What is the integral from 0 to 1 of 6/(sqrt(1-x^2)) ?

The integral from 0 to 1 of 6/(sqrt(1-x^2)) is 3pi

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integral from 0 to 1 of 6/(sqrt(1-x^2))

\int_{0}^{1} \frac{6}{1 - x ^{2}} d x

3 π

Solution steps

\int_{0}^{1} \frac{6}{1 - x ^{2}} d x

Take the constant out:

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

= 6 \cdot \int_{0}^{1} \frac{1}{1 - x ^{2}} d x

Use the common integral:

\int \frac{1}{1 - x ^{2}} d x = arcsin (x)

= 6 [arcsin (x)]_{0}^{1}

Compute the boundaries:

\frac{π}{2}

= 6 \cdot \frac{π}{2}

Simplify

= 3 π

\frac{\partial}{\partial x} (arcsin (\frac{x}{l}))

\frac{d}{d x} (\frac{3}{5 x ^{\frac{2}{5}}})

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

y^{^{'}} = \frac{4 x ^{7} e ^{\frac{y}{x}} + x ^{5} y ^{2}}{x ^{6} y}

What is the integral from 0 to 1 of 6/(sqrt(1-x^2)) ?
The integral from 0 to 1 of 6/(sqrt(1-x^2)) is 3pi