Calculus Examples

\int_{0}^{\frac{π}{3}} \sin (3 t) d t

$\int_{0}^{\frac{π}{3}} \sin (3 t) d t$

Step 1

Let

u = 3 t

$u = 3 t$ . Then

d u = 3 d t

$d u = 3 d t$ , so

\frac{1}{3} d u = d t

$\frac{1}{3} d u = d t$ . Rewrite using

u

$u$ and

d

$d$

u

$u$ .

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Step 1.1

Let

u = 3 t

$u = 3 t$ . Find

\frac{d u}{d t}

$\frac{d u}{d t}$ .

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Step 1.1.1

Differentiate

3 t

$3 t$ .

\frac{d}{d t} [3 t]

$\frac{d}{d t} [3 t]$

Step 1.1.2

Since

3

$3$ is constant with respect to

t

$t$ , the derivative of

3 t

$3 t$ with respect to

t

$t$ is

3 \frac{d}{d t} [t]

$3 \frac{d}{d t} [t]$ .

3 \frac{d}{d t} [t]

$3 \frac{d}{d t} [t]$

Step 1.1.3

Differentiate using the Power Rule which states that

\frac{d}{d t} [t^{n}]

$\frac{d}{d t} [t^{n}]$ is

n t^{n - 1}

$n t^{n - 1}$ where

n = 1

$n = 1$ .

3 \cdot 1

$3 \cdot 1$

Step 1.1.4

Multiply

3

$3$ by

1

$1$ .

3

$3$

3

$3$

Step 1.2

Substitute the lower limit in for

t

$t$ in

u = 3 t

$u = 3 t$ .

u_{lower} = 3 \cdot 0

$u_{lower} = 3 \cdot 0$

Step 1.3

Multiply

3

$3$ by

0

$0$ .

u_{lower} = 0

$u_{lower} = 0$

Step 1.4

Substitute the upper limit in for

t

$t$ in

u = 3 t

$u = 3 t$ .

u_{upper} = 3 \frac{π}{3}

$u_{upper} = 3 \frac{π}{3}$

Step 1.5

Cancel the common factor of

3

$3$ .

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Step 1.5.1

Cancel the common factor.

u_{upper} = 3 \frac{π}{3}

$u_{upper} = 3 \frac{π}{3}$

Step 1.5.2

Rewrite the expression.

u_{upper} = π

$u_{upper} = π$

u_{upper} = π

$u_{upper} = π$

Step 1.6

The values found for

u_{lower}

$u_{lower}$ and

u_{upper}

$u_{upper}$ will be used to evaluate the definite integral.

u_{lower} = 0

$u_{lower} = 0$

u_{upper} = π

$u_{upper} = π$

Step 1.7

Rewrite the problem using

u

$u$ ,

d u

$d u$ , and the new limits of integration.

\int_{0}^{π} \sin (u) \frac{1}{3} d u

$\int_{0}^{π} \sin (u) \frac{1}{3} d u$

\int_{0}^{π} \sin (u) \frac{1}{3} d u

$\int_{0}^{π} \sin (u) \frac{1}{3} d u$

Step 2

Combine

\sin (u)

$\sin (u)$ and

\frac{1}{3}

$\frac{1}{3}$ .

\int_{0}^{π} \frac{\sin (u)}{3} d u

$\int_{0}^{π} \frac{\sin (u)}{3} d u$

Step 3

Since

\frac{1}{3}

$\frac{1}{3}$ is constant with respect to

u

$u$ , move

\frac{1}{3}

$\frac{1}{3}$ out of the integral.

\frac{1}{3} \int_{0}^{π} \sin (u) d u

$\frac{1}{3} \int_{0}^{π} \sin (u) d u$

Step 4

The integral of

\sin (u)

$\sin (u)$ with respect to

u

$u$ is

- \cos (u)

$- \cos (u)$ .

\frac{1}{3} - \cos (u)]_{0}^{π}

$\frac{1}{3} - \cos (u)]_{0}^{π}$

Step 5

Evaluate

- \cos (u)

$- \cos (u)$ at

π

$π$ and at

0

$0$ .

\frac{1}{3} (- \cos (π) + \cos (0))

$\frac{1}{3} (- \cos (π) + \cos (0))$

Step 6

The exact value of

\cos (0)

$\cos (0)$ is

1

$1$ .

\frac{1}{3} (- \cos (π) + 1)

$\frac{1}{3} (- \cos (π) + 1)$

Step 7

Simplify.

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Step 7.1

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because cosine is negative in the second quadrant.

\frac{1}{3} (- - \cos (0) + 1)

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

Step 7.2

The exact value of

\cos (0)

$\cos (0)$ is

1

$1$ .

\frac{1}{3} (- (- 1 \cdot 1) + 1)

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

Step 7.3

Multiply

- 1

$- 1$ by

1

$1$ .

\frac{1}{3} (- - 1 + 1)

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

Step 7.4

Multiply

- 1

$- 1$ by

- 1

$- 1$ .

\frac{1}{3} (1 + 1)

$\frac{1}{3} (1 + 1)$

Step 7.5

Add

1

$1$ and

1

$1$ .

\frac{1}{3} \cdot 2

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

Step 7.6

Combine

\frac{1}{3}

$\frac{1}{3}$ and

2

$2$ .

\frac{2}{3}

$\frac{2}{3}$

\frac{2}{3}

$\frac{2}{3}$

Step 8

The result can be shown in multiple forms.

Exact Form:

\frac{2}{3}

$\frac{2}{3}$

Decimal Form:

0.\overline{6}

$0.\overline{6}$