## Introduction to integral of sin(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 sin(ln x) integral by using different integration techniques.

## What is the integral of sin(ln x)?

The sin(lnx) integration is an antiderivative of sine 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 opposite side to the hypotenuse of a triangle which is written as:

Sin = opposite side / hypotenuse

### Integral of sin(ln(x)) formula

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

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

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

## How to calculate the sin ln x integral?

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

- Integration by parts
- Substitution method
- Definite integral

## Integral sin(lnx) 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 sin squared x by using integration by parts.

### Proof of integration of sin(lnx) by using integration by parts

Since we know that the function sine lnx can be written as the product of two functions. Therefore, we can calculate the integral of sin(ln x) by using integration by parts. To integrate sin(lnx), suppose that:

$I = x\sin(\ln x).\left(\frac{1}{x}\right)$

Applying the integral we get,

$I = ∫(x\sin(\ln x).(1/x))dx$

Since the method of integration by parts is:

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

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

$I = - x.\cos(\ln x). + ∫\cos(\ln x)dx$

It can be written as:

$I = - x.\cos(\ln x) + ∫x\cos(\ln x)\left(1/x\right)dx$

Now by using integration by parts again,

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

Since

$I = ∫\sin(\ln x)dx$

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:

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

Try our integral by parts calculator to verify the above calculations.

## Sin(lnx) integral 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 sin(ln x) by using the substitution method.

### Proof of Integral of sin(lnx) by using substitution method

To integrate sin(lnx) by using substitution method, suppose that:

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

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,

$x = e^u$

Now substitution the value of u in the integral,

$I = ∫e^u.\sin(u)du$

Now integrate it by using integration by parts.

$I = -e^u.\cos(u)+∫e^u.\cos(u)du$

Again applying integration by parts,

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

Since

$I = ∫e^u.\sin(u)du$

then

$I = -e^u.\cos(u)+e^u\sin(u)–I$

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

It implies that,

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

Now substituting the value of u,

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

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

$I=\frac{1}{2}\left[x\sin(\ln x)–x\cos(\ln x)\right]{2}nbsp;

If the integrand is a nonlinear algebraic function, the trig-substitution calculator is a best way to evaluate such integrals.

## Integration of sin(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 sin ln x integral by using the definite integral.

### Proof of sin(lnx) integration by using definite integral

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

$∫^π_0 \sin(\ln x)dx=\left|\frac{1}{2}\left[x\sin(\ln x) – x\cos(\ln x)\right]\right|^π_0$

Now, substituting the limit in the given function.

$∫^π_0 \sin(\ln x)dx =\frac{1}{2}\left[π\sin(\ln π)–π\cos(\ln π)\right]-\frac{1}{2}\left[0\sin(\ln 0)–0\cos(\ln 0)\right]$

It implies that

$∫^π_0 \sin(\ln x)dx≈0$

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

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

Now,

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

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

$∫^{\frac{π}{2}}_0 \sin(\ln x)dx≈0$

Similarly, the definite integral of sin(pi x) represents the area under the curve sin(pi x).