: [x arccos (x) - \int - \frac{x}{1 - x ^{2}} d x]_{0}^{\frac{1}{2}}

= 5 [x arccos (x) - \int - \frac{x}{1 - x ^{2}} d x]_{0}^{\frac{1}{2}}

\int - \frac{x}{1 - x ^{2}} d x = 1 - x^{2}

= 5 [x arccos (x) - 1 - x^{2}]_{0}^{\frac{1}{2}}

Compute the boundaries:

\frac{π}{6} - \frac{3}{2} + 1

= 5 (\frac{π}{6} - \frac{3}{2} + 1)

\frac{d}{d x} (e^{a x^{6}})

f (x) = (x)^{3} \cdot e^{(3 x)}

2 y^{2} = \frac{d y}{d x}

\int_{1}^{e} \frac{( x ^{2} - 1 )}{x} d x

What is the integral from 0 to 1/2 of 5arccos(x) ?
The integral from 0 to 1/2 of 5arccos(x) is 5(pi/6-(sqrt(3))/2+1)