Calculus Examples

e^{x} \cos (x)

$e^{x} \cos (x)$

Step 1

Write

e^{x} \cos (x)

$e^{x} \cos (x)$ as a function.

f (x) = e^{x} \cos (x)

$f (x) = e^{x} \cos (x)$

Step 2

The function

F (x)

$F (x)$ can be found by finding the indefinite integral of the derivative

f (x)

$f (x)$ .

F (x) = \int f (x) d x

$F (x) = \int f (x) d x$

Step 3

Set up the integral to solve.

F (x) = \int e^{x} \cos (x) d x

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

Step 4

Reorder

e^{x}

$e^{x}$ and

\cos (x)

$\cos (x)$ .

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

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

Step 5

Integrate by parts using the formula

\int u d v = u v - \int v d u

$\int u d v = u v - \int v d u$ , where

u = \cos (x)

$u = \cos (x)$ and

d v = e^{x}

$d v = e^{x}$ .

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

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

Step 6

Since

- 1

$- 1$ is constant with respect to

x

$x$ , move

- 1

$- 1$ out of the integral.

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

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

Step 7

Simplify the expression.

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Step 7.1

Multiply

- 1

$- 1$ by

- 1

$- 1$ .

\cos (x) e^{x} + 1 \int e^{x} (\sin (x)) d x

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

Step 7.2

Multiply

\int e^{x} (\sin (x)) d x

$\int e^{x} (\sin (x)) d x$ by

1

$1$ .

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

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

Step 7.3

Reorder

e^{x}

$e^{x}$ and

\sin (x)

$\sin (x)$ .

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

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

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

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

Step 8

Integrate by parts using the formula

\int u d v = u v - \int v d u

$\int u d v = u v - \int v d u$ , where

u = \sin (x)

$u = \sin (x)$ and

d v = e^{x}

$d v = e^{x}$ .

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

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

Step 9

Solving for

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

$\int e^{x} \cos (x) d x$ , we find that

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

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

\frac{\cos (x) e^{x} + \sin (x) e^{x}}{2}

$\frac{\cos (x) e^{x} + \sin (x) e^{x}}{2}$ .

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

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

Step 10

Rewrite

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

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

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

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

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

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

Step 11

The answer is the antiderivative of the function

f (x) = e^{x} \cos (x)

$f (x) = e^{x} \cos (x)$ .

F (x) =

$F (x) =$

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

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