Calculus Examples

$\frac{1}{\sqrt{1 + x^{2}}}$

Step 1

Let $x = \tan (t)$ , where $- \frac{π}{2} \leq t \leq \frac{π}{2}$ . Then $d x = \sec^{2} (t) d t$ . Note that since $- \frac{π}{2} \leq t \leq \frac{π}{2}$ , $\sec^{2} (t)$ is positive.

$\int \frac{1}{\sqrt{1 + \tan^{2} (t)}} \sec^{2} (t) d t$

Step 2

Simplify terms.

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Step 2.1

Simplify $\sqrt{1 + \tan^{2} (t)}$ .

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Step 2.1.1

Rearrange terms.

$\int \frac{1}{\sqrt{\tan^{2} (t) + 1}} \sec^{2} (t) d t$

Step 2.1.2

Apply pythagorean identity.

$\int \frac{1}{\sqrt{\sec^{2} (t)}} \sec^{2} (t) d t$

Step 2.1.3

Pull terms out from under the radical, assuming positive real numbers.

$\int \frac{1}{\sec (t)} \sec^{2} (t) d t$

Step 2.2

Cancel the common factor of $\sec (t)$ .

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Step 2.2.1

Factor $\sec (t)$ out of $\sec^{2} (t)$ .

$\int \frac{1}{\sec (t)} (\sec (t) \sec (t)) d t$

Step 2.2.2

Cancel the common factor.

$\int \frac{1}{\sec (t)} (\sec (t) \sec (t)) d t$

Step 2.2.3

Rewrite the expression.

$\int \sec (t) d t$

Step 3

The integral of $\sec (t)$ with respect to $t$ is $\ln (| \sec (t) + \tan (t) |)$ .

$\ln (| \sec (t) + \tan (t) |) + C$

Step 4

Replace all occurrences of $t$ with $\arctan (x)$ .

$\ln (| \sec (\arctan (x)) + \tan (\arctan (x)) |) + C$