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Integral of Sin 2x
The integral of sin 2x and the integral of sin^{2}x have different values.
 ∫ sin 2x dx = (cos 2x) / 2 + C
 ∫ sin^{2}dx = x/2  (sin 2x)/4 + C
To find the integral of sin^{2}x, 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 sin^{2}x 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.  Integration 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 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.
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) [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.
Integration of Sin^2x dx
The integral of sin^{2}x is denoted by ∫ sin^{2}x 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 sin^{2}x, we use the double angle formula of cos. One of the cos 2x formulas is cos 2x = 1  2 sin^{2}x. By solving this for sin^{2}x, we get sin^{2}x = (1  cos 2x) / 2. We use this to find ∫ sin^{2}x dx. Then we get
∫ sin^{2}x 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
∫ sin^{2}x dx = (1/2) x  (1/2) (sin 2x)/2 + C (or)
∫ sin^{2}x 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 sin^{2}x as sin x · sin x. To find the integral of a product, we can use the integration by parts.
∫ sin^{2}x 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
∫ sin^{2}x dx = (1/2) (2 sin x cos x) + ∫ cos^{2}x dx
By the double angle formula of sin, 2 sin x cos x = sin 2x and by a trigonometric identity, cos^{2}x = 1  sin^{2}x. So
∫ sin^{2}x dx = (1/2) sin 2x + ∫ (1  sin^{2}x) dx
∫ sin^{2}x dx = (1/2) sin 2x + ∫ 1 dx  ∫ sin^{2}x dx
∫ sin^{2}x dx + ∫ sin^{2}x dx = (1/2) sin 2x + x + C_{1}
2 ∫ sin^{2}x dx = x  (1/2) sin 2x + C_{1}
∫ sin^{2}x dx = x/2  (sin 2x)/4 + C (where C = C_{1}/2)
Hence proved.
Definite Integral of Sin^2x
To evaluate the definite integral of sin^{2}x, we just substitute the upper and lower bounds in the value of the integral of sin^{2}x and subtract the resultant values. Let us evaluate some definite integrals of integral sin^{2}x dx here.
Integral of Sin^2x From 0 to 2pi
∫\(_0^{2\pi}\) sin^{2}x 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 sin^{2}x from 0 to 2π is π.
Integral of Sin^2x From 0 to pi
∫\(_0^{\pi}\) sin^{2}x 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 sin^{2}x from 0 to π is π/2.
Important Notes Related to Integral of Sin 2x and Integral of Sin^{2}x:
 ∫ sin 2x dx = (cos 2x)/2 + C
 ∫ sin^{2}x dx = x/2  (sin 2x)/4 + C
☛ Related Topics:
Examples on Integration of Sin 2x and Integral of Sin^2x

Example 1: Evaluate the integral ∫ sin 2x e^{cos 2x} dx.
Solution:
We are going to solve this using the integration by substitution.
Let cos 2x = u. Then 2 sin 2x dx = du (or) sin 2x dx = (1/2) du.
Then the given integral becomes,
∫ e^{u} (1/2) du = (1/2) e^{u} + C.
Let us substitute the value of u = cos 2x back here. Then we get
∫ sin 2x e^{cos 2x} dx = (1/2) e^{cos 2x} + C.
Answer: The integral of sin 2x e^cos 2x dx is (1/2) e^{cos 2x} + C.

Example 2: Evaluate the integral ∫ sin^{2}x cos^{2}x dx.
Solution:
By double angle formula of sin, 2 sin A cos A = sin 2A. Using this,
∫ sin^{2}x cos^{2}x dx = (1/4) ∫ (2 sin x cos x)^{2} dx
= (1/4) ∫ sin^{2}(2x) dx
We have sin^{2}x = (1  cos 2x)/2. From this, sin^{2}2x = (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
= (1/8) (x)  (1/8) ∫ cos u (du/4)
= x/8  (1/32) ∫ cos u du
= x/8  (sin 4x) / 32 + C
Answer: The integral of integral sin^2x cos^2x dx is x/8  (sin 4x) / 32 + C.

Example 3: Evaluate the integral ∫ sin^{2}x cos^{3}x dx.
Solution:
The given integral can be written as ∫ sin^{2}x cos^{2}x (cos x dx)
We know that cos^{2}x = 1  sin^{2}x. Then the above integral becomes
∫ sin^{2}x (1  sin^{2}x) (cos x dx) = ∫ (sin^{2}x  sin^{4}x) (cos x dx)
Assume that sin x = u. Then cos x dx = du.
= ∫ (u^{2}  u^{4}) du
= u^{3}/3  u^{5}/5 + C
Substituting u = sin x back,
= sin^{3}x/3  sin^{5}x/5 + C
Answer: The integral of sin^2x cos^3x dx is sin^{3}x/3  sin^{5}x/5 + C.
FAQs on Integral of Sin 2x
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
 ∫ sin^{2}dx = 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 ∫ sin^{2}x dx and ∫ sin^{2}dx = 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?
∫ sin^{3}x dx = ∫ sin^{2}x sin x dx = ∫ (1  cos^{2}x) sin x dx. Let us substitute cos x = u. Then sin x dx = du. Then the above integral becomes, ∫ (1  u^{2}) ( du) = u + u^{3}/3 + C. Substituting u = sin x back here, ∫ sin^{3}x dx = cos x + cos^{3}x/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 ∫ sin^{2}dx = 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.
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