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

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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.

1.What is the Integral of sin 2x dx?
2.Definite Integral of sin 2x
3.What is the Integral of sin^2x dx?
4.Definite Integral of sin^2x
5.FAQs on Integral of sin 2x and sin^2x

What is the Integral of Sin 2x dx?


The integral of sin 2xis 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 integral of sin 2x formula.

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Definite Integral of Sin 2x


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)

= (-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)

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

= 0

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

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What is the Integral of Sin^2x dx?


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 formulaBy using the integration by parts

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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: Integral 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.


Definite Integral of Sin^2x


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 = \(\left. \right|_0^{2\pi}\)

= <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 = \(\left. \right|_0^{\pi}\)

= <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.

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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

Topics Related to Integral of Sin2x and Integral of Sin 2x:


Example 2: Evaluate the integral ∫ sin2x cos2x dx.

Solution:

By double angle formula of sin, 2 sin A cos A = sin 2A. Using this,

∫ sin2x cos2x dx = (1/4) ∫ (2 sin x cos x)2 dx

= (1/4) ∫ sin2(2x) dx

We have sin2x = (1 - cos 2x)/2. From this, sin22x = (1 - cos 4x)/2. So the above integral becomes

= (1/4) ∫ (1 - cos 4x)/2 dx

= (1/8) ∫ 1 dx - (1/8) ∫ cos 4x dx

To solve the second integral, we assume 4x = u, from which 4dx = du (or) dx = du/4. Then we get