Table of Contents

**What Is The Integral of Sin²x?** It is like finding the inverse of a derivative. In this sense, integrations are anti-derivatives. Integration is adding parts to find the whole. The whole pizza is integrated, whereas the slices are the individual functions that can be integrated. Suppose f(x) is any function and f′(x) is its derivative. The integration of f′(x) with respect to dx is given by

$$∫ f′(x) dx = f(x) + C.$$

Integrals can be expressed in two ways. Integrals of an indefinite function are integrals of a function when there is no limit to the integration. They contain an arbitrary constant. Definite Integrals: An integral of a function with limits of integration. The interval of integration has two values. The lower limit is one and the upper limit is the other. There is no constant integration.

**Integral of sin²x**

Integrating sin²(x) is not an option, so we have to change it into another form, which we can easily do by using trig identities. Integration of sin square will be based on the concept of cos double angles.

$$sin²(x)$$

$$cos(2x) = 1 – 2sin²(x)$$ [From cos double angle Trigonometric identities]
$$sin²(x) = (1/2)(1 – cos(2x)$$

Integrate on both the sides

$$∫sin²(x)dx = ∫(1/2)(1 – cos(2x) dx$$

$$=(1/2) × ∫(1 – cos(2x)) dx$$

$$= 1/2 × (x – 1/2sin(2x)) + C$$

Thus, $$∫sin2(x) dx = (1/2)x – (1/4)sin(2x) + C$$

**Hence,the final integral of sin2x is $$(1/2)x – (1/4)sin(2x) + C$$**

**Step By Step Guide To Find Integral of sin²x**

$$∫sin²(x)dx$$

• Use Pythagorean Identities: $$sin²x=1/2−cos2x/2$$

$$∫(1/2−cos2x/2)dx$$

• Use Sum Rule: $$∫f(x)+g(x)dx=∫f(x)dx+∫g(x)dx.$$

$$∫(1/2)dx−∫(cos2x/2)dx$$

• Use this rule: $$∫adx=ax+C$$

$$x/2−∫(cos2x/2)dx$$

• Use Constant Factor Rule: $$∫cf(x)dx=c∫f(x)dx.$$

$$x/2−1/2∫(cos2x)dx$$

Let u=2x, du=2dx, then , dx= (1/2) du

• Using u and du above, rewrite$$∫cos2xdx$$.

$$∫(cosu)/2du$$

• Use Constant Factor Rule: $$∫cf(x)dx=c∫f(x)dx.$$

$$1/2∫cosudu$$

• Use Trigonometric Integration: the integral of cosu is sinu.

$$(sinu)/2$$

• Substitute u=2x back into the original integral.

$$sin2x/2$$

• The integral should be rewritten after the substitution has been completed.

$$x/2−sin2x/4$$

• Add constant.

$$x/2−sin2x/4+C$$

**How do you integrate sin2x?**

So, now we have to integrate sin 2x

$$∫sin 2x dx = ½ ∫2 × sin(2x) dx (i)$$

Let us assume u = 2x. Then, du = 2dx.

We know that $$∫sin x = − cos x + C$$

Hence on substituting, equation (i) becomes

$$∫sin 2x dx = ½ ∫sin(u) du$$

$$∫sin 2x dx = ½ (– cos u du) + C$$

$$∫sin 2x dx = −½ cos(2x) + C$$

Thus, $$∫sin 2xdx = −½ cos (2x) + C$$

**What can sin 2x equal?**

The Sin2x formula is another double angle formula. With this formula, we can find the sine of an angle whose value has been doubled. We know that sin is one of the primary trigonometric ratios that define the length of the opposite side of an angle to that of the hypotenuse in a right-angled triangle. With the use of basic trigonometric formulas, there are many formulas associated with sin 2x. The range of the sin function is [-1, 1], so the range of sin2x is also [-1, 1].

