What is the integral from 0 to 1/3 of 2arcsin(3y) ?

The integral from 0 to 1/3 of 2arcsin(3y) is 2(pi/6-1/3)

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integral from 0 to 1/3 of 2arcsin(3y)

\int_{0}^{\frac{1}{3}} 2 arcsin (3 y) d y

2 (\frac{π}{6} - \frac{1}{3})

Decimal Notation

0.38053 \dots

Solution steps

\int_{0}^{\frac{1}{3}} 2 arcsin (3 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}{3}} arcsin (3 y) d y

Apply Integration By Parts

: [y arcsin (3 y) - \int \frac{3 y}{1 - 9 y ^{2}} d y]_{0}^{\frac{1}{3}}

= 2 [y arcsin (3 y) - \int \frac{3 y}{1 - 9 y ^{2}} d y]_{0}^{\frac{1}{3}}

\int \frac{3 y}{1 - 9 y ^{2}} d y = - \frac{1}{3} 1 - 9 y^{2}

= 2 [y arcsin (3 y) - (- \frac{1}{3} 1 - 9 y^{2})]_{0}^{\frac{1}{3}}

Simplify

2 [y arcsin (3 y) - (- \frac{1}{3} 1 - 9 y^{2})]_{0}^{\frac{1}{3}} : 2 [\frac{1}{3} (3 y arcsin (3 y) + 1 - 9 y^{2})]_{0}^{\frac{1}{3}}

= 2 [\frac{1}{3} (3 y arcsin (3 y) + 1 - 9 y^{2})]_{0}^{\frac{1}{3}}

Compute the boundaries:

\frac{π}{6} - \frac{1}{3}

= 2 (\frac{π}{6} - \frac{1}{3})

\int 9 x e^{8 x} d x

\int (e^{- 3 x} - 1)^{2} d x

\frac{d y}{d x} - 2 y = 2 x

\frac{d}{d x} (x^{\frac{8}{5}})

\frac{d}{d x} (e^{x} (x^{2} - 3 x + 2))

What is the integral from 0 to 1/3 of 2arcsin(3y) ?
The integral from 0 to 1/3 of 2arcsin(3y) is 2(pi/6-1/3)