Calculus Examples

$\frac{d s}{d t} = 8 \sin^{2} (t + \frac{π}{12})$ , $s (0) = 3$

Step 1

Rewrite the equation.

$d s = 8 \sin^{2} (t + \frac{π}{12}) d t$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d s = \int 8 \sin^{2} (t + \frac{π}{12}) d t$

Step 2.2

Apply the constant rule.

$s + C_{1} = \int 8 \sin^{2} (t + \frac{π}{12}) d t$

Step 2.3

Integrate the right side.

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Step 2.3.1

Since $8$ is constant with respect to $t$ , move $8$ out of the integral.

$s + C_{1} = 8 \int \sin^{2} (t + \frac{π}{12}) d t$

Step 2.3.2

Let $u_{1} = t + \frac{π}{12}$ . Then $d u_{1} = d t$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 2.3.2.1

Let $u_{1} = t + \frac{π}{12}$ . Find $\frac{d u_{1}}{d t}$ .

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Step 2.3.2.1.1

Differentiate $t + \frac{π}{12}$ .

$\frac{d}{d t} [t + \frac{π}{12}]$

Step 2.3.2.1.2

By the Sum Rule, the derivative of $t + \frac{π}{12}$ with respect to $t$ is $\frac{d}{d t} [t] + \frac{d}{d t} [\frac{π}{12}]$ .

$\frac{d}{d t} [t] + \frac{d}{d t} [\frac{π}{12}]$

Step 2.3.2.1.3

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 1$ .

$1 + \frac{d}{d t} [\frac{π}{12}]$

Step 2.3.2.1.4

Since $\frac{π}{12}$ is constant with respect to $t$ , the derivative of $\frac{π}{12}$ with respect to $t$ is $0$ .

$1 + 0$

Step 2.3.2.1.5

Add $1$ and $0$ .

$1$

Step 2.3.2.2

Rewrite the problem using $u_{1}$ and $d u_{1}$ .

$s + C_{1} = 8 \int \sin^{2} (u_{1}) d u_{1}$

Step 2.3.3

Use the half-angle formula to rewrite $\sin^{2} (u_{1})$ as $\frac{1 - \cos (2 u_{1})}{2}$ .

$s + C_{1} = 8 \int \frac{1 - \cos (2 u_{1})}{2} d u_{1}$

Step 2.3.4

Since $\frac{1}{2}$ is constant with respect to $u_{1}$ , move $\frac{1}{2}$ out of the integral.

$s + C_{1} = 8 (\frac{1}{2} \int 1 - \cos (2 u_{1}) d u_{1})$

Step 2.3.5

Simplify.

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Step 2.3.5.1

Combine $\frac{1}{2}$ and $8$ .

$s + C_{1} = \frac{8}{2} \int 1 - \cos (2 u_{1}) d u_{1}$

Step 2.3.5.2

Cancel the common factor of $8$ and $2$ .

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Step 2.3.5.2.1

Factor $2$ out of $8$ .

$s + C_{1} = \frac{2 \cdot 4}{2} \int 1 - \cos (2 u_{1}) d u_{1}$

Step 2.3.5.2.2

Cancel the common factors.

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Step 2.3.5.2.2.1

Factor $2$ out of $2$ .

$s + C_{1} = \frac{2 \cdot 4}{2 (1)} \int 1 - \cos (2 u_{1}) d u_{1}$

Step 2.3.5.2.2.2

Cancel the common factor.

$s + C_{1} = \frac{2 \cdot 4}{2 \cdot 1} \int 1 - \cos (2 u_{1}) d u_{1}$

Step 2.3.5.2.2.3

Rewrite the expression.

$s + C_{1} = \frac{4}{1} \int 1 - \cos (2 u_{1}) d u_{1}$

Step 2.3.5.2.2.4

Divide $4$ by $1$ .

$s + C_{1} = 4 \int 1 - \cos (2 u_{1}) d u_{1}$

Step 2.3.6

Split the single integral into multiple integrals.

$s + C_{1} = 4 (\int d u_{1} + \int - \cos (2 u_{1}) d u_{1})$

Step 2.3.7

Apply the constant rule.

$s + C_{1} = 4 (u_{1} + C_{2} + \int - \cos (2 u_{1}) d u_{1})$

Step 2.3.8

Since $- 1$ is constant with respect to $u_{1}$ , move $- 1$ out of the integral.

$s + C_{1} = 4 (u_{1} + C_{2} - \int \cos (2 u_{1}) d u_{1})$

Step 2.3.9

Let $u_{2} = 2 u_{1}$ . Then $d u_{2} = 2 d u_{1}$ , so $\frac{1}{2} d u_{2} = d u_{1}$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 2.3.9.1

Let $u_{2} = 2 u_{1}$ . Find $\frac{d u_{2}}{d u_{1}}$ .

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Step 2.3.9.1.1

Differentiate $2 u_{1}$ .

$\frac{d}{d u_{1}} [2 u_{1}]$

Step 2.3.9.1.2

Since $2$ is constant with respect to $u_{1}$ , the derivative of $2 u_{1}$ with respect to $u_{1}$ is $2 \frac{d}{d u_{1}} [u_{1}]$ .

