Calculus Examples

$\int \sqrt{9 - x^{2}} d x$

Let $x = 3 \sin (t)$ , where $- \frac{π}{2} \leq t \leq \frac{π}{2}$ . Then $d x = 3 \cos (t) (d x)$ . Note that since $- \frac{π}{2} \leq t \leq \frac{π}{2}$ , $3 \cos (t)$ is positive.

$\int \sqrt{9 - {(3 \sin (t))}^{2}} (3 \cos (t)) d t$

Rewrite $9$ as $3^{2}$ .

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

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 3$ and $b = 3 \sin (t)$ .

$\int \sqrt{(3 + 3 \sin (t)) (3 - (3 \sin (t)))} (3 \cos (t)) d t$

Multiply $3$ by $- 1$ .

$\int \sqrt{(3 + 3 \sin (t)) (3 - 3 \sin (t))} (3 \cos (t)) d t$

Expand $(3 + 3 \sin (t)) (3 - 3 \sin (t))$ using the FOIL Method.

$\int \sqrt{3 \cdot 3 + 3 (- 3 \sin (t)) + 3 \sin (t) \cdot 3 + 3 \sin (t) (- 3 \sin (t))} (3 \cos (t)) d t$

Simplify and combine like terms.

$\int \sqrt{9 - 9 \sin^{2} (t)} (3 \cos (t)) d t$

Simplify with factoring out.

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Factor $9$ out of $9$ .

$\int \sqrt{9 (1) - 9 \sin^{2} (t)} (3 \cos (t)) d t$

Factor $9$ out of $- 9 \sin^{2} (t)$ .

$\int \sqrt{9 (1) + 9 (- \sin^{2} (t))} (3 \cos (t)) d t$

Factor $9$ out of $9 (1) + 9 (- \sin^{2} (t))$ .

$\int \sqrt{9 (1 - \sin^{2} (t))} (3 \cos (t)) d t$

Apply pythagorean identity.

$\int \sqrt{9 \cos^{2} (t)} (3 \cos (t)) d t$

Simplify with factoring out.

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Rewrite $9 \cos^{2} (t)$ as $9 \cos^{2} (t)$ .

$\int \sqrt{9 \cos^{2} (t)} (3 \cos (t)) d t$

Rewrite $9 \cos^{2} (t)$ as ${(3 \cos (t))}^{2}$ .

$\int \sqrt{{(3 \cos (t))}^{2}} (3 \cos (t)) d t$

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

$\int 3 \cos (t) (3 \cos (t)) d t$

Multiply $3$ by $3$ .

$\int 9 \cos (t) \cos (t) d t$

Raise $\cos (t)$ to the power of $1$ .

$\int 9 (\cos^{1} (t) \cos (t)) d t$

Raise $\cos (t)$ to the power of $1$ .

$\int 9 (\cos^{1} (t) \cos^{1} (t)) d t$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\int 9 {\cos (t)}^{1 + 1} d t$

Add $1$ and $1$ .

$\int 9 \cos^{2} (t) d t$

Since $9$ is constant with respect to $t$ , the integral of $9 \cos^{2} (t)$ with respect to $t$ is $9 \int \cos^{2} (t) d t$ .

$9 \int \cos^{2} (t) d t$

Use the half-angle formula to rewrite $\cos^{2} (t)$ as $\frac{1 + \cos (2 t)}{2}$ .

$9 \int \frac{1 + \cos (2 t)}{2} d t$

Since $\frac{1}{2}$ is constant with respect to $t$ , the integral of $\frac{1 + \cos (2 t)}{2}$ with respect to $t$ is $\frac{1}{2} \int 1 + \cos (2 t) d t$ .

$9 (\frac{1}{2} \int 1 + \cos (2 t) d t)$

Combine fractions.

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Write $9$ as a fraction with denominator $1$ .

$\frac{1}{2} \cdot \frac{9}{1} \int 1 + \cos (2 t) d t$

Multiply $\frac{1}{2}$ and $\frac{9}{1}$ .

$\frac{9}{2} \int 1 + \cos (2 t) d t$

Since integration is linear, the integral of $1 + \cos (2 t)$ with respect to $t$ is $\int 1 d t + \int \cos (2 t) d t$ .

$\frac{9}{2} (\int 1 d t + \int \cos (2 t) d t)$

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

$\frac{9}{2} (t + C + \int \cos (2 t) d t)$

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

$\frac{9}{2} (t + C + \int \cos (u) \frac{1}{2} d u)$

Combine fractions.

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Write $\cos (u)$ as a fraction with denominator $1$ .

$\frac{9}{2} (t + C + \int \frac{\cos (u)}{1} \frac{1}{2} d u)$

Multiply $\frac{\cos (u)}{1}$ and $\frac{1}{2}$ .

$\frac{9}{2} (t + C + \int \frac{\cos (u)}{2} d u)$

Since $\frac{1}{2}$ is constant with respect to $u$ , the integral of $\frac{\cos (u)}{2}$ with respect to $u$ is $\frac{1}{2} \int \cos (u) d u$ .

$\frac{9}{2} (t + C + \frac{1}{2} \int \cos (u) d u)$

The integral of $\cos (u)$ with respect to $u$ is $\sin (u)$ .

$\frac{9}{2} (t + C + \frac{1}{2} (\sin (u) + C))$

Simplify the answer.

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Simplify.

$\frac{9}{2} (t + \frac{\sin (u)}{2}) + C$

Replace all occurrences of $t$ with $\arcsin (\frac{x}{3})$ .

$\frac{9}{2} (\arcsin (\frac{x}{3}) + \frac{\sin (u)}{2}) + C$

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

$\frac{9}{2} (\arcsin (\frac{x}{3}) + \frac{\sin (2 t)}{2}) + C$

Replace all occurrences of $t$ with $\arcsin (\frac{x}{3})$ .

$\frac{9}{2} (\arcsin (\frac{x}{3}) + \frac{\sin (2 \arcsin (\frac{x}{3}))}{2}) + C$

Simplify each term.

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Apply the distributive property.

$\frac{9}{2} \arcsin (\frac{x}{3}) + \frac{9}{2} \frac{\sin (2 \arcsin (\frac{x}{3}))}{2} + C$

Simplify $\frac{9}{2} \arcsin (\frac{x}{3})$ .

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Write $\arcsin (\frac{x}{3})$ as a fraction with denominator $1$ .

$\frac{9}{2} \frac{\arcsin (\frac{x}{3})}{1} + \frac{9}{2} \frac{\sin (2 \arcsin (\frac{x}{3}))}{2} + C$

Multiply $\frac{9}{2}$ and $\frac{\arcsin (\frac{x}{3})}{1}$ .

$\frac{9 \arcsin (\frac{x}{3})}{2} + \frac{9}{2} \frac{\sin (2 \arcsin (\frac{x}{3}))}{2} + C$

Simplify $\frac{9}{2} \frac{\sin (2 \arcsin (\frac{x}{3}))}{2}$ .

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Multiply $\frac{9}{2}$ and $\frac{\sin (2 \arcsin (\frac{x}{3}))}{2}$ .

$\frac{9 \arcsin (\frac{x}{3})}{2} + \frac{9 \sin (2 \arcsin (\frac{x}{3}))}{2 \cdot 2} + C$

Multiply $2$ by $2$ .

$\frac{9 \arcsin (\frac{x}{3})}{2} + \frac{9 \sin (2 \arcsin (\frac{x}{3}))}{4} + C$

Reorder terms.

$\frac{9 \arcsin (\frac{1}{3} x)}{2} + \frac{9}{4} \sin (2 \arcsin (\frac{1}{3} x)) + C$

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