## Introduction to integral of sin(2x)

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 antiderivative of sin 2x by using different integration techniques.

## What is the integral of sin 2x?

The integral of sin(2x) is an antiderivative of sine function which is equal to –cos 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

### Integration of sin2x formula

The integral of sin2x formula contains integral sign, coefficient of integration and the function as sine. It is denoted by ∫(sin(2x))dx. In mathematical form, the sin2x integration is:

$∫\sin(2x)dx = -\frac{\cos 2x}{2} + c$

Where c is any constant involved, dx is the coefficient of integration and ∫ is the symbol of integral.

## How to calculate the integral of sin(2x)?

The integral of sin 2x 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:

- Derivatives
- Substitution method
- Definite integral

## Integral of sin(2x) by using derivatives

The derivative of a function calculates the rate of change, and integration is the process of finding the antiderivative of a function. Therefore, we can use the derivative to calculate the integral of a function. Let’s discuss calculating the sin2x integration by using derivatives.

### Proof of sin2x integral by using derivatives

Since we know that the integration is the reverse of the derivative. Therefore, we can calculate the integration of sin2x by using its derivative. For this, we have to look for some derivatives formulas or a formula that gives sin(2x) as the derivative of any function.

In derivative, we know that,

$\frac{d}{dx}(\cos 2x) = -2\sin(2x)$

It means that the derivative of cos x gives us sin(2x). But it has negative sign. Therefore, to obtain the integral of sine, we have to multiply above equation by negative sign, that is:

$-\frac{d}{dx}(\cos 2x) = 2\sin(2x)$

Hence the antiderivative of sin 2x is equal to the negative of cos x. It is written as:

$∫\sin(2x)dx = -\frac{\cos 2x}{2} + c$

## Integral of sin2x 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 by using the substitution method calculator.

### Proof of Integral of sin 2x by using substitution method

To proof the integral of sin(2x) by using substitution method, suppose that:

$y = \sin(2x)$

Differentiating with respect to x,

$\frac{dy}{dx} = 2\cos 2x$

To calculate integral, we can write the above equation as:

$dy = 2\cos 2x dx$

By trigonometric identities, we know that cos 2x = √1 - 2sin²x. Then the above equation becomes,

$dy = \sqrt{1 - 2\sin^2x}.dx$

Now, substituting the value of sin2 x, such as:

$dy = \sqrt{1 – 2y^2}dx$

Multiplying both sides by sin(2x),

$\frac{\sin(2x)dy}{\sqrt{1 - 2y^2}} = \sin(2x)dx$

Again substitute sin(2x) = y on the left side.

$\frac{ydy}{\sqrt{1 - 2y^2}} = \sin(2x) dx$

Integrating on both sides by applying integral,

$∫\frac{y dy}{\sqrt{1 - 2y^2}} = ∫\sin(2x)dx$

Let 1 - 2y² = u. Then -4y dy = du (or) y dy = -1/4 du.

Then the above left-hand side integral becomes,

$-\frac{1}{4}∫\frac{1}{\sqrt{u}}du=∫\sin(2x) dx$

$-\frac{1}{4}∫u^{-\frac{1}{2}}du=∫\sin(2x) dx$

Since the power rule of integration is

$∫x^ndx=\frac{x^{n+1}}{n+1}+C$

Therefore, by using this formula we get,

$-\frac{1}{4}\left(\frac{u^{\frac{1}{2}}}{1/2}\right) + C = ∫\sin(2x) dx$

$-\frac{u^{1/2}}{2}+C=∫\sin(2x)dx$

Again substituting u = 1 - y², we get

$-\frac{(1 - y²)^{\frac{1}{2}}}{2} + C = ∫\sin(2x)dx$

And again Substitute y = sin(2x) here,

$-\frac{(1 - \sin^2 2x)^{\frac{1}{2}}}{2}+C =∫\sin(2x) dx$

$-\frac{(\cos^2 2x)^{1/2}}{2}+C=\int\sin(2x) dx$

$-\frac{\cos 2x}{2}+C = \int \sin(2x) dx$

Hence the integral of sin2x is –cos 2x/2.

## Integration of sin2x 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 2x integral by using the indefinite integral.

### Proof of integral of sin(2x) by using definite integral

To compute the antiderivative of sin2x by using a definite integral finder, we can use the interval from 0 to π or 0 to π/2. Let’s compute the integral sin 2x from 0 to π. For this we can write the integral as:

$∫^π_0 \sin(2x) dx=-\left|\frac{\cos 2x}{2}\right|^π_0$

Now, substituting the limit in the given function.

$∫^π_0 \sin(2x)dx=-\frac{\cos 2(π)}{2}+\frac{\cos 2(0)}{2}$

Since cos 0 is equal to 1 and cos π is equal to -1, therefore,

$∫^π_0 \sin(2x)dx = -\frac{1}{2}+\frac{1}{2}= 0$

Which is the calculation of the definite integral of sin 2x. Now to calculate the sin2x integral between the interval 0 to π/2, we just have to replace π by π/2. Therefore,

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

Now,

$∫^{\frac{π}{2}}_0 \sin(2x) dx=-\frac{\cos 2(π/2)}{2} + \frac{\cos 2(0)}{2}$

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

$∫^{\frac{π}{2}}_0 \sin(2x) dx = \frac{1}{2} + \frac{1}{2}=1$

Therefore, the definite integral of sin(2x) is equal to 1.