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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{1}{2}

Decimal

0.5

Solution steps

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

Apply Integration By Parts

: [x arccos (2 x) - \int - \frac{2 x}{1 - 4 x ^{2}} d x]_{0}^{\frac{1}{2}}

= [x arccos (2 x) - \int - \frac{2 x}{1 - 4 x ^{2}} d x]_{0}^{\frac{1}{2}}

\int - \frac{2 x}{1 - 4 x ^{2}} d x = \frac{1}{2} 1 - 4 x^{2}

= [x arccos (2 x) - \frac{1}{2} 1 - 4 x^{2}]_{0}^{\frac{1}{2}}

Simplify

[x arccos (2 x) - \frac{1}{2} 1 - 4 x^{2}]_{0}^{\frac{1}{2}} : [\frac{1}{2} (2 x arccos (2 x) - 1 - 4 x^{2})]_{0}^{\frac{1}{2}}

= [\frac{1}{2} (2 x arccos (2 x) - 1 - 4 x^{2})]_{0}^{\frac{1}{2}}

Compute the boundaries:

\frac{1}{2}

= \frac{1}{2}

\int_{0}^{s i n (y)} cos (y) d y

\int_{0}^{1} - x^{2} + x d x

\int_{2}^{x} 5 t + 3 d t

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

\int_{- 1}^{0} (4 + 1 - x^{2}) d x

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