Introduction to integral of cos^5

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 cubic sine. You will also understand how to compute cos^5x integration by using different integration techniques.

What is the integral of cos⁵x?

The integral of cos⁵x is an antiderivative of the sine function which is equal to sinx + sin⁵x/5 – 2sin³x/3 + c. It is also known as the reverse derivative of the cosine function which is a trigonometric identity.

The cosine function is the ratio of the opposite side to the hypotenuse of a triangle which is written as:

cos = adjacent side/hypotenuse

The integration of cos^5(x) is a complex integral in calculus. It involves a higher power of a trigonometric function cos x. This power of cos x makes it's integral more complex than solving other integral problems. 

Integral of cos⁵x formula

The formula of the integral of cos⁵x contains the integral sign, coefficient of integration, and the function as sine. It is denoted by ∫(cos⁵x)dx. In mathematical form, the integral of cos⁵x is:

$\int (\cos^5 x)dx = \sin x+ \frac{\sin^5x}{5} – 2\frac{\sin^3x}{3} +c$

Where c is any constant involved, dx is the coefficient of integration and ∫ is the symbol of the integral. In the above formula, we can reduce the power of cos^5x to calculate the integration of cos^4x.

How to calculate the integral of cos^5x?

The cos^5x integration 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. Trigonometric formulas
2. Substitution method
3. Definite integral

Integral of cos^5x by using Trigonometric formulas

The substitution method also involves trigonometric formulas that help to solve integrals easily. Let’s understand how to calculate the integral cos^5 by using different trigonometric formulas.

Proof of integral cos^5x by using Trigonometric formulas

To prove the integral of cos^5, we use different trigonometric formulas. Therefore, suppose that

$I = \cos^5x = \cos^3 x.\cos^2 x$

Applying the integral we get,

$I = \int (\cos^3x.\cos^2x)dx$

Since,

$\cos^2x = 1 – \sin^2x$

$I = \int (\cos^3x)(1 – \sin^2x)dx$

Now,

$I = \int[\cos^3x –\cos^3x.\sin^2x]dx$

Moreover,

$I = \int [\cos x.\cos^2x –\cos^3x.\sin^2x]dx$

$I = \int [\cos x.(1 – \sin^2x) –\cos^3x.\sin^2x]dx$

Again,

$I = \int [\cos x – \cos x.\sin^2x – \cos^3x.\sin^2x]dx$

Now separating integrals,

$I = \int \cos xdx – \int \cos x.\sin^2xdx –\int \cos^3x.\sin^2xdx$

Suppose that sin x = t and dt = cos x.dx, then

$I =\int \cos xdx – \int t^2.dt – \int t^2 \cos^3 xdx$

Again,

$I = \sin x – \frac{t^3}{3} – \int (1 – \sin^2x)\cos x.t^2 dx$

The integral will become after substituting the value of t,

$I = \sin x – \frac{t^3}{3} – \int (1 – t^2).t^2dt$

Integrating,

$I = \sin x – \frac{t^3}{3} – \frac{t^3}{3} + \frac{t^5}{5}$

Substituting the value of t,

$I = \sin x – \frac{2\sin^3}{3} + \frac{sin^5x}{5} + c$

Hence we have verified the cos^5x integration by using trigonometric formulas. Using a trig-substitution calculator online may help you in solving integrals more easily than by hand.

Integral of cos^5x 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 squared by using the substitution method.

Proof of Integral of cos^5x by using substitution method

To prove the integral of cos^5x by using the substitution method calculator, suppose that:

$I = \int (\cos^5x)dx$

Suppose that we can write the above integral as:

$I = \int (\cos^4x.\cos x)dx$

By using trigonometric identities, we can write the above equation by using cos²x = 1 – sin2x, therefore,

$I = \int (1 - \sin^2x)2\cos x dx$

Now suppose that,

u = sin x and du = cos x dx

$I = \int (1 – u^2)^2 du$

Now to simplify more,

$I = \int (1 + u^4 – 2u^2)du$

Now integrating each term with respect to u,

$I = u + \frac{u^5}{5} – \frac{2u^3}{3} + c$

Now substituting the value of u back here,

$I = \sin x + \frac{\sin^5x}{5} – \frac{2\sin^3x}{3} + c$

Hence the integration of cos^5x is verified by using substitution method.

Integral of cos^5x 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:

$\int^b_af(x) dx = F(b) – F(a)$

Let’s understand the verification of the integral of cos⁵x by using a definite integral calculator.

Proof of integral of cos^5x by using definite integral

To compute the integral of cos⁵x by using a definite integral, we can use the interval from 0 to π/2 or –π/2 to 0. Let’s compute the cos^5x integration from 0 to π/2.

The idefinite integral of cos^5x can be written as:

$\int^{\frac{\pi}{2}}_0 \cos^5x dx = \sin x – \frac{2\sin^3 x}{3} + \frac{\sin^5x}{5}|^\frac{\pi}{2}_0$

Substituting the value of limit we get,

$\int^{\frac{\pi}{2}}_0 \cos^5x dx = [\sin \frac{\pi}{2} – \frac{2\sin^3 \frac{\pi}{2}}{3} + \frac{\sin^5 \frac{\pi}{2}}{5}] – [\sin 0 – \frac{2\sin^30}{3} + \frac{\sin^50}{5}]$

Therefore, the integral of cos5 x from 0 to 2π is

$\int^{\frac{\pi}{2}}_0 \cos^5x dx= 1 – \frac{2}{3} + \frac{1}{5} – (0)$

$\int^{\frac{\pi}{2}}_0 cos^5x dx = 1 – \frac{2}{3} + \frac{1}{5}$

$\int^{\frac{\pi}{2}}_0 \cos^5x dx = \frac{4}{15}$

Which is the calculation of the definite integral of cos^5x. Now to calculate the integrate cos^5x between the interval 0 to π, we just have to replace π/2 by π. Therefore,

$\int^0_{-\frac{π}{2}} \cos^5 x dx = \sin x – \frac{2\sin^3x}{3} + \frac{\sin^5x}{5}|^0_{\frac{\pi}{2}}$

$\int^0_{-\frac{π}{2}} \cos^5 x dx = \sin 0 – \frac{2\sin^3 0}{3} + \frac{\sin^5 0}{5} ] – [\sin\frac {-\pi}{2} – \frac{2\cos^3 \frac {-\pi}{2}}{3} + \frac{\sin^5\frac{-π}{2}}{5}]$

So,

$\int^0_{-\frac{π}{2}} \cos^5 x dx = - \left(- 1 + \frac{2}{3} – \frac{1}{5}\right) = -\frac{24}{15}$

Therefore, the cos^5x integration from -π/2 to 0 is -24/15. Similarly, to calculate the integral of cos^4(2x), we can use the same method as discussed above.