Calculus Examples

$\int_{0}^{π} 1 - x \cos (x) d x$

Step 1

Split the single integral into multiple integrals.

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

Step 2

Apply the constant rule.

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

Step 3

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

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

Step 4

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

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

Step 5

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

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

Step 6

Simplify the answer.

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Step 6.1

Substitute and simplify.

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Step 6.1.1

Evaluate $x$ at $π$ and at $0$ .

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

Step 6.1.2

Evaluate $x \sin (x)$ at $π$ and at $0$ .

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

Step 6.1.3

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

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

Step 6.1.4

Simplify.

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Step 6.1.4.1

Add $π$ and $0$ .

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

Step 6.1.4.2

Multiply $0$ by $\sin (0)$ .

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

Step 6.1.4.3

Add $π \sin (π)$ and $0$ .

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

Step 6.2

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

$π - (π \sin (π) - (- \cos (π) + 1))$

Step 6.3

Simplify.

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Step 6.3.1

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

$π - (π \sin (0) - (- \cos (π) + 1))$

Step 6.3.2

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

$π - (π \cdot 0 - (- \cos (π) + 1))$

Step 6.3.3

Multiply $π$ by $0$ .

$π - (0 - (- \cos (π) + 1))$

Step 6.3.4

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.

$π - (0 - (- - \cos (0) + 1))$

Step 6.3.5

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

$π - (0 - (- (- 1 \cdot 1) + 1))$

Step 6.3.6

Multiply $- 1$ by $1$ .

$π - (0 - (- - 1 + 1))$

Step 6.3.7

Multiply $- 1$ by $- 1$ .

$π - (0 - (1 + 1))$

Step 6.3.8

Add $1$ and $1$ .

$π - (0 - 1 \cdot 2)$

Step 6.3.9

Multiply $- 1$ by $2$ .

$π - (0 - 2)$

Step 6.3.10

Subtract $2$ from $0$ .

$π - - 2$

Step 6.3.11

Multiply $- 1$ by $- 2$ .

$π + 2$

Step 7

The result can be shown in multiple forms.

Exact Form:

$π + 2$

Decimal Form:

$5.14159265 \dots$