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

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

Simplify terms.

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Simplify each term.

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

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.

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

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 $t$ , move $9$ out of the integral.

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

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 $t$ , move $\frac{1}{2}$ out of the integral.

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

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

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

Split the single integral into multiple integrals.

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

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

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

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

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

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Rewrite.

$\frac{2}{1}$

Divide $2$ by $1$ .

$2$

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

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

Combine $\cos (u)$ 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 $u$ , move $\frac{1}{2}$ out of the integral.

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

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

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

Simplify.

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

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

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

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

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

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

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

Simplify the answer.

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Combine $\frac{1}{2}$ and $\sin (2 \arcsin (\frac{x}{3}))$ .

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

Apply the distributive property.

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

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

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

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

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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.

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

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