$2 \frac{d}{d u_{1}} [u_{1}]$

Step 2.3.9.1.3

Differentiate using the Power Rule which states that $\frac{d}{d u_{1}} [{u_{1}}^{n}]$ is $n {u_{1}}^{n - 1}$ where $n = 1$ .

$2 \cdot 1$

Step 2.3.9.1.4

Multiply $2$ by $1$ .

$2$

Step 2.3.9.2

Rewrite the problem using $u_{2}$ and $d u_{2}$ .

$s + C_{1} = 4 (u_{1} + C_{2} - \int \cos (u_{2}) \frac{1}{2} d u_{2})$

Step 2.3.10

Combine $\cos (u_{2})$ and $\frac{1}{2}$ .

$s + C_{1} = 4 (u_{1} + C_{2} - \int \frac{\cos (u_{2})}{2} d u_{2})$

Step 2.3.11

Since $\frac{1}{2}$ is constant with respect to $u_{2}$ , move $\frac{1}{2}$ out of the integral.

$s + C_{1} = 4 (u_{1} + C_{2} - (\frac{1}{2} \int \cos (u_{2}) d u_{2}))$

Step 2.3.12

The integral of $\cos (u_{2})$ with respect to $u_{2}$ is $\sin (u_{2})$ .

$s + C_{1} = 4 (u_{1} + C_{2} - \frac{1}{2} (\sin (u_{2}) + C_{3}))$

Step 2.3.13

Simplify.

$s + C_{1} = 4 (u_{1} - \frac{1}{2} \sin (u_{2})) + C_{4}$

Step 2.3.14

Substitute back in for each integration substitution variable.

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Step 2.3.14.1

Replace all occurrences of $u_{1}$ with $t + \frac{π}{12}$ .

$s + C_{1} = 4 (t + \frac{π}{12} - \frac{1}{2} \sin (u_{2})) + C_{4}$

Step 2.3.14.2

Replace all occurrences of $u_{2}$ with $2 u_{1}$ .

$s + C_{1} = 4 (t + \frac{π}{12} - \frac{1}{2} \sin (2 u_{1})) + C_{4}$

Step 2.3.14.3

Replace all occurrences of $u_{1}$ with $t + \frac{π}{12}$ .

$s + C_{1} = 4 (t + \frac{π}{12} - \frac{1}{2} \sin (2 (t + \frac{π}{12}))) + C_{4}$

Step 2.3.15

Simplify.

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Step 2.3.15.1

Apply the distributive property.

$s + C_{1} = 4 (t + \frac{π}{12} - \frac{1}{2} \sin (2 t + 2 \frac{π}{12})) + C_{4}$

Step 2.3.15.2

Cancel the common factor of $2$ .

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Step 2.3.15.2.1

Factor $2$ out of $12$ .

$s + C_{1} = 4 (t + \frac{π}{12} - \frac{1}{2} \sin (2 t + 2 \frac{π}{2 (6)})) + C_{4}$

Step 2.3.15.2.2

Cancel the common factor.

$s + C_{1} = 4 (t + \frac{π}{12} - \frac{1}{2} \sin (2 t + 2 \frac{π}{2 \cdot 6})) + C_{4}$

Step 2.3.15.2.3

Rewrite the expression.

$s + C_{1} = 4 (t + \frac{π}{12} - \frac{1}{2} \sin (2 t + \frac{π}{6})) + C_{4}$

Step 2.3.15.3

Combine $\sin (2 t + \frac{π}{6})$ and $\frac{1}{2}$ .

$s + C_{1} = 4 (t + \frac{π}{12} - \frac{\sin (2 t + \frac{π}{6})}{2}) + C_{4}$

Step 2.3.15.4

Apply the distributive property.

$s + C_{1} = 4 t + 4 \frac{π}{12} + 4 (- \frac{\sin (2 t + \frac{π}{6})}{2}) + C_{4}$

Step 2.3.15.5

Simplify.

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Step 2.3.15.5.1

Cancel the common factor of $4$ .

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Step 2.3.15.5.1.1

Factor $4$ out of $12$ .

$s + C_{1} = 4 t + 4 \frac{π}{4 (3)} + 4 (- \frac{\sin (2 t + \frac{π}{6})}{2}) + C_{4}$

Step 2.3.15.5.1.2

Cancel the common factor.

$s + C_{1} = 4 t + 4 \frac{π}{4 \cdot 3} + 4 (- \frac{\sin (2 t + \frac{π}{6})}{2}) + C_{4}$

Step 2.3.15.5.1.3

Rewrite the expression.

$s + C_{1} = 4 t + \frac{π}{3} + 4 (- \frac{\sin (2 t + \frac{π}{6})}{2}) + C_{4}$

Step 2.3.15.5.2

Cancel the common factor of $2$ .

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Step 2.3.15.5.2.1

Move the leading negative in $- \frac{\sin (2 t + \frac{π}{6})}{2}$ into the numerator.

$s + C_{1} = 4 t + \frac{π}{3} + 4 \frac{- \sin (2 t + \frac{π}{6})}{2} + C_{4}$

Step 2.3.15.5.2.2

Factor $2$ out of $4$ .

