What is the integral from 0 to 1/5 of 4arcsin(5y) ?

The integral from 0 to 1/5 of 4arcsin(5y) is 4(pi/(10)-1/5)

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integral from 0 to 1/5 of 4arcsin(5y)

\int_{0}^{\frac{1}{5}} 4 arcsin (5 y) d y

4 (\frac{π}{10} - \frac{1}{5})

Decimal Notation

0.45663 \dots

Solution steps

\int_{0}^{\frac{1}{5}} 4 arcsin (5 y) d y

Take the constant out:

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

= 4 \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}}

= 4 [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}

= 4 [y arcsin (5 y) - (- \frac{1}{5} 1 - 25 y^{2})]_{0}^{\frac{1}{5}}

Simplify

4 [y arcsin (5 y) - (- \frac{1}{5} 1 - 25 y^{2})]_{0}^{\frac{1}{5}} : 4 [\frac{1}{5} (5 y arcsin (5 y) + 1 - 25 y^{2})]_{0}^{\frac{1}{5}}

= 4 [\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}

= 4 (\frac{π}{10} - \frac{1}{5})

y = 3 x - 4 x^{2}, (- 2, - 22)

\frac{\partial}{\partial x} (e^{- x y} cos (y))

\frac{1}{s ^{2} ( s ^{2} + 3 s + 10 )}

\frac{d}{d x} (\frac{4 e ^{2 x} - 2 e ^{x} + 1}{2 e ^{2 x}})

\int 3 x^{2} e^{x} d x

What is the integral from 0 to 1/5 of 4arcsin(5y) ?
The integral from 0 to 1/5 of 4arcsin(5y) is 4(pi/(10)-1/5)