Calculus Examples

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

Step 1

Expand ${(1 + \sin (x))}^{2}$ .

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Step 1.1

Rewrite ${(1 + \sin (x))}^{2}$ as $(1 + \sin (x)) (1 + \sin (x))$ .

$\int_{0}^{π} (1 + \sin (x)) (1 + \sin (x)) d x$

Step 1.2

Apply the distributive property.

$\int_{0}^{π} 1 (1 + \sin (x)) + \sin (x) (1 + \sin (x)) d x$

Step 1.3

Apply the distributive property.

$\int_{0}^{π} 1 \cdot 1 + 1 \sin (x) + \sin (x) (1 + \sin (x)) d x$

Step 1.4

Apply the distributive property.

$\int_{0}^{π} 1 \cdot 1 + 1 \sin (x) + \sin (x) \cdot 1 + \sin (x) \sin (x) d x$

Step 1.5

Reorder $\sin (x)$ and $1$ .

$\int_{0}^{π} 1 \cdot 1 + 1 \sin (x) + 1 \cdot \sin (x) + \sin (x) \sin (x) d x$

Step 1.6

Multiply $1$ by $1$ .

$\int_{0}^{π} 1 + 1 \sin (x) + 1 \sin (x) + \sin (x) \sin (x) d x$

Step 1.7

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

$\int_{0}^{π} 1 + \sin (x) + 1 \sin (x) + \sin (x) \sin (x) d x$

Step 1.8

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

$\int_{0}^{π} 1 + \sin (x) + \sin (x) + \sin (x) \sin (x) d x$

Step 1.9

Raise $\sin (x)$ to the power of $1$ .

$\int_{0}^{π} 1 + \sin (x) + \sin (x) + \sin^{1} (x) \sin (x) d x$

Step 1.10

Raise $\sin (x)$ to the power of $1$ .

$\int_{0}^{π} 1 + \sin (x) + \sin (x) + \sin^{1} (x) \sin^{1} (x) d x$

Step 1.11

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\int_{0}^{π} 1 + \sin (x) + \sin (x) + {\sin (x)}^{1 + 1} d x$

Step 1.12

Add $1$ and $1$ .

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

Step 1.13

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

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

Step 2

Split the single integral into multiple integrals.

$\int_{0}^{π} d x + \int_{0}^{π} 2 \sin (x) d x + \int_{0}^{π} \sin^{2} (x) d x$

Step 3

Apply the constant rule.

$x]_{0}^{π} + \int_{0}^{π} 2 \sin (x) d x + \int_{0}^{π} \sin^{2} (x) d x$

Step 4

Since $2$ is constant with respect to $x$ , move $2$ out of the integral.

$x]_{0}^{π} + 2 \int_{0}^{π} \sin (x) d x + \int_{0}^{π} \sin^{2} (x) d x$

Step 5

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

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

Step 6

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

$x]_{0}^{π} + 2 (- \cos (x)]_{0}^{π}) + \int_{0}^{π} \frac{1 - \cos (2 x)}{2} d x$

Step 7

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

$x]_{0}^{π} + 2 (- \cos (x)]_{0}^{π}) + \frac{1}{2} \int_{0}^{π} 1 - \cos (2 x) d x$

Step 8

Split the single integral into multiple integrals.

$x]_{0}^{π} + 2 (- \cos (x)]_{0}^{π}) + \frac{1}{2} (\int_{0}^{π} d x + \int_{0}^{π} - \cos (2 x) d x)$

Step 9

Apply the constant rule.

$x]_{0}^{π} + 2 (- \cos (x)]_{0}^{π}) + \frac{1}{2} (x]_{0}^{π} + \int_{0}^{π} - \cos (2 x) d x)$

Step 10

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

$x]_{0}^{π} + 2 (- \cos (x)]_{0}^{π}) + \frac{1}{2} (x]_{0}^{π} - \int_{0}^{π} \cos (2 x) d x)$

Step 11

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

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Step 11.1

Let $u = 2 x$ . Find $\frac{d u}{d x}$ .

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Step 11.1.1

Differentiate $2 x$ .

$\frac{d}{d x} [2 x]$

Step 11.1.2

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

$2 \frac{d}{d x} [x]$

Step 11.1.3

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

$2 \cdot 1$

Step 11.1.4

Multiply $2$ by $1$ .

$2$

Step 11.2

Substitute the lower limit in for $x$ in $u = 2 x$ .

$u_{lower} = 2 \cdot 0$

Step 11.3

Multiply $2$ by $0$ .

$u_{lower} = 0$

Step 11.4

Substitute the upper limit in for $x$ in $u = 2 x$ .

$u_{upper} = 2 π$

Step 11.5

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

$u_{lower} = 0$

$u_{upper} = 2 π$

Step 11.6

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

$x]_{0}^{π} + 2 (- \cos (x)]_{0}^{π}) + \frac{1}{2} (x]_{0}^{π} - \int_{0}^{2 π} \cos (u) \frac{1}{2} d u)$

Step 12

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

$x]_{0}^{π} + 2 (- \cos (x)]_{0}^{π}) + \frac{1}{2} (x]_{0}^{π} - \int_{0}^{2 π} \frac{\cos (u)}{2} d u)$

Step 13

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

$x]_{0}^{π} + 2 (- \cos (x)]_{0}^{π}) + \frac{1}{2} (x]_{0}^{π} - (\frac{1}{2} \int_{0}^{2 π} \cos (u) d u))$

Step 14

The integral of $\cos (u)$ with respect to $u$ is $\sin (u)$ .