$s + C_{1} = 4 t + \frac{π}{3} + 2 (2) \frac{- \sin (2 t + \frac{π}{6})}{2} + C_{4}$

Step 2.3.15.5.2.3

Cancel the common factor.

$s + C_{1} = 4 t + \frac{π}{3} + 2 \cdot 2 \frac{- \sin (2 t + \frac{π}{6})}{2} + C_{4}$

Step 2.3.15.5.2.4

Rewrite the expression.

$s + C_{1} = 4 t + \frac{π}{3} + 2 (- \sin (2 t + \frac{π}{6})) + C_{4}$

Step 2.3.15.5.3

Multiply $- 1$ by $2$ .

$s + C_{1} = 4 t + \frac{π}{3} - 2 \sin (2 t + \frac{π}{6}) + C_{4}$

Step 2.4

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

$s = 4 t + \frac{π}{3} - 2 \sin (2 t + \frac{π}{6}) + K$

Step 3

Use the initial condition to find the value of $K$ by substituting $0$ for $t$ and $3$ for $s$ in $s = 4 t + \frac{π}{3} - 2 \sin (2 t + \frac{π}{6}) + K$ .

$3 = 4 \cdot 0 + \frac{π}{3} - 2 \sin (2 \cdot 0 + \frac{π}{6}) + K$

Step 4

Solve for $K$ .

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Step 4.1

Rewrite the equation as $4 \cdot 0 + \frac{π}{3} - 2 \sin (2 \cdot 0 + \frac{π}{6}) + K = 3$ .

$4 \cdot 0 + \frac{π}{3} - 2 \sin (2 \cdot 0 + \frac{π}{6}) + K = 3$

Step 4.2

Simplify the left side.

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Step 4.2.1

Simplify $4 \cdot 0 + \frac{π}{3} - 2 \sin (2 \cdot 0 + \frac{π}{6}) + K$ .

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Step 4.2.1.1

Simplify each term.

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Step 4.2.1.1.1

Multiply $4$ by $0$ .

$0 + \frac{π}{3} - 2 \sin (2 \cdot 0 + \frac{π}{6}) + K = 3$

Step 4.2.1.1.2

Multiply $2$ by $0$ .

$0 + \frac{π}{3} - 2 \sin (0 + \frac{π}{6}) + K = 3$

Step 4.2.1.1.3

Add $0$ and $\frac{π}{6}$ .

$0 + \frac{π}{3} - 2 \sin (\frac{π}{6}) + K = 3$

Step 4.2.1.1.4

The exact value of $\sin (\frac{π}{6})$ is $\frac{1}{2}$ .

$0 + \frac{π}{3} - 2 (\frac{1}{2}) + K = 3$

Step 4.2.1.1.5

Cancel the common factor of $2$ .

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Step 4.2.1.1.5.1

Factor $2$ out of $- 2$ .

$0 + \frac{π}{3} + 2 (- 1) \frac{1}{2} + K = 3$

Step 4.2.1.1.5.2

Cancel the common factor.

$0 + \frac{π}{3} + 2 \cdot - 1 \frac{1}{2} + K = 3$

Step 4.2.1.1.5.3

Rewrite the expression.

$0 + \frac{π}{3} - 1 + K = 3$

Step 4.2.1.2

Add $0$ and $\frac{π}{3}$ .

$\frac{π}{3} - 1 + K = 3$

Step 4.3

Move all terms not containing $K$ to the right side of the equation.

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Step 4.3.1

Subtract $\frac{π}{3}$ from both sides of the equation.

$- 1 + K = 3 - \frac{π}{3}$

Step 4.3.2

Add $1$ to both sides of the equation.

$K = 3 - \frac{π}{3} + 1$

Step 4.3.3

Add $3$ and $1$ .

$K = 4 - \frac{π}{3}$

Step 5

Substitute $4 - \frac{π}{3}$ for $K$ in $s = 4 t + \frac{π}{3} - 2 \sin (2 t + \frac{π}{6}) + K$ and simplify.

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Step 5.1

Substitute $4 - \frac{π}{3}$ for $K$ .

$s = 4 t + \frac{π}{3} - 2 \sin (2 t + \frac{π}{6}) + 4 - \frac{π}{3}$

Step 5.2

Combine the opposite terms in $4 t + \frac{π}{3} - 2 \sin (2 t + \frac{π}{6}) + 4 - \frac{π}{3}$ .

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Step 5.2.1

Combine the numerators over the common denominator.

$s = 4 t + \frac{π - π}{3} - 2 \sin (2 t + \frac{π}{6}) + 4$

Step 5.2.2

Subtract $π$ from $π$ .

$s = 4 t + \frac{0}{3} - 2 \sin (2 t + \frac{π}{6}) + 4$

Step 5.3

Divide $0$ by $3$ .

$s = 4 t + 0 - 2 \sin (2 t + \frac{π}{6}) + 4$

Step 5.4

Add $4 t$ and $0$ .

$s = 4 t - 2 \sin (2 t + \frac{π}{6}) + 4$