What is the integral from 0 to 1/5 of 2arcsin(5y) ?

The integral from 0 to 1/5 of 2arcsin(5y) is 2(pi/(10)-1/5)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 1/5 of 2arcsin(5y)

\int_{0}^{\frac{1}{5}} 2 arcsin (5 y) d y

2 (\frac{π}{10} - \frac{1}{5})

Decimal Notation

0.22831 \dots

Solution steps

\int_{0}^{\frac{1}{5}} 2 arcsin (5 y) d y

Take the constant out:

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

= 2 \cdot \int_{0}^{\frac{1}{5}} arcsin (5 y) d y

Apply Integration By Parts

: [y arcsin (5 y) - \int \frac{5 y}{1 - 25 y ^{2}} d y]_{0}^{\frac{1}{5}}

= 2 [y arcsin (5 y) - \int \frac{5 y}{1 - 25 y ^{2}} d y]_{0}^{\frac{1}{5}}

\int \frac{5 y}{1 - 25 y ^{2}} d y = - \frac{1}{5} 1 - 25 y^{2}

= 2 [y arcsin (5 y) - (- \frac{1}{5} 1 - 25 y^{2})]_{0}^{\frac{1}{5}}

Simplify

2 [y arcsin (5 y) - (- \frac{1}{5} 1 - 25 y^{2})]_{0}^{\frac{1}{5}} : 2 [\frac{1}{5} (5 y arcsin (5 y) + 1 - 25 y^{2})]_{0}^{\frac{1}{5}}

= 2 [\frac{1}{5} (5 y arcsin (5 y) + 1 - 25 y^{2})]_{0}^{\frac{1}{5}}

Compute the boundaries:

\frac{π}{10} - \frac{1}{5}

= 2 (\frac{π}{10} - \frac{1}{5})

f (x) = sin (x) csc (x)

(sec (x) + tan (x)) (sec (x) - tan (x))

\int \frac{1}{64 e ^{- 3 x} + e ^{3 x}} d x

n = 0 \sum \infty sin (nπ)

(2 x^{3} - x y^{2} - 2 y + 3) d x - (x^{2} y + 2 x) d y = 0

What is the integral from 0 to 1/5 of 2arcsin(5y) ?
The integral from 0 to 1/5 of 2arcsin(5y) is 2(pi/(10)-1/5)