Calculus Examples

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

Step 1

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

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

Step 2

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

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

Step 3

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

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

Step 4

Simplify the expression.

Tap for more steps...

Step 4.1

Multiply $- 1$ by $- 2$ .

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

Step 4.2

Multiply $2$ by $- 1$ .

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

Step 4.3

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

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

Step 5

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

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

Step 6

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

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

Step 7

Simplify by multiplying through.

Tap for more steps...

Step 7.1

Multiply $- 1$ by $2$ .

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

Step 7.2

Multiply $- 2$ by $- 1$ .

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

Step 7.3

Apply the distributive property.

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

Step 7.4

Multiply.

Tap for more steps...

Step 7.4.1

Multiply $- 1$ by $- 2$ .

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

Step 7.4.2

Multiply $2$ by $- 2$ .

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

Step 8

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

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

Step 9

Simplify the answer.

Tap for more steps...

Step 9.1

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

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

Step 9.2

Simplify.

Tap for more steps...

Step 9.2.1

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

Tap for more steps...

Step 9.2.1.1

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

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

Step 9.2.1.2

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

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

Step 9.2.1.3

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

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

Step 9.2.2

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

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

Step 9.2.3

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

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

Step 9.2.4

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

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

Step 9.2.5

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

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

Step 9.2.6

Move the negative in front of the fraction.

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