Introduction to integral of sinx^3

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 calculator 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 sin's integral by using different integration techniques.

What is the integral of sin(x³)?

The integral of sin(x^3) is an antiderivative of sine function which is done by using Taylor’s series expansion. 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 sinx^3 formula

The formula of integral of sin contains integral sign, coefficient of integration and the function as sine. It is denoted by ∫(sin x³)dx. In mathematical form, the integral of sinx^3 is:

$∫\sin x^3dx=\frac{x^4}{4}+\frac{x^{10}}{10×3!}-\frac{x^{16}}{16×5!}+ ...+C$

Where c is any constant involved, dx is the coefficient of integration and ∫ is the symbol of integral. The integral of sin(x2) can be calculated by replacing x by x^2 in the Taylor's series expansion.

How to calculate the integration of sin x cube?

The integral of sin(x3) 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. Taylor's Series
2. Definite integral

Integral of sin x3 by using Taylor’s Series

Taylor’s series is an infinite sum of terms that are expressed in terms of a function’s derivative. It can be used to calculate derivative of a function that is complex to solve. Since sin(x^3) is impossible to integrate by using formal integration. Therefore, we will use Taylor’s series to find the integral of sin(x^3).

Proof of sinx^3 integral by using Taylor’s Series

Since we know that the integration is the reverse of the derivative. Therefore, we can calculate the integration of sinx^3 by using Taylor’s series. For this, we have to first assume the sine series that is,

$\sin x^3=x^3–\frac{x^9}{3!}+\frac{x^{15}}{5!}–\frac{x^{21}}{7!}+…$

We can use the above series in the integral of sin x to calculate the integral of sin^2x cos^2x. Then,

$I=∫\sin x^3dx$

Substituting the series of sinx, we get,

$I=∫\left[x^3–\frac{(x^3)^3}{3!}+\frac{(x^3)^5}{5!}–\frac{(x^3)^7}{7!}+ …\right]dx$

Now we can easily integrate these terms to get the integral sin(x^3). Therefore,

$∫\sin x^3dx=\frac{x^4}{4}+\frac{x^{10}}{10×3!}-\frac{x^{16}}{16×5!}+...+C$

Hence the above equation is the integration of sin x3 by using Taylor’s series.

Integral of (sinx)^3 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 formula can be written as:

$∫^b_a f(x) dx = F(b) – F(a)$

Let’s understand the verification of the integral of sin x^3 by using the indefinite integral.

Proof of integration of sin x cube by using definite integral

To integrate sin(x^3) 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:

$∫^π_0 \sin x^3dx=\left|\frac{x^4}{4}+\frac{x^{10}}{10×3!}-\frac{x^{14}}{16×5!}+ ...\right|^π_0$

Now, substituting the limit in the given function.

$∫^π_0 \sin x^3dx=\frac{π^4}{4}+\frac{π^{10}}{10×3!}-\frac{π^{16}}{16×5!}-0$

The remaining terms are:

$∫^π_0 \sin x^3dx=\frac{π^4}{4}+\frac{π^{10}}{10×3!}-\frac{π^{16}}{16×5!}+ ...$

Which is the calculation of the definite integral of sinx^3. Now to calculate the integration of sinx^3 between the interval 0 to π/2, we just have to replace π by π/2. Therefore,

$∫^{\frac{π}{2}}_0 \sin x^3dx=\left|\frac{x^4}{4}+\frac{x^{10}}{10×3!}-\frac{x^{16}}{16×5!} +...\right|^{\frac{\pi}{2}}_0$

$∫^{\frac{π}{2}}_0 \sin x^3dx=\frac{π^4}{24×4}+\frac{π^{10}}{210×10×3!}-\frac{π^{16}}{216×16×5!}+ ...– 0$

The remaining terms are:

$∫^{\frac{π}{2}}_0 \sin x^3dx=\frac{π^4}{24×4}+\frac{π^{10}}{210×10×3!}-\frac{π^{16}}{216×16×5!}+...$

Hence it is the calculation of integral of sin(x^3) by using definite integral. The above formula is also used to evaluate the integral of sin^2x.

FAQ's

How to find the antiderivative (sin(x))^3?

The antiderivaitve or the integral of sin x cube can be calculated by using different integration techniques such as Taylors series and definite integral. The integral of sin(x)^3 is written as:

$∫\sin x^3dx=\frac{x^4}{4}+\frac{x^{10}}{10×3!}-\frac{x^{16}}{16×5!}+ ...+C$