Sin 2x is the double angle identity for sine in trigonometry. Angular relationships are studied in trigonometry by comparing angles and sides of right-angled triangles. There are two simple formulas for sin 2x:

$$sin(2x) = 2 sin x cos x$$ (in terms of sin and cos)

$$sin(2x) = (2tan x)/(1 + tan2x)$$ (in terms of tan)

These are the main formulas of sin 2x. But we can write this formula in terms of sin x (or) cos x alone using the trigonometric identity sin2x + cos2x = 1. They are

$$sin 2x = 2 √(1 – cos2x) cos x$$ (sin 2x formula in terms of cos)

$$sin 2x = 2 sin x √(1 – sin2x)$$ (sin 2x formula in terms of sin)

**Why is sin 2pi 0?**

To find the value of sin of 2pi, let us first recall the sine function for different standard angles from the trigonometric table. sin 0 = 0, sin π/6 = 1/2, sin π/4 = √2/2, sin π/3 = √3/2, and sin π/2 = 1. This table does not contain the value of sin 2pi. Various methods will be used here to find that sin of 2pi is 0. A few examples will also be given to illustrate how this works.

Sin of 2pi is 0. i.e., sin 2π = 0. Using the trigonometric table, we can find the trigonometric ratios of the standard angles 0, π/6, π/4, π/3, and π/2. This table does not give us the value of sin of 2pi. Trigonometric ratios of nonstandard angles are most commonly found by using the reference angles and quadrant where the angle lies. 2pi can also be found this way. Several other methods can be used to find the value of sin of 2pi, including

- Using double angle formula
- Using reference angle
- Using unit circle

We will prove that sin 2π = 0 in each of these methods.

The sin of 2pi is 0.

**The sin of 2pi Using Double Angle Formula**

We can find the value of sin of 2pi using the double angle sine formula, which is sin 2x = 2 sin x cos x. Because we must determine the value of sin(2π), we should substitute x = π in the above formula. It then gives us:

sin 2π = 2 sin π cos π -(1)

Since π is also a non-standard angle, we find the values of sin π and cos π using the sum and difference formulas. Then we get

sin π = sin (π/2 + π/2) = sin π/2 cos π/2 + cos π/2 sin π/2 = (1)(0) + (0)(1) = 0

cos π = cos (π/2 + π/2) = cos π/2 cos π/2 – sin π/2 sin π/2 = (0)(0) – (1)(1) = -1

Substitute these values in (1),

sin 2π = 2 (0) (-1) = 0

Hence, sin of 2pi = 0.

**Sin of 2pi Using Reference Angles**

When we convert 2π into degrees, we get 360°. 360° lies in the interval [0°, 360°], so its coterminal angle itself is the reference angle. If we subtract 360° from it, we find the coterminal angle. If we divide 360° by 360°, we get 0°. Therefore, the coterminal angle of 360° is 0°. Also, 360° means one full rotation, so it comes either in the first quadrant or the fourth quadrant. So, we’ll take both cases into consideration.

- First Quadrant: We know that in the first quadrant, sin is positive.

Then sin 360° = + sin 0° = 0 (because sin 0° = 0) - Fourth Quadrant: We know that in the fourth quadrant, sin is negative.

Then sin 360° = – sin 0° = 0 (because sin 0° = 0)

From both the cases, sin 360° = sin 2π = 0.

Hence, sin 2π = 0.

**The sin of 2pi Using Unit Circle**

Let us recall a few points about the unit circle before finding the value of sin of 2pi.

- A unit circle is a circle of radius centered at its origin.
- A point on the unit circle corresponds to an angle.
- The angle is formed by the line joining the origin and the point with the positive direction of the x-axis in an anticlockwise direction.
- If P(x, y) corresponds to some angle θ, then x = cos θ and y = sin θ. In other words, the sine of the angle represents the y-coordinate of the point.

A circle of 0° on the unit circle consists of 2π, which is just 360° and represents one full rotation, so it is nothing but the angle made by the x-axis and itself. The unit circle’s point (1, 0) is also on the x-axis, so we know that 0° corresponds to the same point. Thus,

sin 2π = sin 0° = y-coordinate of (1, 0) = 0.

