Calculus Examples

$\int_{1}^{3} x \sin (π x^{2}) d x$

Step 1

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

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Step 1.1

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

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Step 1.1.1

Differentiate $π x^{2}$ .

$\frac{d}{d x} [π x^{2}]$

Step 1.1.2

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

$π \frac{d}{d x} [x^{2}]$

Step 1.1.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$π (2 x)$

Step 1.1.4

Move $2$ to the left of $π$ .

$2 π x$

Step 1.1.5

Reorder $2 π$ and $x$ .

$x (2 π)$

Step 1.1.6

Reorder $x$ and $2$ .

$2 \cdot x π$

Step 1.2

Substitute the lower limit in for $x$ in $u = π x^{2}$ .

$u_{lower} = π \cdot 1^{2}$

Step 1.3

Simplify.

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Step 1.3.1

One to any power is one.

$u_{lower} = π \cdot 1$

Step 1.3.2

Multiply $π$ by $1$ .

$u_{lower} = π$

Step 1.4

Substitute the upper limit in for $x$ in $u = π x^{2}$ .

$u_{upper} = π \cdot 3^{2}$

Step 1.5

Simplify.

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Step 1.5.1

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

$u_{upper} = π \cdot 9$

Step 1.5.2

Move $9$ to the left of $π$ .

$u_{upper} = 9 π$

Step 1.6

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

$u_{lower} = π$

$u_{upper} = 9 π$

Step 1.7

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

$\int_{π}^{9 π} \sin (u) \frac{1}{2 π} d u$

Step 2

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

$\int_{π}^{9 π} \frac{\sin (u)}{2 π} d u$

Step 3

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

$\frac{1}{2 π} \int_{π}^{9 π} \sin (u) d u$

Step 4

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

$\frac{1}{2 π} - \cos (u)]_{π}^{9 π}$

Step 5

Evaluate $- \cos (u)$ at $9 π$ and at $π$ .

$\frac{1}{2 π} (- \cos (9 π) + \cos (π))$

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 $2 π$ .

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

Step 6.1.2

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.

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

Step 6.1.3

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

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

Step 6.1.4

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

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Step 6.1.4.1

Multiply $- 1$ by $1$ .

$\frac{1}{2 π} (- - 1 + \cos (π))$

Step 6.1.4.2

Multiply $- 1$ by $- 1$ .

$\frac{1}{2 π} (1 + \cos (π))$

Step 6.1.5

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.

$\frac{1}{2 π} (1 - \cos (0))$

Step 6.1.6

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

$\frac{1}{2 π} (1 - 1 \cdot 1)$

Step 6.1.7

Multiply $- 1$ by $1$ .

$\frac{1}{2 π} (1 - 1)$

Step 6.2

Subtract $1$ from $1$ .

$\frac{1}{2 π} \cdot 0$

Step 6.3

Multiply $\frac{1}{2 π}$ by $0$ .

$0$