Calculus Examples

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

Let $x = 1 \sin (t)$ , where $- \frac{π}{2} \leq t \leq \frac{π}{2}$ . Then $d x = \cos (t) (d x)$ . Note that since $- \frac{π}{2} \leq t \leq \frac{π}{2}$ , $\cos (t)$ is positive.

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

Rewrite $1$ as $1^{2}$ .

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

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 1$ and $b = 1 \sin (t)$ .

$\int \sqrt{(1 + 1 \sin (t)) (1 - (1 \sin (t)))} \cos (t) d t$

Simplify.

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Multiply $\sin (t)$ by $1$ .

$\int \sqrt{(1 + \sin (t)) (1 - (1 \sin (t)))} \cos (t) d t$

Multiply $\sin (t)$ by $1$ .

$\int \sqrt{(1 + \sin (t)) (1 - \sin (t))} \cos (t) d t$

Expand $(1 + \sin (t)) (1 - \sin (t))$ using the FOIL Method.

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Apply the distributive property.

$\int \sqrt{1 (1 - \sin (t)) + \sin (t) (1 - \sin (t))} \cos (t) d t$

Apply the distributive property.

$\int \sqrt{1 \cdot 1 + 1 (- \sin (t)) + \sin (t) (1 - \sin (t))} \cos (t) d t$

Apply the distributive property.

$\int \sqrt{1 \cdot 1 + 1 (- \sin (t)) + (\sin (t) \cdot 1 + \sin (t) (- \sin (t)))} \cos (t) d t$

Remove parentheses.

$\int \sqrt{1 \cdot 1 + 1 (- \sin (t)) + \sin (t) \cdot 1 + \sin (t) (- \sin (t))} \cos (t) d t$

Simplify and combine like terms.

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Simplify each term.

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Multiply $1$ by $1$ .

$\int \sqrt{1 + 1 (- \sin (t)) + \sin (t) \cdot 1 + \sin (t) (- \sin (t))} \cos (t) d t$

Multiply $- \sin (t)$ by $1$ .

$\int \sqrt{1 - \sin (t) + \sin (t) \cdot 1 + \sin (t) (- \sin (t))} \cos (t) d t$

Multiply $\sin (t)$ by $1$ .

$\int \sqrt{1 - \sin (t) + \sin (t) + \sin (t) (- \sin (t))} \cos (t) d t$

Simplify $\sin (t) (- \sin (t))$ .

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

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

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

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

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

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

Add $1$ and $1$ .

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

Add $- \sin (t)$ and $\sin (t)$ .

$\int \sqrt{1 + 0 - \sin^{2} (t)} \cos (t) d t$

Add $1$ and $0$ .

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

Apply pythagorean identity.

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

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

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

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

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

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

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

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

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

Add $1$ and $1$ .

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

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

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

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

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

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

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

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

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

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

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

Combine fractions.

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

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

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

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

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

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

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

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

Simplify the answer.

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Simplify.

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

Replace all occurrences of $t$ with $\arcsin (\frac{x}{1})$ .

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

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

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

Replace all occurrences of $t$ with $\arcsin (\frac{x}{1})$ .

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

Simplify each term.

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Simplify each term.

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Divide $x$ by $1$ .

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

Divide $x$ by $1$ .

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

Apply the distributive property.

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

Simplify $\frac{1}{2} \arcsin (x)$ .

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Write $\arcsin (x)$ as a fraction with denominator $1$ .

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

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

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

Simplify $\frac{1}{2} \frac{\sin (2 \arcsin (x))}{2}$ .

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

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

Multiply $2$ by $2$ .

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

Reorder terms.

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

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