Integral of Sin 2x

The integral of sin 2x and the integral of sin²x have different values.

• ∫ sin 2x dx = -(cos 2x) / 2 + C
• ∫ sin²dx = x/2 - (sin 2x)/4 + C

To find the integral of sin²x, we use the cos 2x formula and the substitution method whereas we use just the substitution method to find the integral of sin 2x. Let us identify the difference between the integral of sin 2x and the integral of sin²x by finding their values using appropriate methods and also we will solve some problems related to these integrals.

The integral of sin 2x is denoted by ∫ sin 2x dx and its value is -(cos 2x) / 2 + C, where 'C' is the integration constant. For proving this, we use the integration by substitution method. For this, we assume that 2x = u. Then 2 dx = du (or) dx = du/2. Substituting these values in the integral ∫ sin 2x dx,

∫ sin 2x dx = ∫ sin u (du/2)

= (1/2) ∫ sin u du

We know that the integral of sin x is -cos x + C. So,

= (1/2) (-cos u) + C

Substituting u = 2x back here,

∫ sin 2x dx = -(cos 2x) / 2 + C

This is the integration of sin 2x formula.

A definite integral is an indefinite integral with some lower and upper bounds. By the fundamental theorem of Calculus, to evaluate a definite integral, we substitute the upper bound and the lower bound in the value of the indefinite integral and then subtract them in the same order. While evaluating a definite integral, we can ignore the integration constant. Let us calculate some definite integrals of integral sin 2x dx here.

Integral of Sin 2x From 0 to pi/2

∫\(_0^{\pi/2}\) sin 2x dx = (-1/2) cos (2x) \(\left. \right|_0^{\pi/2}\)

= (-1/2) [cos 2(π/2) - cos 2(0)]

= (-1/2) (-1 - 1)

= (-1/2) (-2)

= 1

Therefore, the integral of sin 2x from 0 to pi/2 is 1.

Integral of Sin 2x From 0 to pi

∫\(_0^{\pi}\) sin 2x dx = (-1/2) cos (2x) \(\left. \right|_0^{\pi}\)

= (-1/2) [cos 2(π) - cos 2(0)]

= (-1/2) (1 - 1)

= 0

Therefore, the integral of sin 2x from 0 to pi is 0.

The integral of sin²x is denoted by ∫ sin²x dx and its value is (x/2) - (sin 2x)/4 + C. We can prove this in the following two methods.

• By using the cos 2x formula
• By using the integration by parts

Method 1: Integral of Sin^2x Using Double Angle Formula of Cos

To find the integral of sin²x, we use the double angle formula of cos. One of the cos 2x formulas is cos 2x = 1 - 2 sin²x. By solving this for sin²x, we get sin²x = (1 - cos 2x) / 2. We use this to find ∫ sin²x dx. Then we get

∫ sin²x dx = ∫ (1 - cos 2x) / 2 dx

= (1/2) ∫ (1 - cos 2x) dx

= (1/2) ∫ 1 dx - (1/2) ∫ cos 2x dx

We know that ∫ cos 2x dx = (sin 2x)/2 + C. So

∫ sin²x dx = (1/2) x - (1/2) (sin 2x)/2 + C (or)

∫ sin²x dx = x/2 - (sin 2x)/4 + C

This is the integral of sin^2 x formula. Let us prove the same formula in another method.

Method 2: Integration of Sin^2x Using Integration by Parts

We know that we can write sin²x as sin x · sin x. To find the integral of a product, we can use the integration by parts.

∫ sin²x dx = ∫ sin x · sin x dx = ∫ u dv

Here, u = sin x and dv = sin x dx.

Then du = cos x dx and v = -cos x.

