Calculus Examples

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

Step 1

Split the single integral into multiple integrals.

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

Step 2

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

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

Step 3

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

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

Step 4

Simplify the answer.

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Step 4.1

Combine $- \cos (x)]_{0}^{2 π}$ and $\sin (x)]_{0}^{2 π}$ .

$- \cos (x) + \sin (x)]_{0}^{2 π}$

Step 4.2

Evaluate $- \cos (x) + \sin (x)$ at $2 π$ and at $0$ .

$- \cos (2 π) + \sin (2 π) - (- \cos (0) + \sin (0))$

Step 4.3

Simplify.

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Step 4.3.1

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

$- \cos (2 π) + \sin (2 π) - (- 1 \cdot 1 + \sin (0))$

Step 4.3.2

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

$- \cos (2 π) + \sin (2 π) - (- 1 \cdot 1 + 0)$

Step 4.3.3

Multiply $- 1$ by $1$ .

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

Step 4.3.4

Add $- 1$ and $0$ .

$- \cos (2 π) + \sin (2 π) - - 1$

Step 4.3.5

Multiply $- 1$ by $- 1$ .

$- \cos (2 π) + \sin (2 π) + 1$

Step 4.4

Simplify.

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Step 4.4.1

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

$- \cos (0) + \sin (2 π) + 1$

Step 4.4.2

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

$- 1 \cdot 1 + \sin (2 π) + 1$

Step 4.4.3

Multiply $- 1$ by $1$ .

$- 1 + \sin (2 π) + 1$

Step 4.4.4

Add $- 1$ and $1$ .

$0 + \sin (2 π)$

Step 4.4.5

Subtract $- \sin (2 π)$ from $0$ .

$\sin (2 π)$

Step 5

Simplify.

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Step 5.1

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

$\sin (0)$

Step 5.2

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

$0$