Calculus Examples

cos (2 y)

$\cos (2 y)$

Step 1

Write

cos (2 y)

$\cos (2 y)$ as a function.

f (y) = cos (2 y)

$f (y) = \cos (2 y)$

Step 2

The function

F (y)

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

f (y)

$f (y)$ .

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

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

Step 3

Set up the integral to solve.

F (y) = \int cos (2 y) d y

$F (y) = \int \cos (2 y) d y$

Step 4

Let

u = 2 y

$u = 2 y$ . Then

d u = 2 d y

$d u = 2 d y$ , so

\frac{1}{2} d u = d y

$\frac{1}{2} d u = d y$ . Rewrite using

u

$u$ and

d

$d$

$u$ .

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Step 4.1

Let

$u = 2 y$ . Find

$\frac{d u}{d y}$ .

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Step 4.1.1

Differentiate

$2 y$ .

$\frac{d}{d y} [2 y]$

Step 4.1.2

Since

$2$ is constant with respect to

$y$ , the derivative of

$2 y$ with respect to

$y$ is

$2 \frac{d}{d y} [y]$ .

$2 \frac{d}{d y} [y]$

Step 4.1.3

Differentiate using the Power Rule which states that

$\frac{d}{d y} [y^{n}]$ is

$n y^{n - 1}$ where

$n = 1$ .

$2 \cdot 1$

Step 4.1.4

Multiply

$2$ by

$1$ .

$2$

Step 4.2

Rewrite the problem using

$u$ and

$d u$ .

$\int \cos (u) \frac{1}{2} d u$

Step 5

Combine

$\cos (u)$ and

$\frac{1}{2}$ .

$\int \frac{\cos (u)}{2} d u$

Step 6

Since

$\frac{1}{2}$ is constant with respect to

$u$ , move

$\frac{1}{2}$ out of the integral.

$\frac{1}{2} \int \cos (u) d u$

Step 7

The integral of

$\cos (u)$ with respect to

$u$ is

$\sin (u)$ .

$\frac{1}{2} (\sin (u) + C)$

Step 8

Simplify.

$\frac{1}{2} \sin (u) + C$

Step 9

Replace all occurrences of

$u$ with

$2 y$ .

$\frac{1}{2} \sin (2 y) + C$

Step 10

The answer is the antiderivative of the function

$f (y) = \cos (2 y)$ .

$F (y) =$

$\frac{1}{2} \sin (2 y) + C$