Calculus Examples

$\int e^{- θ} \cos (2 θ) d θ$

Step 1

Reorder $e^{- θ}$ and $\cos (2 θ)$ .

$\int \cos (2 θ) e^{- θ} d θ$

Step 2

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

$\cos (2 θ) (- e^{- θ}) - \int - e^{- θ} (- 2 \sin (2 θ)) d θ$

Step 3

Since $- - 2$ is constant with respect to $θ$ , move $- - 2$ out of the integral.

$\cos (2 θ) (- e^{- θ}) - (- - 2 \int e^{- θ} (\sin (2 θ)) d θ)$

Step 4

Simplify the expression.

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Step 4.1

Multiply $- 1$ by $- 2$ .

$\cos (2 θ) (- e^{- θ}) - (2 \int e^{- θ} (\sin (2 θ)) d θ)$

Step 4.2

Multiply $2$ by $- 1$ .

$\cos (2 θ) (- e^{- θ}) - 2 \int e^{- θ} (\sin (2 θ)) d θ$

Step 4.3

Reorder $e^{- θ}$ and $\sin (2 θ)$ .

$\cos (2 θ) (- e^{- θ}) - 2 \int \sin (2 θ) e^{- θ} d θ$

Step 5

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

$\cos (2 θ) (- e^{- θ}) - 2 (\sin (2 θ) (- e^{- θ}) - \int - e^{- θ} (2 \cos (2 θ)) d θ)$

Step 6

Since $- 1 \cdot 2$ is constant with respect to $θ$ , move $- 1 \cdot 2$ out of the integral.

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

Step 7

Simplify by multiplying through.

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Step 7.1

Multiply $- 1$ by $2$ .

$\cos (2 θ) (- e^{- θ}) - 2 (\sin (2 θ) (- e^{- θ}) - (- 2 \int e^{- θ} (\cos (2 θ)) d θ))$

Step 7.2

Multiply $- 2$ by $- 1$ .

$\cos (2 θ) (- e^{- θ}) - 2 (\sin (2 θ) (- e^{- θ}) + 2 \int e^{- θ} (\cos (2 θ)) d θ)$

Step 7.3

Apply the distributive property.

$\cos (2 θ) (- e^{- θ}) - 2 (\sin (2 θ) (- e^{- θ})) - 2 (2 \int e^{- θ} (\cos (2 θ)) d θ)$

Step 7.4

Multiply.

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Step 7.4.1

Multiply $- 1$ by $- 2$ .

$\cos (2 θ) (- e^{- θ}) + 2 (\sin (2 θ) (e^{- θ})) - 2 (2 \int e^{- θ} (\cos (2 θ)) d θ)$

Step 7.4.2

Multiply $2$ by $- 2$ .

$\cos (2 θ) (- e^{- θ}) + 2 (\sin (2 θ) (e^{- θ})) - 4 \int e^{- θ} (\cos (2 θ)) d θ$

Step 8

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

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

Step 9

Simplify the answer.

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Step 9.1

Rewrite $\frac{\cos (2 θ) (- e^{- θ}) + 2 \sin (2 θ) e^{- θ}}{5} + C$ as $\frac{- \cos (2 θ) e^{- θ} + 2 \sin (2 θ) e^{- θ}}{5} + C$ .

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

Step 9.2

Simplify.

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Step 9.2.1

Factor $e^{- θ}$ out of $- \cos (2 θ) e^{- θ} + 2 \sin (2 θ) e^{- θ}$ .

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Step 9.2.1.1

Factor $e^{- θ}$ out of $- \cos (2 θ) e^{- θ}$ .

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

Step 9.2.1.2

Factor $e^{- θ}$ out of $2 \sin (2 θ) e^{- θ}$ .

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

Step 9.2.1.3

Factor $e^{- θ}$ out of $e^{- θ} (- \cos (2 θ)) + e^{- θ} (2 \sin (2 θ))$ .

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

Step 9.2.2

Factor $- 1$ out of $- \cos (2 θ)$ .

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

Step 9.2.3

Factor $- 1$ out of $2 \sin (2 θ)$ .

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

Step 9.2.4

Factor $- 1$ out of $- (\cos (2 θ)) - (- 2 \sin (2 θ))$ .

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

Step 9.2.5

Rewrite $- (\cos (2 θ) - 2 \sin (2 θ))$ as $- 1 (\cos (2 θ) - 2 \sin (2 θ))$ .

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

Step 9.2.6

Move the negative in front of the fraction.

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