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

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

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

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

\frac{π}{6} - \frac{1}{3}

Decimal

0.19026 \dots

Solution steps

\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}}

= [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}

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

Simplify

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

= \frac{π}{6} - \frac{1}{3}

x \to π lim (\frac{3 x + 1}{1 - x})

\frac{d}{d x} ((ln (3 x))^{2})

x^{3}, x^{2} + x, - 1, 2

\frac{d}{d x} (\frac{80 sin ( 0.6 x )}{x} + 15)

(7, 2), (8, y)

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