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

The Mathematician — Tue, 24 Jan 2023 13:00:54 +0000

The integral of

\cot^2(x)

- x - \cot(x)

.

Proof. We want to determine the integral of

\cot^2(x)

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

We have seen here that

\cot^2(x) = \csc^2(x) - 1

. So we get:

\begin{align*} \int \cot^2(x) dx &= \int (\csc^2(x) - 1)dx \\ &= \int \csc^2(x)dx - \int 1\cdot dx. \end{align*}

Further, we saw here that the integral of

\csc^2(x)

-\cot(x)

plus some constant. So, we get:

\begin{align*} \int \cot^2(x) dx &= \int (\csc^2(x) - 1)dx \\ &= \int \csc^2(x)dx - \int 1\cdot dx \\ &= -\cot(x) - x + C. \end{align*}

Therefore, the integral of

\cot^2(x)

-\cot(x) - x

What is the Derivative of cot^2(x)?

The Mathematician — Thu, 10 Nov 2022 13:00:24 +0000

The derivative of

\cot^2(x)

-2\cot(x)\csc^2(x)

.

Solution. Let

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

f(u) = u^2

, and

g(x) = \cot(x)

such that

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

. We will use the chain rule to determine the derivative of

\cot^2(x)

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

Earlier, we saw here that

g'(x) = \frac{d}{dx} \cot(x) = -\csc^2(x)

, and

f'(u) = 2u

. Therefore, we get:

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

So, we have:

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

Therefore, the derivative of

\cot^2(x)

-2\cot(x)\csc^2(x)