Calculus Examples

$\int_{0}^{\frac{π}{12}} \sin^{3} (6 x) d x$

Step 1

Let $u_{1} = 6 x$ . Then $d u_{1} = 6 d x$ , so $\frac{1}{6} d u_{1} = d x$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 1.1

Let $u_{1} = 6 x$ . Find $\frac{d u_{1}}{d x}$ .

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Step 1.1.1

Differentiate $6 x$ .

$\frac{d}{d x} [6 x]$

Step 1.1.2

Since $6$ is constant with respect to $x$ , the derivative of $6 x$ with respect to $x$ is $6 \frac{d}{d x} [x]$ .

$6 \frac{d}{d x} [x]$

Step 1.1.3

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

$6 \cdot 1$

Step 1.1.4

Multiply $6$ by $1$ .

$6$

Step 1.2

Substitute the lower limit in for $x$ in $u_{1} = 6 x$ .

$u_{lower} = 6 \cdot 0$

Step 1.3

Multiply $6$ by $0$ .

$u_{lower} = 0$

Step 1.4

Substitute the upper limit in for $x$ in $u_{1} = 6 x$ .

$u_{upper} = 6 \frac{π}{12}$

Step 1.5

Cancel the common factor of $6$ .

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Step 1.5.1

Factor $6$ out of $12$ .

$u_{upper} = 6 \frac{π}{6 (2)}$

Step 1.5.2

Cancel the common factor.

$u_{upper} = 6 \frac{π}{6 \cdot 2}$

Step 1.5.3

Rewrite the expression.

$u_{upper} = \frac{π}{2}$

Step 1.6

The values found for $u_{lower}$ and $u_{upper}$ will be used to evaluate the definite integral.

$u_{lower} = 0$

$u_{upper} = \frac{π}{2}$

Step 1.7

Rewrite the problem using $u_{1}$ , $d u_{1}$ , and the new limits of integration.

$\int_{0}^{\frac{π}{2}} \sin^{3} (u_{1}) \frac{1}{6} d u_{1}$

Step 2

Combine $\sin^{3} (u_{1})$ and $\frac{1}{6}$ .

$\int_{0}^{\frac{π}{2}} \frac{\sin^{3} (u_{1})}{6} d u_{1}$

Step 3

Since $\frac{1}{6}$ is constant with respect to $u_{1}$ , move $\frac{1}{6}$ out of the integral.

$\frac{1}{6} \int_{0}^{\frac{π}{2}} \sin^{3} (u_{1}) d u_{1}$

Step 4

Factor out $\sin^{2} (u_{1})$ .

$\frac{1}{6} \int_{0}^{\frac{π}{2}} \sin^{2} (u_{1}) \sin (u_{1}) d u_{1}$

Step 5

Using the Pythagorean Identity, rewrite $\sin^{2} (u_{1})$ as $1 - \cos^{2} (u_{1})$ .

$\frac{1}{6} \int_{0}^{\frac{π}{2}} (1 - \cos^{2} (u_{1})) \sin (u_{1}) d u_{1}$

Step 6

Let $u_{2} = \cos (u_{1})$ . Then $d u_{2} = - \sin (u_{1}) d u_{1}$ , so $- \frac{1}{\sin (u_{1})} d u_{2} = d u_{1}$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 6.1

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

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Step 6.1.1

Differentiate $\cos (u_{1})$ .

$\frac{d}{d u_{1}} [\cos (u_{1})]$

Step 6.1.2

The derivative of $\cos (u_{1})$ with respect to $u_{1}$ is $- \sin (u_{1})$ .

$- \sin (u_{1})$

Step 6.2

Substitute the lower limit in for $u_{1}$ in $u_{2} = \cos (u_{1})$ .

$u_{lower} = \cos (0)$

Step 6.3

The exact value of $\cos (0)$ is $1$ .

$u_{lower} = 1$

Step 6.4

Substitute the upper limit in for $u_{1}$ in $u_{2} = \cos (u_{1})$ .

$u_{upper} = \cos (\frac{π}{2})$

Step 6.5

The exact value of $\cos (\frac{π}{2})$ is $0$ .

$u_{upper} = 0$

Step 6.6

The values found for $u_{lower}$ and $u_{upper}$ will be used to evaluate the definite integral.

$u_{lower} = 1$

$u_{upper} = 0$

Step 6.7

Rewrite the problem using $u_{2}$ , $d u_{2}$ , and the new limits of integration.

$\frac{1}{6} \int_{1}^{0} - 1 + {u_{2}}^{2} d u_{2}$

Step 7

Split the single integral into multiple integrals.

$\frac{1}{6} (\int_{1}^{0} - 1 d u_{2} + \int_{1}^{0} {u_{2}}^{2} d u_{2})$

Step 8

Apply the constant rule.

