Calculus Examples

$\int e^{2 x} \cos (x) d x$

Reorder $e^{2 x}$ and $\cos (x)$ .

$\int \cos (x) e^{2 x} d x$

Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = \cos (x)$ and $d v = e^{2 x}$ .

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

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

$\cos (x) (\frac{1}{2} e^{2 x}) - (\frac{1}{2} \cdot - 1 \int e^{2 x} (\sin (x)) d x)$

Combine fractions.

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Combine $\frac{1}{2}$ and $e^{2 x}$ .

$\cos (x) \frac{e^{2 x}}{2} - (\frac{1}{2} \cdot - 1 \int e^{2 x} (\sin (x)) d x)$

Combine $\cos (x)$ and $\frac{e^{2 x}}{2}$ .

$\frac{\cos (x) e^{2 x}}{2} - (\frac{1}{2} \cdot - 1 \int e^{2 x} (\sin (x)) d x)$

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

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

Reorder $e^{2 x}$ and $\sin (x)$ .

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

Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = \sin (x)$ and $d v = e^{2 x}$ .

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

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

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

Simplify terms.

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Combine $\frac{1}{2}$ and $e^{2 x}$ .

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

Combine $\sin (x)$ and $\frac{e^{2 x}}{2}$ .

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

Apply the distributive property.

$\frac{\cos (x) e^{2 x}}{2} - \frac{- 1}{2} \cdot \frac{\sin (x) e^{2 x}}{2} - \frac{- 1}{2} (- (\frac{1}{2} \int e^{2 x} \cos (x) d x))$

Multiply $\frac{\sin (x) e^{2 x}}{2}$ by $\frac{- 1}{2}$ .

$\frac{\cos (x) e^{2 x}}{2} - \frac{\sin (x) e^{2 x} \cdot - 1}{2 \cdot 2} - \frac{- 1}{2} (- (\frac{1}{2} \int e^{2 x} \cos (x) d x))$

Multiply.

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Multiply $2$ by $2$ .

$\frac{\cos (x) e^{2 x}}{2} - \frac{\sin (x) e^{2 x} \cdot - 1}{4} - \frac{- 1}{2} (- (\frac{1}{2} \int e^{2 x} \cos (x) d x))$

Multiply $- 1$ by $- 1$ .

$\frac{\cos (x) e^{2 x}}{2} - \frac{\sin (x) e^{2 x} \cdot - 1}{4} + 1 (\frac{- 1}{2}) (\frac{1}{2} \int e^{2 x} \cos (x) d x)$

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

$\frac{\cos (x) e^{2 x}}{2} - \frac{\sin (x) e^{2 x} \cdot - 1}{4} + \frac{- 1}{2} (\frac{1}{2} \int e^{2 x} \cos (x) d x)$

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

$\frac{\cos (x) e^{2 x}}{2} - \frac{\sin (x) e^{2 x} \cdot - 1}{4} + \frac{- 1}{2 \cdot 2} \int e^{2 x} \cos (x) d x$

Multiply $2$ by $2$ .

$\frac{\cos (x) e^{2 x}}{2} - \frac{\sin (x) e^{2 x} \cdot - 1}{4} + \frac{- 1}{4} \int e^{2 x} \cos (x) d x$

Solving for $\int e^{2 x} \cos (x) d x$ , we find that $\int e^{2 x} \cos (x) d x$ = $\frac{\frac{\cos (x) e^{2 x}}{2} - \frac{\sin (x) e^{2 x} \cdot - 1}{4}}{\frac{5}{4}}$ .

$\frac{\frac{\cos (x) e^{2 x}}{2} - \frac{\sin (x) e^{2 x} \cdot - 1}{4}}{\frac{5}{4}} + C$

Simplify the answer.

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Simplify.

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Move $- 1$ to the left of $\sin (x) e^{2 x}$ .

$\frac{\frac{\cos (x) e^{2 x}}{2} - \frac{- 1 \cdot (\sin (x) e^{2 x})}{4}}{\frac{5}{4}} + C$

Rewrite $- 1 \sin (x)$ as $- \sin (x)$ .

$\frac{\frac{\cos (x) e^{2 x}}{2} - \frac{- \sin (x) e^{2 x}}{4}}{\frac{5}{4}} + C$

Move the negative in front of the fraction.

$\frac{\frac{\cos (x) e^{2 x}}{2} - - \frac{\sin (x) e^{2 x}}{4}}{\frac{5}{4}} + C$

Multiply $- 1$ by $- 1$ .

$\frac{\frac{\cos (x) e^{2 x}}{2} + 1 \frac{\sin (x) e^{2 x}}{4}}{\frac{5}{4}} + C$

Multiply $\frac{\sin (x) e^{2 x}}{4}$ by $1$ .

$\frac{\frac{\cos (x) e^{2 x}}{2} + \frac{\sin (x) e^{2 x}}{4}}{\frac{5}{4}} + C$

Multiply by the reciprocal of the fraction to divide by $\frac{5}{4}$ .

$(\frac{\cos (x) e^{2 x}}{2} + \frac{\sin (x) e^{2 x}}{4}) \frac{4}{5} + C$

Rewrite $(\frac{\cos (x) e^{2 x}}{2} + \frac{\sin (x) e^{2 x}}{4}) \frac{4}{5} + C$ as $(\frac{1}{2} \cos (x) e^{2 x} + \frac{1}{4} \sin (x) e^{2 x}) \frac{4}{5} + C$ .

$(\frac{1}{2} \cos (x) e^{2 x} + \frac{1}{4} \sin (x) e^{2 x}) \frac{4}{5} + C$