Hence, sin 2π = 0.

**Integral Of Sin^2X Cos^2X:**

To solve this integral, we can use the double-angle identity for sine: sin^2(x) = (1/2)(1 – cos(2x)). By substituting this identity into the integral, we get:

∫sin^2(x)cos^2(x) dx = ∫(1/2)(1 – cos(2x))cos^2(x) dx.

Expanding the expression, we have:

∫(1/2)(cos^2(x) – cos^2(x)cos(2x)) dx.

Now, we can split this integral into two separate integrals:

∫(1/2)cos^2(x) dx – ∫(1/2)cos^2(x)cos(2x) dx.

The first integral, ∫(1/2)cos^2(x) dx, can be easily evaluated using a trigonometric identity:

∫(1/2)cos^2(x) dx = (1/2)∫(1 + cos(2x)) dx = (1/2)(x/2 + (1/4)sin(2x)) + C1,

where C1 is the constant of integration.

For the second integral, ∫(1/2)cos^2(x)cos(2x) dx, we can use the product-to-sum identity for cosine:

∫(1/2)cos^2(x)cos(2x) dx = (1/4)∫cos(2x) + cos(4x) dx.

Using the same process as above, we can evaluate this integral:

(1/4)(1/2)(1/2)sin(2x) + (1/4)(1/4)(1/4)sin(4x) + C2,

where C2 is the constant of integration.

Therefore, the overall solution to the integral ∫sin^2(x)cos^2(x) dx is:

(1/2)(x/2 + (1/4)sin(2x)) – (1/8)sin(2x) – (1/64)sin(4x) + C,

where C = C1 + C2 is the constant of integration.

**Integral Of Sin(2X):**

The integral of sin(2x) can be found by using a simple substitution. Let u = 2x, then du = 2dx. Rearranging this equation, we get dx = (1/2)du. Substituting these values into the integral, we have:

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

The integral of sin(u) with respect to u is -cos(u), so we have:

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

where C is the constant of integration.

**Integral Of Sin 2 2X:**

It seems that there might be a typo or ambiguity in the notation “Sin 2 2X.” If you intended to write sin^2(2x), please refer to the solution for question 6. Otherwise, if you meant sin(2 * 2x), we can proceed as follows:

sin(2 * 2x) = sin(4x), which represents a periodic function with a period of 2π. Therefore, when integrating sin(4x) over a specific interval, the result will depend on the limits of integration.

**Integral of sin^2x:**

The integral of sin^2x represents the definite integral of the function sin^2x with respect to x. It can be denoted as ∫sin^2x dx. This integral is a fundamental problem in calculus and often appears in various applications.

To solve the integral of sin^2x, we can use the trigonometric identity: sin^2x = (1 – cos2x)/2. Applying this identity, we have:

∫sin^2x dx = ∫(1 – cos2x)/2 dx

Splitting the integral into two parts, we can rewrite it as:

∫(1/2 – cos2x/2) dx

Now, we can integrate each term separately:

∫(1/2) dx – ∫(cos2x/2) dx

Integrating, we get:

(1/2)x – (1/4)sin2x + C

where C is the constant of integration.

**Integral of 1/sin 2x:**

The integral of 1/sin 2x represents the indefinite integral of the function 1/sin 2x with respect to x. It can be denoted as ∫(1/sin 2x) dx.

To solve this integral, we can use the reciprocal identity for sine: 1/sin θ = csc θ. Applying this identity, we have:

∫(1/sin 2x) dx = ∫csc 2x dx

Using a trigonometric identity, we can express csc 2x as 1/sin 2x = 1/(2sin x cos x).

Now, we can rewrite the integral as:

∫(1/(2sin x cos x)) dx

Applying a trigonometric substitution, let’s set u = sin x, and du = cos x dx. The integral becomes:

∫(1/(2u(1 – u^2))) du

By decomposing the integrand into partial fractions, we can express it as:

(1/2)∫(1/(u(1 – u)(1 + u))) du

Using partial fraction decomposition, we get:

(1/2)∫(A/u + B/(1 – u) + C/(1 + u)) du

Solving for A, B, and C, we find that A = 1/2, B = -1/2, and C = 1/2.

