Calculus Examples

$\frac{1}{\sin (2 x)}$

Step 1

Write $\frac{1}{\sin (2 x)}$ as a function.

$f (x) = \frac{1}{\sin (2 x)}$

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.

$F (x) = \int \frac{1}{\sin (2 x)} d x$

Step 4

Convert from $\frac{1}{\sin (2 x)}$ to $\csc (2 x)$ .

$\int \csc (2 x) d x$

Step 5

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

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Step 5.1

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

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Step 5.1.1

Differentiate $2 x$ .

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

Step 5.1.2

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 5.1.3

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 5.1.4

Multiply $2$ by $1$ .

$2$

Step 5.2

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

$\int \csc (u) \frac{1}{2} d u$

Step 6

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

$\int \frac{\csc (u)}{2} d u$

Step 7

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

$\frac{1}{2} \int \csc (u) d u$

Step 8

The integral of $\csc (u)$ with respect to $u$ is $\ln (| \csc (u) - \cot (u) |)$ .

$\frac{1}{2} (\ln (| \csc (u) - \cot (u) |) + C)$

Step 9

Simplify.

$\frac{1}{2} \ln (| \csc (u) - \cot (u) |) + C$

Step 10

Replace all occurrences of $u$ with $2 x$ .

$\frac{1}{2} \ln (| \csc (2 x) - \cot (2 x) |) + C$

Step 11

The answer is the antiderivative of the function $f (x) = \frac{1}{\sin (2 x)}$ .

$F (x) =$ $\frac{1}{2} \ln (| \csc (2 x) - \cot (2 x) |) + C$