cos^2(x) Archives - Epsilonify

What is the integral of cos^2(x)?

The Mathematician — Thu, 12 Jan 2023 13:00:00 +0000

The integral of

\cos^2(x)

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

.

Proof. We want to determine the integral of

\cos^2(x)

, that is:

\begin{align*} \int \cos^2(x) dx. \end{align*}

Notice that we saw here that we have the next identity equality

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

, and here that

\sin(2x) = 2\sin(x)\cos(x)

. Both identities will be applied to find the integral of

\cos^2(x)

\begin{align*} \int \cos^2(x) dx &= \int \frac{1}{2}(1 + \cos(2x)) dx \\ &= \frac{1}{2} \int (1 + \cos(2x)) dx, \quad \text{ }\cos^2(x) = \frac{1}{2}(1 + \cos(2x)) \\ &= \frac{1}{2} \bigg(x + \frac{1}{2}\sin(2x)\bigg) + C \\ &= \frac{x}{2} + \frac{1}{4}\sin(2x) + C \\ &= \frac{x}{2} + \frac{1}{4}\cdot 2\sin(x)\cos(x) + C, \quad \text{ } \sin(2x) = 2\sin(x)\cos(x) \\ &= \frac{1}{2}(x + \sin(x)\cos(x)) + C. \end{align*}

Therefore, the integral of

\cos^2(x)

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

The Mathematician — Sat, 10 Sep 2022 13:00:47 +0000

We will determine the derivative of

\cos(2x)

.

Proof. Let

F(x) = \cos(2x)

f(u) = \cos(u)

, and

g(x) = 2x

such that

F(x) = f(g(x))

. To prove the derivative of

\cos(2x)

, we will use the chain rule:

\begin{align*} F'(x) = f'(g(x))g'(x). \end{align*}

We have seen here that

f'(u) = \frac{d}{du} \cos(u) = -\sin(u)

, and

\frac{d}{dx} 2x = 2

. Therefore, we get

\begin{align*} f'(g(x)) = -\sin(g(x)) = \sin(2x) \quad \text{ and } \quad g'(x) = 2. \end{align*}

Substituting everything together, we get

\begin{align*} F'(x) &= f'(g(x))g'(x) \\ &= -\sin(2x)\cdot 2 \\ &= -2\sin(2x). \end{align*}

Therefore, we see that the derivative of

\cos(2x)

-\sin(2x)

The Mathematician — Wed, 07 Sep 2022 15:00:26 +0000

We will determine the derivative of

\cos^2(x)

. We will do that in two different ways: chain rule and product rule

Proof 1. Let

F(x) = \cos^2(x)

f(u) = u^2

and

g(x) = \cos(x)

such that

F(x) = f(g(x))

. We will use the chain rule:

\begin{align*} F'(x) = f'(g(x))g'(x). \end{align*}

We know from this article that

(\cos(x))' = -\sin(x)

. Therefore, we get

\begin{align*} f'(g(x)) = 2g(x) = 2\cos(x) \quad \text{and} \quad g'(x) = -\sin(x). \end{align*}

So we get

\begin{align*} F'(x) = f'(g(x))g'(x) = -2\cos(x)\sin(x). \end{align*}

Proof 2. Let

f(x) = \cos^2(x)

. Then we want to show

\begin{align*} f'(x) = 2\cos(x)\sin(x). \end{align*}

The first step we apply is that

\begin{align*} \cos^2(x) = \cos(x)\cos(x) \end{align*}

We can apply the product rule now and, therefore

\begin{align*} (\cos(x)\cos(x))' = (\cos(x))'\cos(x) + \cos(x)(\cos(x))' \end{align*}

We have seen earlier that this article that

(\cos(x))' = -\sin(x)

. Therefore, we get

\begin{align*} (\cos(x))'\cos(x) + \cos(x)(\cos(x))' = -\sin(x)\cos(x) - \cos(x)\sin(x). \end{align*}

-\sin(x)\cos(x) - \cos(x)\sin(x) = -2\cos(x)\sin(x)

. In a complete picture, we get

\begin{align*} f'(x) &= (\cos(x)\cos(x))' \\ &= (\cos(x))'\cos(x) + \cos(x)(\cos(x))' \\ &= -\sin(x)\cos(x) - \cos(x)\sin(x) \\ &= -2\cos(x)\sin(x). \end{align*}

Therefore,

f'(x) = \frac{d}{dx} \cos^2(x) = -2\cos(x)\sin(x)