Introduction to integral of sin(2t)
In calculus, the integral is a fundamental concept that assigns numbers to functions to define displacement, area, volume of revolution, 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.
Integral calculators 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 integration of sin 2t by using different integration techniques.
What is the integral of sin2t?
The integral of sin(2t) is an antiderivative of sine function which is equal to –cos(2x)/2. 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 2t formula
The formula of integration of sin2t contains integral sign, coefficient of integration and the function as sine. It is denoted by ∫(sin(2t))dx. In mathematical form, the integral of sin(2t) is:
$∫\sin(2t)dx=-\frac{\cos(2t)}{2}+c$
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(2t)?
The integral of sin(2t) is its antiderivative that can be calculated by using different integration techniques. In this article, we will discuss how to calculate sin2t integration by using:
- Derivatives
- Substitution method
- Definite integral
Integral of sin(2t) 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 sin2t by using derivatives.
Proof of integration of sin 2t by using derivatives
Since we know that the integration is the reverse of the derivative. Therefore, we can calculate the integral of sin(2t) by using its derivative. For this, we have to look for some derivatives formulas or a formula that gives sin(2t) as the derivative of any function.
In derivative, we know that,
$\frac{d}{dx}(\cos(2t))=-2\sin(2t)$
It means that the derivative of cos x gives us sin(2t). But it has negative sign. Therefore, to obtain the integral of sine, we have to multiply above equation by negative sign, that is:
$-\frac{d}{dx}(\cos(2t))=2\sin(2t)$
Hence the integral of sin(2t) is equal to the negative of cos 2t. It is written as:
$∫\sin(2t)dx=-\frac{\cos(2t)}{2}+c$
Integral of sin2t 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 integration of sin2t by using substitution method
To proof the integral of sin(2t) by using the u-substitution method, suppose that:
$y=\sin(2t)$
Differentiating with respect to x,
$\frac{dy}{dx}=2\cos(2t)$
To calculate integral you can also use the trig substitution calculator, we can write the above equation as:
$dy=2\cos(2t)dt$
By trigonometric identities, we know that cos(2t) = √1 - 2sin²t. Then the above equation becomes,
$dy=\sqrt{1 - 2\sin^2x}dt$
Now, substituting the value of sin2 t, such as:
$dy=\sqrt{1 – 2y^2}.dt$
Multiplying both sides by sin(2t),
$\frac{\sin(2t)dy}{\sqrt{1 - 2y^2}}=\sin(2t)dt$
Again substitute sin(2t) = y on the left side.
$\frac{ydy}{\sqrt{1 - 2y^2}}=\sin(2t)dt$
Integrating on both sides by applying integral,
$∫\frac{ydy}{\sqrt{1 - 2y^2}}=∫\sin(2t)dt$
Let 1 - 2y² = u. Then -4y dy = du (or) y dy = -1/4 du.
Then the above left-hand side integral becomes,
$-\frac{1}{4}∫\frac{1}{\sqrt{u}}du=∫\sin(2t)dt$
$-\frac{1}{4}∫u^{-\frac{1}{2}}du=∫\sin(2t)dt$
Since the power rule of integration is
$∫t^ndt=\frac{t^{n+1}}{n+1}+C$
Therefore, by using this formula we get,
$-\frac{1}{4}\left(\frac{u^{1/2}}{1/2}\right)+C=∫\sin(2t)dt$
$-\frac{u^{\frac{1}{2}}}{2}+C= ∫\sin(2t) dt$
$Again substituting u = 1 - y², we get
$-\frac{1}{2}(1 - y^2)^{\frac{1}{2}}+C=∫\sin(2t)dt$
And again Substitute y = sin(2t) here,
$-\frac{1}{2}(1-\sin^2 2t)^{\frac{1}{2}}+C=∫\sin(2t)dt$
$-\frac{(\cos^2 2t)^{\frac{1}{2}}}{2}+C =∫\sin(2t)dt$
$-\frac{\cos(2t)}{2}+C=∫\sin(2t)dt$
Hence the sin2t integration is –cos(2t)/2.
Integral of sin 2t 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(2t) by using the indefinite integral.
Proof of integral of sin(2t) by using definite integral
To compute the integral of sin(2t) by using a definite integral solver, we can use the interval from 0 to π or 0 to π/2. Let’s compute the integral of sin(2t) from 0 to π. For this we can write the integral as:
$∫^π_0 \sin(2t)dt=-\left|\frac{\cos(2t)}{2}\right|^π_0$
Now, substituting the limit in the given function.
$∫^π_0 \sin(2t)dt=-\frac{\cos 2(π)}{2}+\frac{\cos2(0)}{2}$
Since cos 0 is equal to 1 and cos π is equal to -1, therefore,
$∫^π_0 \sin(2t) dt =-\frac{1}{2}+\frac{1}{2}= 0$
Which is the calculation of the definite integral of sin 2t. Now to calculate the integral of sin(2t) between the interval 0 to π/2, we just have to replace π by π/2. Therefore,
$∫^{\frac{π}{2}}_0 \sin(2t)dt=-\left|\frac{\cos(2t)}{2}\right|^{\frac{π}{2}}_0$
Now,
$∫^{\frac{π}{2}}_0 \sin(2t)dt=-\frac{\cos 2(π/2)}{2} + \frac{\cos 2(0)}{2}$
Since cos 0 is equal to 1 and cos π/2 is equal to 0, therefore,
$∫^{\frac{π}{2}}_0 \sin(2t)dt=\frac{1}{2}+\frac{1}{2}=1$
Therefore, the definite integral of sin2t is equal to 1. Moreover, for any value of n, you can integrate sin(nx).