Calculus Examples

$\int_{0}^{\sqrt{\frac{2}{2}}} \frac{1}{\sqrt{1 - r^{2}}} d r d r$

Step 1

Rewrite $1$ as $1^{2}$ .

$\int_{0}^{\sqrt{\frac{2}{2}}} \frac{1}{\sqrt{1^{2} - r^{2}}} d r d r$

Step 2

The integral of $\frac{1}{\sqrt{1^{2} - r^{2}}}$ with respect to $r$ is $\arcsin (r)$

$\arcsin (r)]_{0}^{\sqrt{\frac{2}{2}}} d r$

Step 3

Simplify the answer.

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Step 3.1

Substitute and simplify.

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Step 3.1.1

Evaluate $\arcsin (r)$ at $\sqrt{\frac{2}{2}}$ and at $0$ .

$((\arcsin (\sqrt{\frac{2}{2}})) - \arcsin (0)) d r$

Step 3.1.2

Cancel the common factor of $2$ .

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Step 3.1.2.1

Cancel the common factor.

$(\arcsin (\sqrt{\frac{2}{2}}) - \arcsin (0)) d r$

Step 3.1.2.2

Rewrite the expression.

$(\arcsin (\sqrt{1}) - \arcsin (0)) d r$

Step 3.2

Any root of $1$ is $1$ .

$(\arcsin (1) - \arcsin (0)) d r$

Step 3.3

Simplify.

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Step 3.3.1

The exact value of $\arcsin (0)$ is $0$ .

$(\arcsin (1) - 0) d r$

Step 3.3.2

Multiply $- 1$ by $0$ .

$(\arcsin (1) + 0) d r$

Step 3.3.3

Add $\arcsin (1)$ and $0$ .

$\arcsin (1) d r$

Step 4

Simplify.

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Step 4.1

The exact value of $\arcsin (1)$ is $\frac{π}{2}$ .

$\frac{π}{2} d r$

Step 4.2

Combine $\frac{π}{2}$ and $d$ .

$\frac{π d}{2} r$

Step 4.3

Combine $\frac{π d}{2}$ and $r$ .

$\frac{π d r}{2}$

Step 5