Calculus Examples

$\frac{\cos (x)}{1 + \sin (x)}$

Step 1

Write $\frac{\cos (x)}{1 + \sin (x)}$ as a function.

$f (x) = \frac{\cos (x)}{1 + \sin (x)}$

Step 2

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 3

Set up the integral to solve.

$F (x) = \int \frac{\cos (x)}{1 + \sin (x)} d x$

Step 4

Let $u = 1 + \sin (x)$ . Then $d u = \cos (x) d x$ , so $\frac{1}{\cos (x)} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 4.1

Let $u = 1 + \sin (x)$ . Find $\frac{d u}{d x}$ .

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Step 4.1.1

Differentiate $1 + \sin (x)$ .

$\frac{d}{d x} [1 + \sin (x)]$

Step 4.1.2

Differentiate.

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Step 4.1.2.1

By the Sum Rule, the derivative of $1 + \sin (x)$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [\sin (x)]$ .

$\frac{d}{d x} [1] + \frac{d}{d x} [\sin (x)]$

Step 4.1.2.2

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$0 + \frac{d}{d x} [\sin (x)]$

Step 4.1.3

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

$0 + \cos (x)$

Step 4.1.4

Add $0$ and $\cos (x)$ .

$\cos (x)$

Step 4.2

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

$\int \frac{1}{u} d u$

Step 5

The integral of $\frac{1}{u}$ with respect to $u$ is $\ln (| u |)$ .

$\ln (| u |) + C$

Step 6

Replace all occurrences of $u$ with $1 + \sin (x)$ .

$\ln (| 1 + \sin (x) |) + C$

Step 7

The answer is the antiderivative of the function $f (x) = \frac{\cos (x)}{1 + \sin (x)}$ .

$F (x) =$ $\ln (| 1 + \sin (x) |) + C$