Calculus Examples

$\int \sin (2 x) {(\cos (2 x) + 1)}^{\frac{1}{2}} d x$

Let $u_{2} = \cos (2 x) + 1$ . Then $d u_{2} = - 2 \sin (2 x) d x$ , so $- \frac{1}{2} d u_{2} = \sin (2 x) d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Let $u_{2} = \cos (2 x) + 1$ . Find $\frac{d u_{2}}{d x}$ .

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Differentiate $\cos (2 x) + 1$ .

$\frac{d}{d x} [\cos (2 x) + 1]$

By the Sum Rule, the derivative of $\cos (2 x) + 1$ with respect to $x$ is $\frac{d}{d x} [\cos (2 x)] + \frac{d}{d x} [1]$ .

$\frac{d}{d x} [\cos (2 x)] + \frac{d}{d x} [1]$

Evaluate $\frac{d}{d x} [\cos (2 x)]$ .

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Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \cos (x)$ and $g (x) = 2 x$ .

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To apply the Chain Rule, set $u_{1}$ as $2 x$ .

$\frac{d}{d u_{1}} [\cos (u_{1})] \frac{d}{d x} [2 x] + \frac{d}{d x} [1]$

The derivative of $\cos (u_{1})$ with respect to $u_{1}$ is $- \sin (u_{1})$ .

$- \sin (u_{1}) \frac{d}{d x} [2 x] + \frac{d}{d x} [1]$

Replace all occurrences of $u_{1}$ with $2 x$ .

$- \sin (2 x) \frac{d}{d x} [2 x] + \frac{d}{d x} [1]$

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

$- \sin (2 x) (2 \frac{d}{d x} [x]) + \frac{d}{d x} [1]$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$- \sin (2 x) (2 \cdot 1) + \frac{d}{d x} [1]$

Multiply $2$ by $1$ .

$- \sin (2 x) \cdot 2 + \frac{d}{d x} [1]$

Multiply $2$ by $- 1$ .

$- 2 \sin (2 x) + \frac{d}{d x} [1]$

Differentiate using the Constant Rule.

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Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$- 2 \sin (2 x) + 0$

Add $- 2 \sin (2 x)$ and $0$ .

$- 2 \sin (2 x)$

Rewrite the problem using $u_{2}$ and $d u_{2}$ .

$\int {u_{2}}^{\frac{1}{2}} \frac{1}{- 2} d u_{2}$

Simplify.

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Move the negative in front of the fraction.

$\int {u_{2}}^{\frac{1}{2}} (- \frac{1}{2}) d u_{2}$

Combine ${u_{2}}^{\frac{1}{2}}$ and $\frac{1}{2}$ .

$\int - \frac{{u_{2}}^{\frac{1}{2}}}{2} d u_{2}$

Since $- 1$ is constant with respect to $u_{2}$ , move $- 1$ out of the integral.

$- \int \frac{{u_{2}}^{\frac{1}{2}}}{2} d u_{2}$

Since $\frac{1}{2}$ is constant with respect to $u_{2}$ , move $\frac{1}{2}$ out of the integral.

$- (\frac{1}{2} \int {u_{2}}^{\frac{1}{2}} d u_{2})$

By the Power Rule, the integral of ${u_{2}}^{\frac{1}{2}}$ with respect to $u_{2}$ is $\frac{2}{3} {u_{2}}^{\frac{3}{2}}$ .

$- \frac{1}{2} (\frac{2}{3} {u_{2}}^{\frac{3}{2}} + C)$

Simplify.

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Rewrite $- \frac{1}{2} (\frac{2}{3} {u_{2}}^{\frac{3}{2}} + C)$ as $- \frac{1}{2} \cdot \frac{2}{3} {u_{2}}^{\frac{3}{2}} + C$ .

$- \frac{1}{2} \cdot \frac{2}{3} {u_{2}}^{\frac{3}{2}} + C$

Simplify.

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Multiply $\frac{2}{3}$ by $\frac{1}{2}$ .

$- \frac{2}{3 \cdot 2} {u_{2}}^{\frac{3}{2}} + C$

Multiply $3$ by $2$ .

$- \frac{2}{6} {u_{2}}^{\frac{3}{2}} + C$

Cancel the common factor of $2$ and $6$ .

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Factor $2$ out of $2$ .

$- \frac{2 (1)}{6} {u_{2}}^{\frac{3}{2}} + C$

Cancel the common factors.

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Factor $2$ out of $6$ .

$- \frac{2 \cdot 1}{2 \cdot 3} {u_{2}}^{\frac{3}{2}} + C$

Cancel the common factor.

$- \frac{2 \cdot 1}{2 \cdot 3} {u_{2}}^{\frac{3}{2}} + C$

Rewrite the expression.

$- \frac{1}{3} {u_{2}}^{\frac{3}{2}} + C$

Replace all occurrences of $u_{2}$ with $\cos (2 x) + 1$ .

$- \frac{1}{3} {(\cos (2 x) + 1)}^{\frac{3}{2}} + C$