Calculus Examples

$\int \frac{1}{{(16 - x^{2})}^{\frac{3}{2}}} d x$

Step 1

Apply the rule $x^{\frac{m}{n}} = \sqrt[n]{x^{m}}$ to rewrite the exponentiation as a radical.

$\int \frac{1}{\sqrt{{(16 - x^{2})}^{3}}} d x$

Step 2

Let $x = 4 \sin (t)$ , where $- \frac{π}{2} \leq t \leq \frac{π}{2}$ . Then $d x = 4 \cos (t) d t$ . Note that since $- \frac{π}{2} \leq t \leq \frac{π}{2}$ , $4 \cos (t)$ is positive.

$\int \frac{1}{\sqrt{{(16 - {(4 \sin (t))}^{2})}^{3}}} (4 \cos (t)) d t$

Step 3

Simplify terms.

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Step 3.1

Simplify $\sqrt{{(16 - {(4 \sin (t))}^{2})}^{3}}$ .

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Step 3.1.1

Simplify each term.

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Step 3.1.1.1

Apply the product rule to $4 \sin (t)$ .

$\int \frac{1}{\sqrt{{(16 - (4^{2} \sin^{2} (t)))}^{3}}} (4 \cos (t)) d t$

Step 3.1.1.2

Raise $4$ to the power of $2$ .

$\int \frac{1}{\sqrt{{(16 - (16 \sin^{2} (t)))}^{3}}} (4 \cos (t)) d t$

Step 3.1.1.3

Multiply $16$ by $- 1$ .

$\int \frac{1}{\sqrt{{(16 - 16 \sin^{2} (t))}^{3}}} (4 \cos (t)) d t$

Step 3.1.2

Factor $16$ out of $16$ .

$\int \frac{1}{\sqrt{{(16 (1) - 16 \sin^{2} (t))}^{3}}} (4 \cos (t)) d t$

Step 3.1.3

Factor $16$ out of $- 16 \sin^{2} (t)$ .

$\int \frac{1}{\sqrt{{(16 (1) + 16 (- \sin^{2} (t)))}^{3}}} (4 \cos (t)) d t$

Step 3.1.4

Factor $16$ out of $16 (1) + 16 (- \sin^{2} (t))$ .

$\int \frac{1}{\sqrt{{(16 (1 - \sin^{2} (t)))}^{3}}} (4 \cos (t)) d t$

Step 3.1.5

Apply pythagorean identity.

$\int \frac{1}{\sqrt{{(16 \cos^{2} (t))}^{3}}} (4 \cos (t)) d t$

Step 3.1.6

Apply the product rule to $16 \cos^{2} (t)$ .

$\int \frac{1}{\sqrt{16^{3} {(\cos^{2} (t))}^{3}}} (4 \cos (t)) d t$

Step 3.1.7

Raise $16$ to the power of $3$ .

$\int \frac{1}{\sqrt{4096 {(\cos^{2} (t))}^{3}}} (4 \cos (t)) d t$

Step 3.1.8

Multiply the exponents in ${(\cos^{2} (t))}^{3}$ .

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Step 3.1.8.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\int \frac{1}{\sqrt{4096 {\cos (t)}^{2 \cdot 3}}} (4 \cos (t)) d t$

Step 3.1.8.2

Multiply $2$ by $3$ .

$\int \frac{1}{\sqrt{4096 \cos^{6} (t)}} (4 \cos (t)) d t$

Step 3.1.9

Rewrite $4096 \cos^{6} (t)$ as ${(64 \cos^{3} (t))}^{2}$ .

$\int \frac{1}{\sqrt{{(64 \cos^{3} (t))}^{2}}} (4 \cos (t)) d t$

Step 3.1.10

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

$\int \frac{1}{64 \cos^{3} (t)} (4 \cos (t)) d t$

Step 3.2

Cancel the common factor of $4 \cos (t)$ .

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Step 3.2.1

Factor $4 \cos (t)$ out of $64 \cos^{3} (t)$ .

$\int \frac{1}{4 \cos (t) (16 \cos^{2} (t))} (4 \cos (t)) d t$

Step 3.2.2

Cancel the common factor.

$\int \frac{1}{4 \cos (t) (16 \cos^{2} (t))} (4 \cos (t)) d t$

Step 3.2.3

Rewrite the expression.

$\int \frac{1}{16 \cos^{2} (t)} d t$

Step 4

Since $\frac{1}{16}$ is constant with respect to $t$ , move $\frac{1}{16}$ out of the integral.

$\frac{1}{16} \int \frac{1}{\cos^{2} (t)} d t$

Step 5

Convert from $\frac{1}{\cos^{2} (t)}$ to $\sec^{2} (t)$ .

$\frac{1}{16} \int \sec^{2} (t) d t$

Step 6

Since the derivative of $\tan (t)$ is $\sec^{2} (t)$ , the integral of $\sec^{2} (t)$ is $\tan (t)$ .

$\frac{1}{16} (\tan (t) + C)$

Step 7

Simplify.

$\frac{1}{16} \tan (t) + C$

Step 8

Replace all occurrences of $t$ with $\arcsin (\frac{x}{4})$ .

$\frac{1}{16} \tan (\arcsin (\frac{x}{4})) + C$

Step 9

Reorder terms.

$\frac{1}{16} \tan (\arcsin (\frac{1}{4} x)) + C$