Calculus Examples

$\frac{d y}{d x} = \frac{1}{\sqrt{36 - x^{2}}}$ , $y (0) = π$

Step 1

Rewrite the equation.

$d y = \frac{1}{\sqrt{36 - x^{2}}} d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d y = \int \frac{1}{\sqrt{36 - x^{2}}} d x$

Step 2.2

Apply the constant rule.

$y + C_{1} = \int \frac{1}{\sqrt{36 - x^{2}}} d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

Rewrite $36$ as $6^{2}$ .

$y + C_{1} = \int \frac{1}{\sqrt{6^{2} - x^{2}}} d x$

Step 2.3.2

The integral of $\frac{1}{\sqrt{6^{2} - x^{2}}}$ with respect to $x$ is $\arcsin (\frac{x}{6})$

$y + C_{1} = \arcsin (\frac{x}{6}) + C_{2}$

Step 2.3.3

Rewrite $\arcsin (\frac{x}{6}) + C_{2}$ as $\arcsin (\frac{1}{6} x) + C_{2}$ .

$y + C_{1} = \arcsin (\frac{1}{6} x) + C_{2}$

Step 2.4

Group the constant of integration on the right side as $K$ .

$y = \arcsin (\frac{1}{6} x) + K$

Step 3

Use the initial condition to find the value of $K$ by substituting $0$ for $x$ and $π$ for $y$ in $y = \arcsin (\frac{1}{6} x) + K$ .

$π = \arcsin (\frac{1}{6} \cdot 0) + K$

Step 4

Solve for $K$ .

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Step 4.1

Rewrite the equation as $\arcsin (\frac{1}{6} \cdot 0) + K = π$ .

$\arcsin (\frac{1}{6} \cdot 0) + K = π$

Step 4.2

Simplify the left side.

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Step 4.2.1

Simplify $\arcsin (\frac{1}{6} \cdot 0) + K$ .

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Step 4.2.1.1

Simplify each term.

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Step 4.2.1.1.1

Multiply $\frac{1}{6}$ by $0$ .

$\arcsin (0) + K = π$

Step 4.2.1.1.2

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

$0 + K = π$

Step 4.2.1.2

Add $0$ and $K$ .

$K = π$

Step 5

Substitute $π$ for $K$ in $y = \arcsin (\frac{1}{6} x) + K$ and simplify.

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Step 5.1

Substitute $π$ for $K$ .

$y = \arcsin (\frac{1}{6} x) + π$

Step 5.2

Combine $\frac{1}{6}$ and $x$ .

$y = \arcsin (\frac{x}{6}) + π$