## Introduction to integral of sin nx*cos mx

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(nx)cos(mx) integral by using different integration techniques.

## What is the integral of sin(nx)cos(mx)?

The integral of sin(nx) cos(mx) is an antiderivative of sine function which is equal to –sin(m+n)/2(m+n) – sin(m – n)/2(m – n). 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(nx)cos(mx) formula

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

$∫\sin(nx).\cos(mx)dx = -\frac{1}{2}\left[\frac{\sin(m+n)}{m+n} + \frac{\sin(m – n)}{(m – n)}\right]$

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(nx)cos(mx)?

The integral of sin(nx)cos(mx) 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:

- Trigonometric formulas
- Substitution method
- Definite integral

## Integral of sin(nx)cos(mx) by using Trigonometric formulas

The substitution method also involves trigonometric formulas that helps to solve integrals easily. Let’s understand how to calculate the integral of sin(nx)cos(mx) by using different trigonometric formulas.

### Proof of integral of sin(nx)cos(mx) by using Trigonometric formulas

To prove the integral of sin(nx)cos(mx), we use different trigonometric formulas. Therefore,

$I = ∫\sin(nx)\cos(mx)dx$

To calculate the integral of sin(8x)cos(5x), replace n by 8 and m by 5 in the above fomrula.

Since we know the formulas of sum and product in trigonometry,

$\sin(nx)\cos(mx) =\frac{1}{2}\left[\cos(m – n)x – \cos(m + n)x\right]$

Therefore, Using this formula in the integral,

$I = ∫\sin(nx)\cos(mx)dx = ∫\frac{1}{2}\left[\cos(m – n)x – \cos(m + n)x\right]dx$

Now we can integral each term separately, we get

$I =\frac{1}{2}\left[\frac{\sin(m – n)x}{(m – n)} – \frac{\sin(m+n)x}{(m+n)}\right]$

Hence the integral of sin(nx)cos(mx) is verified.

## Integral of sin(nx)cos(mx) 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.

### Proof of Integral of sin(nx)cos(mx) by using substitution method

To proof the integral of sin(nx)cos(mx) by using u-substitution method, suppose that:

$I = ∫\sin nx.cos mx)dx$

Using trigonometric formula,

$\sin nx.\cos mx =\frac{1}{2}\left[\cos(m – n)x – \cos(m+n)x\right]$

Then the above integral can be written as:

$I = \frac{1}{2}∫\left[\cos(m – n)x – \cos(m+n)x\right]dx$

To verfity the sin(nx)cos(mx) integral by using u-substitution calculator, we start by supposing that,

$u = (m – n)x\quad\text{and}\quad du=(m – n)dx$

Similarly,

$t = (m + n)x\quad\text{and}\quad dt=(m + n)dx$

Now substituting the values of u and t in the integral,

$I=\frac{1}{2(m-n)}∫\cos(u)du – \frac{1}{2(m+n)}∫\cos(t)dt$

By integrating with respect to u and t, we get;

$I=\frac{1}{2(m-n)}\sin(u)–\frac{1}{2(m+n)}\sin(t) + c$

Now substituting the value of u and t back here,

$I=\frac{1}{2(m-n)}\sin(m – n)x–\frac{1}{2(m+n)}\sin(m + n)x + c$

Hence we have verified the integral of sin(nx)cos(mx) by using substitution method. This method is also used to find the integration of sin^3xcosx.

## Integral of sin(nx)cos(mx) 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 sin x by using the definite integral.

### Proof of integral of sin(nx)cos(mx) by using definite integral

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

$∫^π_0 \sin(nx)\cos(mx)dx = ½ [\sin(m – n)x – \sin(m+n)x]|^π_0$

Now, substituting the limit in the given function.

$∫^π_0 \sin(nx)\cos(mx) dx=\frac{1}{2}\left[\frac{\sin(m – n)π}{m-n} –\frac{\sin(m+n)π}{m+n}-\left(\frac{\sin0}{m-n}-\frac{\sin0}{m+n}\right)\right]$

Or,

$∫^π_0 \sin(nx)\cos(mx)dx=\frac{1}{2}\left[\frac{\sin(m–n)π}{m-n}–\frac{\sin(m+n)π}{m+n}\right]$

When m = n then,

$∫^π_0 \sin(nx)\cos(mx)dx=\frac{1}{2}[0 – \sin π]$

Since sin 0 is equal to 0 and sin π is also equal to 0, therefore,

$∫^π_0 \sin(nx)\cos(mx)dx=0$

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

$∫^{\frac{π}{2}}_0 \sin(nx)\cos(mx)dx=\frac{1}{2}\left[\frac{\sin(m – n)x}{m-n} –\frac{\sin(m+n)x}{m+n}\right]|^{\frac{π}{2}}_0$

Now,

$∫^{\frac{π}{2}}_0 \sin(nx)\cos(mx)dx=\frac{1}{2}\left[\frac{\sin(m – n)\frac{π}{2}}{m-n} – \frac{\sin(m+n)\frac{π}{2}}{m+n}-\left(\frac{\sin0}{m-n}-\frac{\sin0}{m+n}\right)\right]$

Or,

$∫^{\frac{π}{2}}_0 \sin(nx)\cos(mx)dx=½[\sin(m – n)π/2 – sin(m+n)π/2]$

When m = n

$∫^{\frac{π}{2}}_0 \sin(nx)\cos(mx)dx=\frac{1}{2}\left[\sin 0 – \sin\frac{π}{2}\right]$

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

$∫^{\frac{π}{2}}_0 \sin(nx)\cos(mx)dx=\frac{1}{2}(0 – 1) = -\frac{1}{2}$

Therefore, the definite integral of sin(nx)cos(mx) is equal to -½. Moreover, if both interval points of an integral are infinity, you can use imporper integral calculator to integral such an integral.