Integral of Cos x

Before going to find the integral of cos x, let us recall what is integral. An integral is nothing but the anti-derivative. Anti-derivative, as its name suggests, can be found by using the reverse process of differentiation. i.e.,

• Differentiation is finding f'(x) from f(x).
• Integration is finding f(x) from f'(x).

Thus, the integration of cos x is found by using differentiation. Let us see more about the integral of cos x along with its formula and proof (in different methods).

The integral of cos x dx is sin x. Mathematically, this is written as ∫ cos x dx = sin x + C, where, C is the integration constant. Here,

• '∫' is a symbol of integration and it is known as the "integral"
• cos x is the integrand
• dx means the integration is with respect to x.

But how to do the integration of cos x? We are going to find the integral of cos x in various methods such as using the derivatives and using the substitution method in the upcoming sections.

The integration is the opposite (reverse process) of differentiation. So to find the integral of cos x, we have to see by differentiating what function would give us cos x. i.e., we have to see

d/dx \(\fbox{?}\) = cos x.

Let us recall the formulas of differentiation and search for some formula that gives us cos x as the derivative. Yes, we found one formula which says d/dx (sin x) = cos x.

Thus, the derivative of sin x is cos x. So the integration of cos x (anti-derivative) must be sin x. We have to add an integration constant after integrating any function. Thus, we have

∫ cos x dx = sin x + C

Hence proved.

We can prove that the integral of cos x to be sin x + C using the substitution method. For this, we assume that y = cos x. Then dy/dx = -sin x (or)

dy = -sin x dx

By trigonometric identities, sin x = √1 - cos²x. Then the above equation becomes,

dy = -√1 - cos²x dx

dy = -√1 - y² dx

-dy / √1 - y² = dx

Multiplying both sides by cos x,

(-cos x dy) / √1 - y² = cos x dx

Again substitute cos x = y on the left side.

(-y dy) / √1 - y² = cos x dx

Integrating on both sides,

∫ (-y dy) / √1 - y² = ∫ cos x dx

Let 1 - y² = u. Then -2y dy = du (or) -y dy = 1/2 du.

Then the above left-hand side integral becomes,

(1/2) ∫ 1/√u du = ∫ cos x dx

(1/2) ∫ u^-1/2 du = ∫ cos x dx

Using one of the integration formulas, ∫ xⁿ dx = (xⁿ⁺¹)/(n+1) + C. So we get

(1/2) (u^1/2 / (1/2)) + C = ∫ cos x dx

u^1/2 + C = ∫ cos x dx

Substituting u = 1 - y² back here,

(1 - y²)^1/2 + C = ∫ cos x dx

Substitute y = cos x back here,

(1 - cos²x)^1/2 + C = ∫ cos x dx

(sin²x)^1/2 + C = ∫ cos x dx

sin x + C = ∫ cos x dx

Hence proved.

We know that ∫ cos x dx = sin x + C. So by the fundamental theorem of calculus, ∫ₐ^ᵇ cos x dx = [sin b + C] - [sin a + C] = sin b + C - sin a - C = sin b - sin a. Notice that the integration constant C is NOT present in the definite integral of cos x. So we always ignore C while calculating the definite integral. Let us see some examples.

• ∫₀^π cos x dx = [sin x]₀^π
  = sin π - (sin 0)
  = 0 - 0
  = 0
• ∫₀^π/2 cos x dx = [sin x]₀^π/2
  = sin π/2 - (sin 0)
  = 1 - 0
  = 1

The integral of a function within an interval is nothing but the area occupied by its graph in that interval. Let us now graph the function f(x) = cos x and calculate the approximate area under the curve for some intervals by using geometric formulas. Also, we will calculate the exact areas by using the definite integral of cos x within the same intervals and compare both results.

Now, we will calculate the areas under the curve of y = cos x in each of the intervals [0, π/2] and [0, π]. To calculate the approximate areas, we drew the triangles. They don't give the exact area. From the figure, it is clear that the area calculated like using the triangles is less than the actual area as the triangle(s) doesn't cover the complete area under the curve.

The numbers in the second column and third column are (approximately) equal (anyway, the numbers in the second column are less than the numbers of the third column as the triangles are NOT covering the entire area). Thus, the integral of cos x is sin x and is geometrically proved.

Topics Related to Integral of Cos x:

Here are some topics that you may be interested in while doing integration of cos x.

• Calculus
• Integral Calculator
• Applications of Integrals
• Definite Integral Calculator
• Derivative Calculator
• Calculus Calculator
• Integral of Sin x

What is the Integral of Cos x?

The integral of cos x is sin x + C. i.e., ∫ cos x dx = sin x + C. Here, C is the integration constant.

Is Integral of Cos x Equal to - Sin x?

No, the integral of cos x is NOT -sin x, in fact, the integration of cos x is sin x + C.

How to Find Integral of Cos x?

We have d/dx (sin x) = cos x. Therefore, the integral of cos x, being the anti-derivative of cos x, is sin x + C.

What is the Integral of Cos x From 0 to 2π?

We know that ∫ cos x dx = sin x. If we apply the limits 0 and 2π, we get sin 2π - (sin 0) = 0 - 0 = 0.

Why is the Integral of Cos x Equal to Sin x?

We are aware that the derivative of sin x is cos x. Since integral is nothing but anti-derivative, the integral of cos x is sin x (of course, we add the integration constant C to this).

What is the Anti-derivative of Cos x?

The anti-derivative of cos x is nothing but the integration of cos x and hence it is equal to sin x + C.

What is the Integral of Cos 3x?

Let 3x = u then 3 dx = du. From this, dx = (1/3) du. So ∫ cos 3x dx = ∫ cos u (1/3) du = (1/3) (sin u) + C. Substituting u = 3x back, we get ∫ cos 3x dx = (1/3) (sin (3x)) + C.