Integral Of Cos(Lnx)

Integral of cos(lnx) along with its formula and proof with examples. Also learn how to calculate integration of cos(lnx) with step by step examples.

Alan Walker-

Published on 2023-04-14

Introduction to integral of cos(lnx)

In calculus, the integral is a fundamental concept that assigns numbers to functions to define displacement, area, volume, and all those functions that contain a combination of tiny elements. It is categorized into two parts, definite integral and indefinite integral. The process of integration calculates the integrals. This process is defined as finding an antiderivative of a function.

Integrals can handle almost all functions, such as trigonometric, algebraic, exponential, logarithmic, etc. This article will teach you what is integral to a trigonometric function sine. You will also understand how to compute cos(ln x) integral by using different integration techniques.

What is the integral of cos(ln x)?

The integral of cos(lnx) is an antiderivative of cos(ln x) function which is equal to ½[xsin(ln x) + xcos(ln x)]. It is also known as the reverse derivative of sine function which is a trigonometric identity.

The sine function is the ratio of adjacent side to the hypotenuse of a triangle which is written as:

cos = adjacent side / hypotenuse

Integral of cos(ln x) formula

The formula of integral of cos(lnx) contains integral sign, coefficient of integration and the function as sine. It is denoted by ∫(cos(lnx))dx. In mathematical form, the integral of cos(lnx) is:

$\int \cos(\ln x)dx = \frac{1}{2}[x\sin(\ln x) + x\cos(\ln x)]+c$

Where c is any constant involved, dx is the coefficient of integration and ∫ is the symbol of integral. If we replace cos(ln x) by cos(e^x) in the above integration formula, we can calculate the integral of cos(e^x)

How to calculate the integral of cos(lnx)?

The integral of cos(lnx) is its antiderivative that can be calculated by using different integration techniques. In this article, we will discuss how to calculate integral of cosine by using:

1. Integration by parts
2. Substitution method
3. Definite integral

Integral of cos(lnx) squared by using integration by parts

The derivative of a function calculates the rate of change, and integration is the process of finding the antiderivative of a function. The integration by parts is a method of solving integral of two functions combined together. Let’s discuss calculating the integral of cos ln x by using integration by parts.

Proof of integral of cos(lnx) by using integration by parts

Since we know that the function sine squared x can be written as the product of two functions. Therefore, we can calculate the integral of cos ln x by using the integration by parts. For this, suppose that:

$I = \cos(\ln x)$

Applying the integral we get,

$I =\int \cos(\ln x))dx$

Since the formula of integration by parts calculator is:

$\int[f(x).g(x)]dx = f(x).\int g(x)dx - \int[f’(x).\int g(x)]dx$

Now replacing f(x) and g(x) by sin(lnx)(1/x) and x, we get,

$I = x.\cos(\ln x) + \int \frac{x\sin(\ln x)}{x}dx$

It can be written as:

$I = x.\cos(\ln x) – \int[\sin(\ln x)]dx$

Now by using integration by parts again,

$I = x.\cos(\ln x) + [x\sin(\ln x) - \int \cos(\ln x)dx]$

Since

$I = ∫\cos(\ln x)dx$

Therefore,

$I = x.\cos(\ln x) + [x\sin(\ln x) - I]$

Moreover,

$2I = x.\cos(\ln x) + x\sin(\ln x)$

Hence the integral of sin(ln x) by using integration by parts is:

$I =\frac{1}{2}[x\sin(\ln x) + x\cos(\ln x)] + c$

The method of integration by parts is also helpful to find the integral of cos(x^2).

Integral of cos(lnx) by using substitution method

The substitution method involves many trigonometric formulas. We can use these formulas to verify the integrals of different trigonometric functions such as sine, cosine, tangent, etc. Let’s understand how to prove the integral of cos by using the substitution method.

Proof of Integral of cos(lnx) by using substitution method

To proof the integral of cos(lnx) by using the substitution method, suppose that:

$I = ∫\cos(\ln x)dx = ∫x\cos(\ln x)\left(\frac{1}{x}\right)dx$

We will use u-substitution method to solve the given integral. For this, suppose that,

$u=\ln x\quad\text{and}\quad du =\frac{1}{x}.dx$

Then, x = eu

Now substitution the value of u in the integral,

$I = ∫e^u.\cos(u)du$

Now integrate it by using integration by parts.

$I = e^u.\sin(u) – ∫e^u.\sin(u)du$

Again applying integration by parts,

$I = e^u.\sin(u) + e^u\cos(u) – ∫e^u.\cos(u)du$

Since

$I = ∫e^u.\cos(u)du$

then

$I = e^u.\sin(u) + e^u\cos(u) – I$

$2I = e^u.\sin(u) + e^u\sin(u)$

It implies that,

$I =\frac{1}{2}[e^u\sin(u) + e^u\cos(u)]$

Now substituting the value of u,

$I =\frac{1}{2}[e^{\ln x}\sin(\ln x) + e^{\ln x}\cos(\ln x)]$

Hence the proof of integral of cos(lnx) is:

$I =\frac{1}{2}[x\sin(\ln x) + x\cos(\ln x)]$

Integral of cos(lnx) by using definite integral

The definite integral is a type of integral that calculates the area of a curve by using infinitesimal area elements between two points. The definite integral can be written as:

$∫^b_a f(x) dx = F(b) – F(a)$

Let’s understand the verification of the integral of cos(lnx) by using the indefinite integral.

Proof of integral of cos(lnx) by using definite integral

To compute the integral of cos(lnx) by using a definite integral calculator, we can use the interval from 0 to π or 0 to π/2. Let’s compute the integral of cos(lnx) from 0 to π. For this we can write the integral as:

$∫^\pi_0 \cos(\ln x) dx =\frac{1}{2}\left|x\sin(\ln x) + x\cos(\ln x)\right|^\pi_0$

Now, substituting the limit in the given function.

$∫^\pi_0 \cos(\ln x) dx =\frac{1}{2}[π\sin(\ln π) + π\cos(\ln π)] –\frac{1}{2}[0\sin(\ln 0) + 0\cos(\ln 0)]$

It implies that

$∫^\pi_0 \cos(\ln x) dx ≈ 0$

Which is the calculation of the definite integral of cos(lnx). Now to calculate the integral of cos(lnx) between the interval 0 to π/2, we just have to replace π by π/2. Therefore,

$∫^{\frac{\pi}{2}}_0 \cos(\ln x) dx =\frac{1}{2}\left|x\sin(\ln x)+x\cos(\ln x)\right|^{\frac{\pi}{2}}_0$

Now,

$∫^{\frac{\pi}{2}}_0 \cos(\ln x)dx =\frac{1}{2}\left[\frac{π}{2}\sin(\ln π/2) + \frac{π}{2}\cos(\ln π/2)\right] –\frac{1}{2}[0\sin(\ln0) + 0\cos(\ln0)]$

Since cos 0 is equal to 1 and cos π/2 is equal to 0, therefore,

$∫^{\frac{\pi}{2}}_0 \cos(\ln x)dx ≈ 0$