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

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

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

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

\frac{π}{2} - 1

Decimal Notation

0.57079 \dots

Solution steps

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

Take the constant out:

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

= 3 \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}}

= 3 [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}

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

Simplify

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

= 3 [\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}

= 3 (\frac{π}{6} - \frac{1}{3})

Simplify

= \frac{π}{2} - 1

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