Introduction to integral of sin^2θ

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 calculator 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 squared. You will also understand how to compute sin square theta integration by using different integration techniques.

What is the integral of sin^2θ?

The integral of sin^2θ is an antiderivative of sine function which is equal to θ/2–sin 2θ/4+c. 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 sin square theta formula

The formula of integral of sin contains integral sign, coefficient of integration and the function as sine. It is denoted by ∫(sin²θ)dθ. In mathematical form, the integration of sin squared theta is:

$∫\sin^2θdθ = \frac{θ}{2}-\frac{\sin2θ}{4}+ c$

Where c is any constant involved, dθ is the coefficient of integration and ∫ is the symbol of integral. This integration is helpful to calculate the integral of sin cube x. 

How to calculate the sin square theta integration?

The integral of sin^2θ 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:

1. Integration by parts
2. Substitution method
3. Definite integral

Integral of sin θ squared 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 θ by using integration by parts.

Proof of integral of sin^2θ by using integration by parts

Since we know that the function sine squared θ can be written as the product of two functions. Therefore, we can calculate the integral of sin^2θ by using integration by parts. For this, suppose that:

$I = \sin θ.\sin θ$

Applying the integral we get,

$I = ∫\sin θ.\sin θdθ$

Since the method of integration by parts is:

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

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

$I=-\sin θ.\cos θ + ∫[\cos θ.\cos θ]dθ$

It can be written as:

$I=-\sin θ.\cos θ + ∫\cos^2θdθ$

Now by using a trigonometric identity cos²θ = 1+cos2θ/2. Therefore, substituting the value of cos2θ in the above equation, we get:

$I=-\sinθ.\cos θ+∫\left(\frac{1+\cos2θ}{2}\right)dθ$

Integrating remaining terms,

$I=-\sin θ.\cos θ+\frac{1}{2}(θ)+\frac{\sin2θ}{4}$

Or,

$I=-\frac{\sin 2θ}{2}+ \frac{θ}{2}+\frac{\sin 2θ}{4}$

Or,

$I = \frac{θ}{2}–\frac{\sin 2θ}{4}$

Hence the integral of sin^2x is equal to,

$∫\sin^2θdx = \frac{θ}{2}–\frac{\sin 2θ}{4}{2}lt;/p>

Integral of sin^2θ 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 square theta integration by using the substitution method.

Proof of Integral of sin^2θ by using substitution method

To proof the integration of sin square theta by using substitution method, suppose that:

$I=∫\sin^2θ = ∫(1- \cos^2θ)dθ$

Further we can cos2 θ can be substituted as cos²θ = 1+cos2θ/2. Then the above equation will become.

$I=θ-∫\left(\frac{1+ \cos2θ}{2}\right)dθ$

Integrating,

$I=θ–\frac{θ}{2}-\frac{\sin2θ}{4}$

Moreover,

$I=\frac{θ}{2}-\frac{\sin2θ}{4}$

Hence the integration of sin^2θ is verified by using substitution method. Also learn to calculate the integral of sin^4x by using substitution method.

Integral of sin^2θ 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^2θ by using the definite integral.

Proof of integration of sin squared theta by using definite integral

To compute the integral of sin^2θ by using a definite integral, we can use the interval from 0 to 2π or 0 to π. Let’s compute the integral of sin^2θ from 0 to 2π. The indefinite integral of sin^2θ can be written as:

$∫^{2π}_0 \sin^2θdθ=\left|\frac{θ}{2}-\frac{\sin 2θ}{4}\right|^{2π}_0$

Substituting the value of limit we get,

$∫^2π_0 \sin^2θdθ=\frac{2π}{2}-\frac{\sin4π}{4}-\left[0 - \frac{\sin 0}{4}\right]$

$∫^{2π}_0 \sin^2θdθ = π - \frac{0}{4}$

Therefore, the integral of sin2 θ from 0 to 2π is

$∫^{2π}_0 \sin^2θ dθ = π$

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

$∫^{2π}_0 \sin^2θdθ= \left|\frac{θ}{2}-\frac{\sin 2θ}{4}\right|^{2π}_0$

$∫^{2π}_0 \sin^2θdθ=\left[\frac{π}{2}-\frac{\sin π}{4}\right]-\left[0-\frac{\sin 0}{4}\right]$

$∫^{2π}_0 \sin^2θ dθ = \frac{π}{2}-\frac{0}{4}$

$∫^{2π}_0 sin^2θ dθ = \frac{π}{2}$

Therefore, the integration of sin square theta from 0 to π is π/2.