$\frac{1}{6} (- u_{2}]_{1}^{0} + \int_{1}^{0} {u_{2}}^{2} d u_{2})$

Step 9

By the Power Rule, the integral of ${u_{2}}^{2}$ with respect to $u_{2}$ is $\frac{1}{3} {u_{2}}^{3}$ .

$\frac{1}{6} (- u_{2}]_{1}^{0} + \frac{1}{3} {u_{2}}^{3}]_{1}^{0})$

Step 10

Combine $- u_{2}]_{1}^{0}$ and $\frac{1}{3} {u_{2}}^{3}]_{1}^{0}$ .

$\frac{1}{6} - u_{2} + \frac{1}{3} {u_{2}}^{3}]_{1}^{0}$

Step 11

Substitute and simplify.

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Step 11.1

Evaluate $- u_{2} + \frac{1}{3} {u_{2}}^{3}$ at $0$ and at $1$ .

$\frac{1}{6} ((- 0 + \frac{1}{3} \cdot 0^{3}) - (- 1 \cdot 1 + \frac{1}{3} \cdot 1^{3}))$

Step 11.2

Simplify.

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Step 11.2.1

Multiply $- 1$ by $0$ .

$\frac{1}{6} (0 + \frac{1}{3} \cdot 0^{3} - (- 1 \cdot 1 + \frac{1}{3} \cdot 1^{3}))$

Step 11.2.2

Raising $0$ to any positive power yields $0$ .

$\frac{1}{6} (0 + \frac{1}{3} \cdot 0 - (- 1 \cdot 1 + \frac{1}{3} \cdot 1^{3}))$

Step 11.2.3

Multiply $\frac{1}{3}$ by $0$ .

$\frac{1}{6} (0 + 0 - (- 1 \cdot 1 + \frac{1}{3} \cdot 1^{3}))$

Step 11.2.4

Add $0$ and $0$ .

$\frac{1}{6} (0 - (- 1 \cdot 1 + \frac{1}{3} \cdot 1^{3}))$

Step 11.2.5

Multiply $- 1$ by $1$ .

$\frac{1}{6} (0 - (- 1 + \frac{1}{3} \cdot 1^{3}))$

Step 11.2.6

One to any power is one.

$\frac{1}{6} (0 - (- 1 + \frac{1}{3} \cdot 1))$

Step 11.2.7

Multiply $\frac{1}{3}$ by $1$ .

$\frac{1}{6} (0 - (- 1 + \frac{1}{3}))$

Step 11.2.8

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\frac{1}{6} (0 - (- 1 \cdot \frac{3}{3} + \frac{1}{3}))$

Step 11.2.9

Combine $- 1$ and $\frac{3}{3}$ .

$\frac{1}{6} (0 - (\frac{- 1 \cdot 3}{3} + \frac{1}{3}))$

Step 11.2.10

Combine the numerators over the common denominator.

$\frac{1}{6} (0 - \frac{- 1 \cdot 3 + 1}{3})$

Step 11.2.11

Simplify the numerator.

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Step 11.2.11.1

Multiply $- 1$ by $3$ .

$\frac{1}{6} (0 - \frac{- 3 + 1}{3})$

Step 11.2.11.2

Add $- 3$ and $1$ .

$\frac{1}{6} (0 - \frac{- 2}{3})$

Step 11.2.12

Move the negative in front of the fraction.

$\frac{1}{6} (0 - - \frac{2}{3})$

Step 11.2.13

Multiply $- 1$ by $- 1$ .

$\frac{1}{6} (0 + 1 (\frac{2}{3}))$

Step 11.2.14

Multiply $\frac{2}{3}$ by $1$ .

$\frac{1}{6} (0 + \frac{2}{3})$

Step 11.2.15

Add $0$ and $\frac{2}{3}$ .

$\frac{1}{6} \cdot \frac{2}{3}$

Step 11.2.16

Multiply $\frac{1}{6}$ by $\frac{2}{3}$ .

$\frac{2}{6 \cdot 3}$

Step 11.2.17

Multiply $6$ by $3$ .

$\frac{2}{18}$

Step 11.2.18

Cancel the common factor of $2$ and $18$ .

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Step 11.2.18.1

Factor $2$ out of $2$ .

$\frac{2 (1)}{18}$

Step 11.2.18.2

Cancel the common factors.

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Step 11.2.18.2.1

Factor $2$ out of $18$ .

$\frac{2 \cdot 1}{2 \cdot 9}$

Step 11.2.18.2.2

Cancel the common factor.

$\frac{2 \cdot 1}{2 \cdot 9}$

Step 11.2.18.2.3

Rewrite the expression.

$\frac{1}{9}$

Step 12

The result can be shown in multiple forms.

Exact Form:

$\frac{1}{9}$

Decimal Form:

$0.\overline{1}$