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

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

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

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

\frac{π}{2} - 1

Decimal

0.57079 \dots

Solution steps

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

Take the constant out:

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

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

Apply Integration By Parts

= 5 [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}

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

Simplify

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

= 5 [\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}

= 5 (\frac{π}{10} - \frac{1}{5})

Simplify

= \frac{π}{2} - 1

f (x) = cos (5 x)

x \to - 1 lim (x + 2)

\frac{\partial}{\partial x} (18 x y)

x + y - 1 + (x - y + 3) y^{^{'}} = 0

d er i v a t i v e 5 x^{2} e^{x}

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