Calculus Examples

$\int_{π}^{2 π} x \sin (x) d x$

Step 1

Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = x$ and $d v = \sin (x)$ .

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

Step 2

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

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

Step 3

Simplify.

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Step 3.1

Multiply $- 1$ by $- 1$ .

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

Step 3.2

Multiply $\int_{π}^{2 π} \cos (x) d x$ by $1$ .

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

Step 4

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

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

Step 5

Simplify the answer.

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Step 5.1

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

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

Step 5.2

Substitute and simplify.

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Step 5.2.1

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

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

Step 5.2.2

Multiply $- 1$ by $2$ .

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

Step 6

Simplify.

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Step 6.1

Simplify each term.

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Step 6.1.1

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

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

Step 6.1.2

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

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

Step 6.1.3

Multiply $- 2$ by $1$ .

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

Step 6.1.4

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

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

Step 6.1.5

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

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

Step 6.1.6

Simplify each term.

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Step 6.1.6.1

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.

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

Step 6.1.6.2

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

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

Step 6.1.6.3

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

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Step 6.1.6.3.1

Multiply $- 1$ by $1$ .

$- 2 π + 0 - (π (- - 1) + \sin (π))$

Step 6.1.6.3.2

Multiply $- 1$ by $- 1$ .

$- 2 π + 0 - (π \cdot 1 + \sin (π))$

Step 6.1.6.4

Multiply $π$ by $1$ .

$- 2 π + 0 - (π + \sin (π))$

Step 6.1.6.5

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant.

$- 2 π + 0 - (π + \sin (0))$

Step 6.1.6.6

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

$- 2 π + 0 - (π + 0)$

Step 6.1.7

Add $π$ and $0$ .

$- 2 π + 0 - π$

Step 6.2

Add $- 2 π$ and $0$ .

$- 2 π - π$

Step 6.3

Subtract $π$ from $- 2 π$ .

$- 3 π$

Step 7

The result can be shown in multiple forms.

Exact Form:

$- 3 π$

Decimal Form:

$- 9.42477796 \dots$