Introduction to integral of sin x*cos x

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 integral sin cos by using different integration techniques.

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

The integral of sin xcos x is an antiderivative of sine function which is equal to –cos(2x)/4. 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 sinxcosx formula

The formula of integral of sin 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 cos is:

$∫\sin x.\cos xdx=-\frac{\cos(2x)}{4}+c$

Where c is any constant involved, dx is the coefficient of integration and ∫ is the symbol of integral. The integral of sin^2xcos^2x can also be calculated by using the above formula.

How to calculate the sin cos integral?

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

1. Derivatives
2. Substitution method
3. Definite integral

Integral of sin xcos x 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 integral of sinx cosx by using derivatives.

Proof of sin cos integration by using derivatives

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

In derivative, we know that,

$\frac{1}{2}\frac{d}{dx}(2\sin x.\cos x) =\frac{1}{2}\frac{d}{dx}(\sin(2x))=-2\cos(2x)$

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

$-\frac{1}{2}\frac{d}{dx}(\cos(2x))=2\sin(2x)$

Hence the integral of sin xcos x is equal to the negative of cos(2x)/4. It is written as:

$∫\sin x\cos xdx=-\frac{\cos(2x)}{4}+c$

This method can also be helpful to find the integration of sin x cos^2x.

Integral of sin xcos x 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 sin cos integral by using the u-substitution method calculator.

Proof of integral sincos by using substitution method

To proof the integral sin cos by using substitution method, suppose that:

$I = ∫\sin x.\cos xdx$

Suppose that, u = sin x and du = cos xdx, then above integral will becomes,

$I = ∫udu$

By integrating with respect to u we get;

$I = \frac{u^2}{2}+ c$

Now substituting the value of u back here,

$I=\frac{\sin^2x}{2} + c$

Since sin^2x = 1 – cos(2x), then,

$I =\frac{1 – \cos(2x)}{2}+c$

Then,

$I=\frac{1}{4}-\frac{\cos(2x)}{4}+c$

Hence the integral of sin xcos x is –cos(2x)/4. You can also try our trigonometric substitution calculator to eveluate integrals by writting them in simpler form.

Integral sin cos 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)4

Let’s understand the verification of the integral of sin x by using the definite integral finder.

Proof of integral of sin cos by using definite integral

To compute the sin cos integral 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 x\cos xdx=-\left|\frac{\cos(2x)}{4}\right|^π_0 $

Now, substituting the limit in the given function.

$∫^π_0 \sin x\cos xdx=-\frac{\cos(2π)}{4}+\frac{\cos(0)}{4}$

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

$∫^π_0 \sin x\cos xdx=-\frac{1}{4}-\frac{1}{4}=-\frac{1}{2}$

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

$∫^{\frac{π}{2}}_0 \sin x\cos x dx=-\left|\frac{\cos(2x)}{4}\right|^{\frac{π}{2}}_0$

Now,

$∫^{\frac{π}{2}}_0 \sin x\cos xdx=-\frac{\cos \frac{2π}{2}}{4} + \frac{\cos (0)}{4}$

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

$∫^{\frac{π}{2}}_0 \sin x\cos xdx=-\frac{1}{4} + \frac{1}{4}=0$

Therefore, the definite integral of sin cos is equal to 0. For any value of m and n, the definite integral of sin mxcos nx can be calculated with above calculations.