Calculus Examples

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

Step 1

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

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Step 1.1

Let $u = 1 - \cos (x)$ . Find $\frac{d u}{d x}$ .

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Step 1.1.1

Differentiate $1 - \cos (x)$ .

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

Step 1.1.2

Differentiate.

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Step 1.1.2.1

By the Sum Rule, the derivative of $1 - \cos (x)$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [- \cos (x)]$ .

$\frac{d}{d x} [1] + \frac{d}{d x} [- \cos (x)]$

Step 1.1.2.2

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$0 + \frac{d}{d x} [- \cos (x)]$

Step 1.1.3

Evaluate $\frac{d}{d x} [- \cos (x)]$ .

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Step 1.1.3.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- \cos (x)$ with respect to $x$ is $- \frac{d}{d x} [\cos (x)]$ .

$0 - \frac{d}{d x} [\cos (x)]$

Step 1.1.3.2

The derivative of $\cos (x)$ with respect to $x$ is $- \sin (x)$ .

$0 - - \sin (x)$

Step 1.1.3.3

Multiply $- 1$ by $- 1$ .

$0 + 1 \sin (x)$

Step 1.1.3.4

Multiply $\sin (x)$ by $1$ .

$0 + \sin (x)$

Step 1.1.4

Add $0$ and $\sin (x)$ .

$\sin (x)$

Step 1.2

Substitute the lower limit in for $x$ in $u = 1 - \cos (x)$ .

$u_{lower} = 1 - \cos (0)$

Step 1.3

Simplify.

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Step 1.3.1

Simplify each term.

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Step 1.3.1.1

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

$u_{lower} = 1 - 1 \cdot 1$

Step 1.3.1.2

Multiply $- 1$ by $1$ .

$u_{lower} = 1 - 1$

Step 1.3.2

Subtract $1$ from $1$ .

$u_{lower} = 0$

Step 1.4

Substitute the upper limit in for $x$ in $u = 1 - \cos (x)$ .

$u_{upper} = 1 - \cos (π)$

Step 1.5

Simplify.

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Step 1.5.1

Simplify each term.

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Step 1.5.1.1

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.

$u_{upper} = 1 - - \cos (0)$

Step 1.5.1.2

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

$u_{upper} = 1 - (- 1 \cdot 1)$

Step 1.5.1.3

Multiply $- (- 1 \cdot 1)$ .

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Step 1.5.1.3.1

Multiply $- 1$ by $1$ .

$u_{upper} = 1 - - 1$

Step 1.5.1.3.2

Multiply $- 1$ by $- 1$ .

$u_{upper} = 1 + 1$

Step 1.5.2

Add $1$ and $1$ .

$u_{upper} = 2$

Step 1.6

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

$u_{lower} = 0$

$u_{upper} = 2$

Step 1.7

Rewrite the problem using $u$ , $d u$ , and the new limits of integration.

$\int_{0}^{2} u^{2} d u$

Step 2

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

$\frac{1}{3} u^{3}]_{0}^{2}$

Step 3

Substitute and simplify.

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Step 3.1

Evaluate $\frac{1}{3} u^{3}$ at $2$ and at $0$ .

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

Step 3.2

Simplify.

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Step 3.2.1

Raise $2$ to the power of $3$ .

$\frac{1}{3} \cdot 8 - \frac{1}{3} \cdot 0^{3}$

Step 3.2.2

Combine $\frac{1}{3}$ and $8$ .

$\frac{8}{3} - \frac{1}{3} \cdot 0^{3}$

Step 3.2.3

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

$\frac{8}{3} - \frac{1}{3} \cdot 0$

Step 3.2.4

Multiply $0$ by $- 1$ .

$\frac{8}{3} + 0 (\frac{1}{3})$

Step 3.2.5

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

$\frac{8}{3} + 0$

Step 3.2.6

Add $\frac{8}{3}$ and $0$ .

$\frac{8}{3}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$\frac{8}{3}$

Decimal Form:

$2.\overline{6}$

Mixed Number Form:

$2 \frac{2}{3}$