$x]_{0}^{π} + 2 (- \cos (x)]_{0}^{π}) + \frac{1}{2} (x]_{0}^{π} - \frac{1}{2} \sin (u)]_{0}^{2 π})$

Step 15

Substitute and simplify.

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Step 15.1

Evaluate $x$ at $π$ and at $0$ .

$(π) + 0 + 2 (- \cos (x)]_{0}^{π}) + \frac{1}{2} (x]_{0}^{π} - \frac{1}{2} \sin (u)]_{0}^{2 π})$

Step 15.2

Evaluate $- \cos (x)$ at $π$ and at $0$ .

$π + 0 + 2 (- \cos (π) + \cos (0)) + \frac{1}{2} (x]_{0}^{π} - \frac{1}{2} \sin (u)]_{0}^{2 π})$

Step 15.3

Evaluate $x$ at $π$ and at $0$ .

$π + 0 + 2 (- \cos (π) + \cos (0)) + \frac{1}{2} ((π) + 0 - \frac{1}{2} \sin (u)]_{0}^{2 π})$

Step 15.4

Evaluate $\sin (u)$ at $2 π$ and at $0$ .

$π + 0 + 2 (- \cos (π) + \cos (0)) + \frac{1}{2} ((π) + 0 - \frac{1}{2} ((\sin (2 π)) - \sin (0)))$

Step 15.5

Simplify.

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Step 15.5.1

Add $π$ and $0$ .

$π + 2 (- \cos (π) + \cos (0)) + \frac{1}{2} ((π) + 0 - \frac{1}{2} ((\sin (2 π)) - \sin (0)))$

Step 15.5.2

Add $π$ and $0$ .

$π + 2 (- \cos (π) + \cos (0)) + \frac{1}{2} (π - \frac{1}{2} (\sin (2 π) - \sin (0)))$

Step 16

Simplify.

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Step 16.1

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

$π + 2 (- \cos (π) + 1) + \frac{1}{2} (π - \frac{1}{2} (\sin (2 π) - \sin (0)))$

Step 16.2

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

$π + 2 (- \cos (π) + 1) + \frac{1}{2} (π - \frac{1}{2} (\sin (2 π) - 0))$

Step 16.3

Multiply $- 1$ by $0$ .

$π + 2 (- \cos (π) + 1) + \frac{1}{2} (π - \frac{1}{2} (\sin (2 π) + 0))$

Step 16.4

Add $\sin (2 π)$ and $0$ .

$π + 2 (- \cos (π) + 1) + \frac{1}{2} (π - \frac{1}{2} \sin (2 π))$

Step 16.5

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

$π + 2 (- \cos (π) + 1) + \frac{1}{2} (π - \frac{\sin (2 π)}{2})$

Step 17

Simplify.

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Step 17.1

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.

$π + 2 (- - \cos (0) + 1) + \frac{1}{2} (π - \frac{\sin (2 π)}{2})$

Step 17.2

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

$π + 2 (- (- 1 \cdot 1) + 1) + \frac{1}{2} (π - \frac{\sin (2 π)}{2})$

Step 17.3

Multiply $- 1$ by $1$ .

$π + 2 (- - 1 + 1) + \frac{1}{2} (π - \frac{\sin (2 π)}{2})$

Step 17.4

Multiply $- 1$ by $- 1$ .

$π + 2 (1 + 1) + \frac{1}{2} (π - \frac{\sin (2 π)}{2})$

Step 17.5

Add $1$ and $1$ .

$π + 2 \cdot 2 + \frac{1}{2} (π - \frac{\sin (2 π)}{2})$

Step 17.6

Multiply $2$ by $2$ .

$π + 4 + \frac{1}{2} (π - \frac{\sin (2 π)}{2})$

Step 17.7

Simplify the numerator.

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Step 17.7.1

Subtract full rotations of $2 π$ until the angle is greater than or equal to $0$ and less than $2 π$ .

$π + 4 + \frac{1}{2} (π - \frac{\sin (0)}{2})$

Step 17.7.2

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

$π + 4 + \frac{1}{2} (π - \frac{0}{2})$

Step 17.8

Divide $0$ by $2$ .

$π + 4 + \frac{1}{2} (π - 0)$

Step 17.9

Multiply $- 1$ by $0$ .

$π + 4 + \frac{1}{2} (π + 0)$

Step 17.10

Add $π$ and $0$ .

$π + 4 + \frac{1}{2} π$

Step 17.11

Combine $\frac{1}{2}$ and $π$ .

$π + 4 + \frac{π}{2}$

Step 17.12

To write $π$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$π \cdot \frac{2}{2} + \frac{π}{2} + 4$

Step 17.13

Combine $π$ and $\frac{2}{2}$ .

$\frac{π \cdot 2}{2} + \frac{π}{2} + 4$

Step 17.14

Combine the numerators over the common denominator.

$\frac{π \cdot 2 + π}{2} + 4$

Step 17.15

Add $π \cdot 2$ and $π$ .

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Step 17.15.1

Reorder $π$ and $2$ .

$\frac{2 \cdot π + π}{2} + 4$

Step 17.15.2

Add $2 \cdot π$ and $π$ .

$\frac{3 \cdot π}{2} + 4$

$\frac{3 π}{2} + 4$

Step 18

The result can be shown in multiple forms.

Exact Form:

$\frac{3 π}{2} + 4$

Decimal Form:

$8.71238898 \dots$