By integration by parts formula,

∫ u dv = uv - ∫ v du

∫ sin x · sin x dx = (sin x) (-cos x) - ∫ (-cos x)(cos x) dx

∫ sin²x dx = (-1/2) (2 sin x cos x) + ∫ cos²x dx

By the double angle formula of sin, 2 sin x cos x = sin 2x and by a trigonometric identity, cos²x = 1 - sin²x. So

∫ sin²x dx = (-1/2) sin 2x + ∫ (1 - sin²x) dx

∫ sin²x dx = (-1/2) sin 2x + ∫ 1 dx - ∫ sin²x dx

∫ sin²x dx + ∫ sin²x dx = (-1/2) sin 2x + x + C₁

2 ∫ sin²x dx = x - (1/2) sin 2x + C₁

∫ sin²x dx = x/2 - (sin 2x)/4 + C (where C = C₁/2)

Hence proved.

To evaluate the definite integral of sin²x, we just substitute the upper and lower bounds in the value of the integral of sin²x and subtract the resultant values. Let us evaluate some definite integrals of integral sin²x dx here.

Integral of Sin^2x From 0 to 2pi

∫\(_0^{2\pi}\) sin²x dx = [x/2 - (sin 2x)/4] \(\left. \right|_0^{2\pi}\)

= [2π/2 - (sin 4π)/4] - [0 - (sin 0)/4]

= π - 0/4

= π

Therefore, the integral of sin²x from 0 to 2π is π.

Integral of Sin^2x From 0 to pi

∫\(_0^{\pi}\) sin²x dx = [x/2 - (sin 2x)/4] \(\left. \right|_0^{\pi}\)

= [π/2 - (sin 2π)/4] - [0 - (sin 0)/4]

= π/2 - 0/4

= π/2

Therefore, the integral of sin²x from 0 to π is π/2.

Important Notes Related to Integral of Sin 2x and Integral of Sin²x:

• ∫ sin 2x dx = -(cos 2x)/2 + C
• ∫ sin²x dx = x/2 - (sin 2x)/4 + C

☛ Related Topics:

• Integral Calculator
• Indefinite Integral Calculator
• Definite Integral Calculator

What is the Integration of Sin 2x dx?

The integral of sin 2x dx is written as ∫ sin 2x dx and ∫ sin 2x dx = -(cos 2x)/2 + C, where C is the integration constant.

Is the Integral sin 2x dx Same as Integral sin^2x dx?

No, the values of these two integrals are NOT same. We have

• ∫ sin²dx = x/2 - (sin 2x)/4 + C
• ∫ sin 2x dx = (-cos 2x)/2 + C

What is the Integral of Sin^2x dx?

The integration of sin^2x dx is written as ∫ sin²x dx and ∫ sin²dx = x/2 - (sin 2x)/4 + C. Here, C is the constant of integration.

How to Find the Definite Integral of Sin 2x from 0 to Pi/4?

We know that ∫ sin 2x dx = -(cos 2x)/2. Substituting the limits 0 and π/4, we get (-1/2) cos 2(π/4) - (-1/2) cos 2(0) = (1/2) 0 + (1/2) (1) = 1/2.

What is the Integral of Sin^3x dx?

∫ sin³x dx = ∫ sin²x sin x dx = ∫ (1 - cos²x) sin x dx. Let us substitute cos x = u. Then -sin x dx = du. Then the above integral becomes, ∫ (1 - u²) (- du) = -u + u³/3 + C. Substituting u = sin x back here, ∫ sin³x dx = -cos x + cos³x/3 + C.

What is the Integral of Sin 3x dx?

To find the ∫ sin 3x dx, let that 3x = u. Then 3 dx = du. From this, dx = du/3. Then the above integral becomes, ∫ sin u (1/3) du = (1/3) (-cos u) + C = (-1/3) cos (3x) + C.

How to Find the Definite Integral of sin^2x from 0 to Pi/4?

We know that ∫ sin²dx = x/2 - (sin 2x)/4 + C. Substituting the limits here, we get [π/8 - (sin π/2)/4] - [0 - (sin 0)/4] = π/8 - 1/4.