## Introduction to integral of cos(t^2)

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 the cos(t^2) integral by using different integration techniques.

## What is the integral of cos(t^{2})?

The integral of cos(t)^2 is an antiderivative of the cosine function which is done by using Taylor’s series expansion. It is also known as the reverse derivative of the cos(t2) function which is a trigonometric identity. The sine function is the ratio of the opposite side to the hypotenuse of a triangle which is written as:

cos = adjacent side/hypotenuse

The integral of cos (t2) is a common integral in calculus. It contains a trigonometric function cos with an angle t^2. It is helpful in solving many integration problems involving such functions.

### Integral of cos(t^2) formula

The formula of the integral of sin contains the integral sign, coefficient of integration, and the function as sine. It is denoted by ∫(cos t2)dx. In mathematical form, the integral of cos t^2 is:

$\int (\cos t^2)dx = t - \frac{t^5}{5×2!} + \frac{t^9}{9×4!} + \frac{t^{13}}{13×6!} + ....+C$

Where c is any constant involved, dx is the coefficient of integration and ∫ is the symbol of the integral. Replacing the cos(t^2) by cos(x^2) will give the integral of cos(x^2).

## How to calculate the integral of cos(t^{2})?

The integral of cos(t^{2}) is its antiderivative that can be calculated by using different integration techniques. In this article, we will discuss how to calculate integral of cosine by using:

- Taylor’s series expansion
- Definite integral

## Integral of cos t^{2} 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 the derivative of a function that is complex to solve. Since cos(t^{2}) is impossible to integrate by using formal integration. Therefore, we will use Taylor’s series to find the integral of cos(t^{2}).

### Proof of integral of cos t2 by using Taylor’s Series

Since we know that the integration is the reverse of the derivative. Therefore, we can calculate the integral of cos t 2 by using Taylor’s series. For this, we have to first assume the sine series that is,

$\cos t=t – \frac{t^2}{2!} + \frac{t^4}{4!} – \frac{t^6}{6!} + … $

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

$I = \int (\cos t^2)dx$

Substituting the series of cosx, we get,

$I = \int [1 – \frac{(t^2)^4}{2!} + \frac{(t^2)^4}{4!} – \frac{(t^2)^6}{6!} + … ] dx$

Now we can easily integrate these terms to get the integral cos(t^{2}). Therefore,

$\int \cos t^2dx = t - \frac{t^5}{5×2!} + \frac{t^9}{9×4!} + \frac{t^{13}}{13×6!} + ....+ C$

Hence the above equation is the integration of cos(t)^2 by using Taylor’s series. Similarly, the integral of cos(x^3) can also be obtained by using Taylor's series expansion.

## Integral of cos t2 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:

$\int^b_a f(x) dx = F(b) – F(a)$

Let’s understand the verification of the cos(t^2) integral by using the definite integral.

### Proof of integral of cos(t^2) by using definite integral

To compute the integral of cos t 2 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 cos t^2 from 0 to π. For this we can write the integral as:

$\int^\pi_0 \cos t^2 dx = \left|t - \frac{t^5}{5×2!} + \frac{t^9}{9×4!} + \frac{t^{13}}{13×6!} + ...\right|^\pi_0$

Now, substituting the limit in the given function.

$\int^\pi_0 \cos t^2 dx = \left[π - \frac{π^5}{5×2!} + \frac{π^9}{9×4!} + \frac{π^{13}}{13×6!} + ...\right]–\left[0 - \frac{0^5}{5×2!} + \frac{0^9}{9×4!} + \frac{0^{13}}{13×6!} + ..\right]$

The remaining terms are:

$\int^\pi_0 \cos t^2 dx= π - \frac{π^5}{20} + \frac{π^9}{216} + \frac{π^{13}}{9360} + …$

Which is the calculation of the definite integral of cos t2. Now to calculate the integral of cos t2 between the interval 0 to π/2, we just have to replace π by π/2. Therefore,

$\int^{\frac{\pi}{2}}_0 \cos t^2 dx = \left|t - \frac{t^5}{5×2!} + \frac{t^9}{9×4!} + \frac{t^{13}}{13×6!} + ...\right|^{\frac{\pi}{2}}_0$

$\int^{\frac{\pi}{2}}_0 \cos t^2 dx= π - \frac{(π/2)^5}{5×2!} + \frac{(π/2)^9}{9×4!} + \frac{(π/2)^{13}}{13×6!} +… -{ 0}$

The remaining terms are:

$\int^{\frac{\pi}{2}}_0 \cos t^2 dx = π - \frac{π^5}{160×2!} + \frac{π^9}{4608×4!} + \frac{π^{13}}{106496×6!} + …$

Hence it is the calculation of the integral of cos t^{2} by using a definite integral. Since these calculations are tricky and long-term, we offer you to use our definite integral calculator to evaluate this integral.