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

The integral from 0 to 1/2 of 5arcsin(2y) is 5(pi/4-1/2)

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

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

5 (\frac{π}{4} - \frac{1}{2})

Decimal Notation

1.42699 \dots

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Compute the boundaries:

\frac{π}{4} - \frac{1}{2}

= 5 (\frac{π}{4} - \frac{1}{2})

\int \frac{1}{x ^{2} x ^{2} - 100} d x

f (x) = \frac{sec ( x )}{8 + sec ( x )}

\int \frac{1}{x ^{3} x ^{2} - 100} d x

\frac{d}{d x} (3 x^{3} e^{- x})

f (x) = 3 x e^{2 x}

What is the integral from 0 to 1/2 of 5arcsin(2y) ?
The integral from 0 to 1/2 of 5arcsin(2y) is 5(pi/4-1/2)