Calculus Examples

$\int \frac{1}{\sqrt{x^{2} - 25}} d x$

Step 1

Let $x = 5 \sec (t)$ , where $- \frac{π}{2} \leq t \leq \frac{π}{2}$ . Then $d x = 5 \sec (t) \tan (t) d t$ . Note that since $- \frac{π}{2} \leq t \leq \frac{π}{2}$ , $5 \sec (t) \tan (t)$ is positive.

$\int \frac{1}{\sqrt{{(5 \sec (t))}^{2} - 25}} (5 \sec (t) \tan (t)) d t$

Step 2

Simplify terms.

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Step 2.1

Simplify $\sqrt{{(5 \sec (t))}^{2} - 25}$ .

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Step 2.1.1

Simplify each term.

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Step 2.1.1.1

Apply the product rule to $5 \sec (t)$ .

$\int \frac{1}{\sqrt{5^{2} \sec^{2} (t) - 25}} (5 \sec (t) \tan (t)) d t$

Step 2.1.1.2

Raise $5$ to the power of $2$ .

$\int \frac{1}{\sqrt{25 \sec^{2} (t) - 25}} (5 \sec (t) \tan (t)) d t$

Step 2.1.2

Factor $25$ out of $25 \sec^{2} (t)$ .

$\int \frac{1}{\sqrt{25 (\sec^{2} (t)) - 25}} (5 \sec (t) \tan (t)) d t$

Step 2.1.3

Factor $25$ out of $- 25$ .

$\int \frac{1}{\sqrt{25 \sec^{2} (t) + 25 \cdot - 1}} (5 \sec (t) \tan (t)) d t$

Step 2.1.4

Factor $25$ out of $25 \sec^{2} (t) + 25 \cdot - 1$ .

$\int \frac{1}{\sqrt{25 (\sec^{2} (t) - 1)}} (5 \sec (t) \tan (t)) d t$

Step 2.1.5

Apply pythagorean identity.

$\int \frac{1}{\sqrt{25 \tan^{2} (t)}} (5 \sec (t) \tan (t)) d t$

Step 2.1.6

Rewrite $25 \tan^{2} (t)$ as ${(5 \tan (t))}^{2}$ .

$\int \frac{1}{\sqrt{{(5 \tan (t))}^{2}}} (5 \sec (t) \tan (t)) d t$

Step 2.1.7

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

$\int \frac{1}{5 \tan (t)} (5 \sec (t) \tan (t)) d t$

Step 2.2

Cancel the common factor of $5 \tan (t)$ .

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Step 2.2.1

Factor $5 \tan (t)$ out of $5 \sec (t) \tan (t)$ .

$\int \frac{1}{5 \tan (t)} (5 \tan (t) (\sec (t))) d t$

Step 2.2.2

Cancel the common factor.

$\int \frac{1}{5 \tan (t)} (5 \tan (t) \sec (t)) d t$

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 $arcsec (\frac{x}{5})$ .

$\ln (| \sec (arcsec (\frac{x}{5})) + \tan (arcsec (\frac{x}{5})) |) + C$

Step 5

Reorder terms.

$\ln (| \sec (arcsec (\frac{1}{5} x)) + \tan (arcsec (\frac{1}{5} x)) |) + C$