Calculus Examples

$f (x) = \cos (4 x - 3)$

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.

$F (x) = \int \cos (4 x - 3) d x$

Step 3

Let $u = 4 x - 3$ . Then $d u = 4 d x$ , so $\frac{1}{4} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 3.1

Let $u = 4 x - 3$ . Find $\frac{d u}{d x}$ .

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Step 3.1.1

Differentiate $4 x - 3$ .

$\frac{d}{d x} [4 x - 3]$

Step 3.1.2

By the Sum Rule, the derivative of $4 x - 3$ with respect to $x$ is $\frac{d}{d x} [4 x] + \frac{d}{d x} [- 3]$ .

$\frac{d}{d x} [4 x] + \frac{d}{d x} [- 3]$

Step 3.1.3

Evaluate $\frac{d}{d x} [4 x]$ .

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Step 3.1.3.1

Since $4$ is constant with respect to $x$ , the derivative of $4 x$ with respect to $x$ is $4 \frac{d}{d x} [x]$ .

$4 \frac{d}{d x} [x] + \frac{d}{d x} [- 3]$

Step 3.1.3.2

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

$4 \cdot 1 + \frac{d}{d x} [- 3]$

Step 3.1.3.3

Multiply $4$ by $1$ .

$4 + \frac{d}{d x} [- 3]$

Step 3.1.4

Differentiate using the Constant Rule.

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Step 3.1.4.1

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3$ with respect to $x$ is $0$ .

$4 + 0$

Step 3.1.4.2

Add $4$ and $0$ .

$4$

Step 3.2

Rewrite the problem using $u$ and $d u$ .

$\int \cos (u) \frac{1}{4} d u$

Step 4

Combine $\cos (u)$ and $\frac{1}{4}$ .

$\int \frac{\cos (u)}{4} d u$

Step 5

Since $\frac{1}{4}$ is constant with respect to $u$ , move $\frac{1}{4}$ out of the integral.

$\frac{1}{4} \int \cos (u) d u$

Step 6

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

$\frac{1}{4} (\sin (u) + C)$

Step 7

Simplify.

$\frac{1}{4} \sin (u) + C$

Step 8

Replace all occurrences of $u$ with $4 x - 3$ .

$\frac{1}{4} \sin (4 x - 3) + C$

Step 9

The answer is the antiderivative of the function $f (x) = \cos (4 x - 3)$ .

$F (x) =$ $\frac{1}{4} \sin (4 x - 3) + C$