Calculus Examples

$f (x) = \sin^{2} (x) + \cos^{2} (x)$

Step 1

The function $F (x)$ can be found by finding the indefinite integral of the derivative $f (x)$ .

$F (x) = \int f (x) d x$

Step 2

Set up the integral to solve.

$F (x) = \int \sin^{2} (x) + \cos^{2} (x) d x$

Step 3

Split the single integral into multiple integrals.

$\int \sin^{2} (x) d x + \int \cos^{2} (x) d x$

Step 4

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

$\int \frac{1 - \cos (2 x)}{2} d x + \int \cos^{2} (x) d x$

Step 5

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

$\frac{1}{2} \int 1 - \cos (2 x) d x + \int \cos^{2} (x) d x$

Step 6

Split the single integral into multiple integrals.

$\frac{1}{2} (\int d x + \int - \cos (2 x) d x) + \int \cos^{2} (x) d x$

Step 7

Apply the constant rule.

$\frac{1}{2} (x + C + \int - \cos (2 x) d x) + \int \cos^{2} (x) d x$

Step 8

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

$\frac{1}{2} (x + C - \int \cos (2 x) d x) + \int \cos^{2} (x) d x$

Step 9

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

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Step 9.1

Let $u_{1} = 2 x$ . Find $\frac{d u_{1}}{d x}$ .

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Step 9.1.1

Differentiate $2 x$ .

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

Step 9.1.2

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 9.1.3

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 9.1.4

Multiply $2$ by $1$ .

$2$

Step 9.2

Rewrite the problem using $u_{1}$ and $d u_{1}$ .

$\frac{1}{2} (x + C - \int \cos (u_{1}) \frac{1}{2} d u_{1}) + \int \cos^{2} (x) d x$

Step 10

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

$\frac{1}{2} (x + C - \int \frac{\cos (u_{1})}{2} d u_{1}) + \int \cos^{2} (x) d x$

Step 11

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

$\frac{1}{2} (x + C - (\frac{1}{2} \int \cos (u_{1}) d u_{1})) + \int \cos^{2} (x) d x$

Step 12

The integral of $\cos (u_{1})$ with respect to $u_{1}$ is $\sin (u_{1})$ .

$\frac{1}{2} (x + C - \frac{1}{2} (\sin (u_{1}) + C)) + \int \cos^{2} (x) d x$

Step 13

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

$\frac{1}{2} (x + C - \frac{1}{2} (\sin (u_{1}) + C)) + \int \frac{1 + \cos (2 x)}{2} d x$

Step 14

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

$\frac{1}{2} (x + C - \frac{1}{2} (\sin (u_{1}) + C)) + \frac{1}{2} \int 1 + \cos (2 x) d x$

Step 15

Split the single integral into multiple integrals.

$\frac{1}{2} (x + C - \frac{1}{2} (\sin (u_{1}) + C)) + \frac{1}{2} (\int d x + \int \cos (2 x) d x)$

Step 16

Apply the constant rule.

$\frac{1}{2} (x + C - \frac{1}{2} (\sin (u_{1}) + C)) + \frac{1}{2} (x + C + \int \cos (2 x) d x)$

Step 17

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

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Step 17.1

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

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Step 17.1.1

Differentiate $2 x$ .

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

Step 17.1.2

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 17.1.3

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 17.1.4

Multiply $2$ by $1$ .

$2$

Step 17.2

Rewrite the problem using $u_{2}$ and $d u_{2}$ .

$\frac{1}{2} (x + C - \frac{1}{2} (\sin (u_{1}) + C)) + \frac{1}{2} (x + C + \int \cos (u_{2}) \frac{1}{2} d u_{2})$

Step 18

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

$\frac{1}{2} (x + C - \frac{1}{2} (\sin (u_{1}) + C)) + \frac{1}{2} (x + C + \int \frac{\cos (u_{2})}{2} d u_{2})$

Step 19

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

$\frac{1}{2} (x + C - \frac{1}{2} (\sin (u_{1}) + C)) + \frac{1}{2} (x + C + \frac{1}{2} \int \cos (u_{2}) d u_{2})$

Step 20

The integral of $\cos (u_{2})$ with respect to $u_{2}$ is $\sin (u_{2})$ .

$\frac{1}{2} (x + C - \frac{1}{2} (\sin (u_{1}) + C)) + \frac{1}{2} (x + C + \frac{1}{2} (\sin (u_{2}) + C))$

Step 21

Simplify.

$\frac{1}{2} (x - \frac{1}{2} \sin (u_{1})) + \frac{1}{2} (x + \frac{1}{2} \sin (u_{2})) + C$

Step 22

Substitute back in for each integration substitution variable.

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Step 22.1

Replace all occurrences of $u_{1}$ with $2 x$ .

$\frac{1}{2} (x - \frac{1}{2} \sin (2 x)) + \frac{1}{2} (x + \frac{1}{2} \sin (u_{2})) + C$

Step 22.2

Replace all occurrences of $u_{2}$ with $2 x$ .

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

Step 23

Simplify.

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Step 23.1

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

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

Step 23.2

Apply the distributive property.

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

Step 23.3

Combine $\frac{1}{2}$ and $x$ .

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

Step 23.4

Multiply $\frac{1}{2} (- \frac{\sin (2 x)}{2})$ .

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Step 23.4.1

Multiply $\frac{1}{2}$ by $\frac{\sin (2 x)}{2}$ .

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

Step 23.4.2

Multiply $2$ by $2$ .

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

Step 23.5

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

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

Step 23.6

Apply the distributive property.

$\frac{x}{2} - \frac{\sin (2 x)}{4} + \frac{1}{2} x + \frac{1}{2} \cdot \frac{\sin (2 x)}{2} + C$

Step 23.7

Combine $\frac{1}{2}$ and $x$ .

$\frac{x}{2} - \frac{\sin (2 x)}{4} + \frac{x}{2} + \frac{1}{2} \cdot \frac{\sin (2 x)}{2} + C$

Step 23.8

Multiply $\frac{1}{2} \cdot \frac{\sin (2 x)}{2}$ .

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Step 23.8.1

Multiply $\frac{1}{2}$ by $\frac{\sin (2 x)}{2}$ .

$\frac{x}{2} - \frac{\sin (2 x)}{4} + \frac{x}{2} + \frac{\sin (2 x)}{2 \cdot 2} + C$

Step 23.8.2

Multiply $2$ by $2$ .

$\frac{x}{2} - \frac{\sin (2 x)}{4} + \frac{x}{2} + \frac{\sin (2 x)}{4} + C$

Step 24

Reorder terms.

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

Step 25

The answer is the antiderivative of the function $f (x) = \sin^{2} (x) + \cos^{2} (x)$ .

$F (x) =$ $\frac{1}{2} x - \frac{1}{4} \sin (2 x) + \frac{1}{2} x + \frac{1}{4} \sin (2 x) + C$