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

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

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

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

\frac{π}{6} - \frac{1}{3}

Decimal

0.19026 \dots

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Compute the boundaries:

\frac{π}{6} - \frac{1}{3}

= \frac{π}{6} - \frac{1}{3}

\int_{0}^{3} \frac{x}{x ^{2} - 1} d x

\int_{1}^{2} (\frac{1}{x ^{2}} - \frac{1}{x}) d x

\int_{1}^{4} π (\frac{2}{y})^{2} d y

\int_{0}^{1} 1 + r^{2} r d r

\int_{- 1}^{4} (- x^{2} + 6 x + 8 - x^{2}) d x

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