Substituting these values back into the integral, we have:

(1/2)∫(1/(2u) – 1/(2(1 – u)) + 1/(2(1 + u))) du

Integrating each term, we get:

(1/2)(1/2)ln|u| – (1/2)ln|1 – u| + (1/2)ln|1 + u| + C

Substituting back u = sin x, we have:

(1/4)ln|sin x| – (1/2)ln|1 – sin x| + (1/2)ln|1 + sin x| + C

where C is the constant of integration.

**Integral of sin^2(2x):**

The integral of sin^2(2x) represents the definite integral of the function sin^2(2x) with respect to x. It can be denoted as ∫sin^2(2x) dx.

Using the double-angle formula for sine, sin^2(2x) can be expressed as (1 – cos(

4x))/2. Therefore, the integral becomes:

∫(1 – cos(4x))/2 dx

Splitting the integral into two parts, we have:

(1/2)∫dx – (1/2)∫cos(4x) dx

Integrating, we get:

(1/2)x – (1/8)sin(4x) + C

where C is the constant of integration.

**Integral of sin^2x from 0 to 2π:**

The integral of sin^2x from 0 to 2π represents the definite integral of sin^2x over the interval [0, 2π]. It can be denoted as ∫[0, 2π]sin^2x dx.

To solve this integral, we can use the formula derived earlier: sin^2x = (1 – cos2x)/2. Substituting this into the integral, we have:

∫[0, 2π](1 – cos2x)/2 dx

Splitting the integral and integrating each term separately, we get:

(1/2)∫[0, 2π]dx – (1/2)∫[0, 2π]cos2x dx

The first integral evaluates to (1/2)(2π – 0) = π.

For the second integral, we can use the periodicity of the cosine function to simplify the evaluation over the interval [0, 2π]:

(1/2)∫[0, 2π]cos2x dx = (1/4)∫[0, 2π]cos(4x) dx

Integrating, we get:

(1/4)(1/4)sin(4x)|[0, 2π] = 0

Therefore, the integral of sin^2x from 0 to 2π simplifies to π.

**Integral of sin 2x proof:**

To prove the integral of sin 2x, we need to evaluate the definite integral of sin 2x over a specific interval.

Using the double-angle formula for sine, sin 2x can be expressed as 2sin x cos x. Therefore, the integral becomes:

∫2sin x cos x dx

Using the trigonometric identity for the product of sines, we have:

∫sin 2x dx = -∫sin^2 x d(cos x)

Integrating the right-hand side, we get:

-∫sin^2 x d(cos x) = -sin^2 x/2 + C

where C is the constant of integration.

Therefore, the integral of sin 2x is equal to -sin^2 x/2 + C.

**Integrate sin^2x from 0 to π:**

To evaluate the integral of sin^2x from 0 to π, we can use the formula sin^2x = (1 – cos2x)/2.

The integral becomes:

∫[0, π](1 – cos2x)/2 dx

Splitting the integral and integrating each term separately, we get:

(1/2)∫[0, π]dx – (1/2)∫[0, π]cos2x dx

The first integral evaluates to (1/2)(π – 0) = π/2.

For the second integral, we can use the periodicity of the cosine function to simplify the evaluation over the interval [0, π]:

(1/2)∫[0, π]cos2x dx = (1/4)∫[0, π]

cos(4x) dx

Integrating, we get:

(1/4)(1/4)sin(4x)|[0, π] = 0

Therefore, the integral of sin^2x from 0 to π simplifies to π/2.

**Integral of sinx:**

The integral of sinx represents the indefinite integral of the function sinx with respect to x. It can be denoted as ∫sinx dx.

Integrating sinx, we get:

∫sinx dx = -cosx + C

where C is the constant of integration.

**Integration by parts sin^2x:**

To integrate sin^2x using the integration by parts method, we can express sin^2x as (1 – cos2x)/2. