What is the integral of cos(ln(x))?

The Mathematician — Tue, 04 Apr 2023 13:00:53 +0000

The integral of

\cos(\ln(x))

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

.

Solution. We need to find out what the integral of

\cos(\ln(x))

is, that is:

\begin{align*} I = \int \cos(\ln(x)) dx. \end{align*}

We will apply integration by parts with the following formula:

\begin{align*} \int UdV = UV - \int VdU. \end{align*}

By applying the integration by parts, we will consider the following formulas:

\begin{align*} U = \cos(\ln(x)), \quad &dV = dx\\ dU = \frac{-\sin(\ln(x))}{x}dx, \quad &V = x. \end{align*}

The derivative of

U

, which we see above, can be verified here. So we get the following integral:

\begin{align*} \int \cos(\ln(x)) dx = x\cos(\ln(x)) + \int \sin(\ln(x))dx. \end{align*}

Secondly, we integrate by parts again for

\int \sin(\ln(x))dx

\begin{align*} U = \sin(\ln(x)), \quad &dV = dx\\ dU = \frac{\cos(\ln(x))}{x}dx, \quad &V = x. \end{align*}

To see how we did

dU

, see here. Therefore, combined with the previous integral, we get:

\begin{align*} \int \cos(\ln(x)) dx &= x\sin(\ln(x)) + \int \sin(\ln(x))dx \\ &= x\cos(\ln(x)) + x\sin(\ln(x)) - \int \cos(\ln(x))dx \\ &= x\cos(\ln(x)) + x\sin(\ln(x)) - I. \end{align*}

Now we bring

I

to the left-hand side and divide by 2. So we finally get the integral we wanted:

\begin{align*} \int \cos(\ln(x)) dx = \frac{1}{2}x(\cos(\ln(x)) + \sin(\ln(x))) + C. \end{align*}

Therefore, the integral of

\sin(\ln(x))

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

What is the derivative of cos(ln(x))?

The Mathematician — Mon, 27 Mar 2023 13:00:16 +0000

The derivative of

\cos(\ln(x))

-\sin(\ln(x))/x

.

Solution. We will use the chain rule to find out what the derivative of

F(x) = f(g(x)) = \cos(\ln(x))

is. In other words, we will use the following:

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

where

f(u) = \cos(u)

and

g(x) = \ln(x)

. The derivative of

f(u)

f'(u) = -\sin(u)

, which can be checked here. Also, we see here that

g'(x) = \frac{1}{x}

. So we get:

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

Substituting the results, we get:

\begin{align*} F'(x) &= f'(g(x))g'(x) \\ &= -\sin(\ln(x))\frac{1}{x} \\ &= -\frac{\sin(\ln(x))}{x}. \end{align*}

Therefore, the derivative of

\cos(\ln(x))

-\sin(\ln(x))/x