Calculus Examples

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

Step 1

Let $u_{1} = 3 x$ . Then $d u_{1} = 3 d x$ , so $\frac{1}{3} d u_{1} = d x$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 1.1

Let $u_{1} = 3 x$ . Find $\frac{d u_{1}}{d x}$ .

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Step 1.1.1

Differentiate $3 x$ .

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

Step 1.1.2

Since $3$ is constant with respect to $x$ , the derivative of $3 x$ with respect to $x$ is $3 \frac{d}{d x} [x]$ .

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

Step 1.1.3

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

$3 \cdot 1$

Step 1.1.4

Multiply $3$ by $1$ .

$3$

Step 1.2

Substitute the lower limit in for $x$ in $u_{1} = 3 x$ .

$u_{lower} = 3 \cdot 0$

Step 1.3

Multiply $3$ by $0$ .

$u_{lower} = 0$

Step 1.4

Substitute the upper limit in for $x$ in $u_{1} = 3 x$ .

$u_{upper} = 3 \frac{π}{6}$

Step 1.5

Cancel the common factor of $3$ .

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Step 1.5.1

Factor $3$ out of $6$ .

$u_{upper} = 3 \frac{π}{3 (2)}$

Step 1.5.2

Cancel the common factor.

$u_{upper} = 3 \frac{π}{3 \cdot 2}$

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.

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

Step 2

Simplify.

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Step 2.1

Combine $\frac{1}{3}$ and $\cos (u_{1})$ .

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

Step 2.2

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

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

Step 3

Since $\frac{1}{3}$ is constant with respect to $u_{1}$ , move $\frac{1}{3}$ out of the integral.

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

Step 4

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

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Step 4.1

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

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Step 4.1.1

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

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

Step 4.1.2

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

$\cos (u_{1})$

Step 4.2

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

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

Step 4.3

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

$u_{lower} = 0$

Step 4.4

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

$u_{upper} = \sin (\frac{π}{2})$

Step 4.5

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

$u_{upper} = 1$

Step 4.6

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

$u_{lower} = 0$

$u_{upper} = 1$

Step 4.7

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

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

Step 5

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

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

Step 6

Evaluate $- \cos (u_{2})$ at $1$ and at $0$ .

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

Step 7

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

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

Step 8

Simplify.

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Step 8.1

Evaluate $\cos (1)$ .

$\frac{1}{3} (- 1 \cdot 0.5403023 + 1)$

Step 8.2

Multiply $- 1$ by $0.5403023$ .

$\frac{1}{3} (- 0.5403023 + 1)$

Step 8.3

Add $- 0.5403023$ and $1$ .

$\frac{1}{3} \cdot 0.45969769$

Step 8.4

Combine $\frac{1}{3}$ and $0.45969769$ .

$\frac{0.45969769}{3}$

Step 8.5

Divide $0.45969769$ by $3$ .

$0.15323256$