Integral of Sin x

Before going to find the integral of sin x, let us recall what is integral. An integral is also known as the antiderivative. Antiderivative, as its name suggests, is found by using the reverse process of differentiation. i.e.,

• Finding f'(x) from f(x) is differentiation.
• Finding f(x) from f'(x) is integration.

Thus, the integration of sin x is found by using differentiation. Let us see more about the integral of sin x along with its formula, proof, and graphical intuition.

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

• '∫' represents the "integral"
• sin x is the integrand
• dx is always associated with any integral and it means the small difference in the angle x.

But how to solve the integration of sin x? We can do this in various methods such as using the derivatives and using the substitution method. We will see the proof of integral of sin x in each of these methods in the upcoming sections.

Integral of Sin x Formula

The integral of sin x is cos x. i.e.,

• ∫ sin x dx = -cos x + C, where C is the constant of integration.

We know that integration is the reverse process of differentiation. So to find the integral of sin x, we have to see by differentiating what function would give us sin x. i.e., we have to see

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

Let us recall the formulas of differentiation and search for some formula that gives us sin x as the derivative. Yes, we found one formula which says d/dx (cos x) = -sin x. Oops! This is giving -sin x but we wanted it to be sin x. No worries! Let us make it what we wanted by multiplying both sides by -1. Then we get

-d/dx (cos x) = - (-sin x) (or)

d/dx (-cos x ) = sin x

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

∫ sin x dx = -cos x + C

Hence proved.

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

dy = cos x dx

By trigonometric identities, cos x = √1 - sin²x. Then the above equation becomes,

dy = √1 - sin²x dx

dy = √1 - y² dx

dy / √1 - y² = dx

Multiplying both sides by sin x,

(sin x dy) / √1 - y² = sin x dx

Again substitute sin x = y on the left side.

(y dy) / √1 - y² = sin x dx

Integrating on both sides,

∫ (y dy) / √1 - y² = ∫ sin 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 = ∫ sin x dx

(-1/2) ∫ u^-1/2 du = ∫ sin 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 = ∫ sin x dx

-u^1/2 + C = ∫ sin x dx

Substituting u = 1 - y² back here,

-(1 - y²)^1/2 + C = ∫ sin x dx

Substitute y = sin x back here,

-(1 - sin²x)^1/2 + C = ∫ sin x dx

-(cos²x)^1/2 + C = ∫ sin x dx

-cos x + C = ∫ sin x dx

Hence proved.

We know that ∫ sin x dx = -cos x + C. So by the fundamental theorem of calculus, ∫ₐ^ᵇ sin x dx = [-cos b + C] - [-cos a + C] = -cos b + C + cos a - C = -cos b + cos a. Thus, the integration constant C does NOT play any role while calculating the definitegral of sin x. Let us see some examples.

Integral of Sin x From 0 to π

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

Thus, the integral of sin x from 0 to π is 2.

Integral of Sin x From 0 to π/2

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

Thus, the integral of sin x from 0 to π/2 is 1.

In general, the integral of a function within an interval is the amount of area occupied by the graph of the function within that particular interval. Let us now graph the function f(x) = sin x and calculate the approximate area under the curve for some intervals by using basic geometric formulas. Also, let us calculate the exact areas by using the definite integration of sin x within the same intervals. Let us compare both the results.

Let us calculate the areas under the curve in each of the intervals [0, π/2] and [0, π]. To calculate these areas approximately, we drew the triangles. Of course, this doesn't give the exact area. From the figure, the area calculated like this 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 (of course, the numbers in the second column are less than that of the third column as the triangles are NOT covering the entire area). Thus, the integral of sin x is -cos x and is geometrically proved.

Topics Related to Integral of Sin x:

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

• Calculus
• Calculus Calculator
• Integral Calculator
• Derivative Calculator

What is the Integral of Sin x?

The integral of sin x is -cos x + C. It is mathematically written as ∫ sin x dx = -cos x + C. Here, C is the integration constant.

Is Integral of Sin x Equal to Cos x?

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

How to Find Integral of Sin x?

We know that d/dx (cos x) = - sin x. Multiply both sides by -1, we get d/dx(-cos x) = sin x. Therefore, the integral of sin x, as it is the anti-derivative of sin x, is -cos x + C.

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

We know that ∫ sin x dx = -cos x. If we apply the limits 0 and 2π, we get -cos 2π - (-cos 0) = -1 + 1 = 0.

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

One can easily prove that the derivative of -cos x is sin x. Since integral is nothing but anti-derivative, the integral of sin x is -cos x (of course, we add the integration constant C to this).

What is the Anti-derivative of Sin x?

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

What is the Integral of Sin 3x?

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