Calculus Examples

$\int \frac{\cos (t)}{1 + \sin (t)} d t$

Step 1

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

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Step 1.1

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

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Step 1.1.1

Differentiate $1 + \sin (t)$ .

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

Step 1.1.2

Differentiate.

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Step 1.1.2.1

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

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

Step 1.1.2.2

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

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

Step 1.1.3

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

$0 + \cos (t)$

Step 1.1.4

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

$\cos (t)$

Step 1.2

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

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

Step 2

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

$\ln (| u |) + C$

Step 3

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

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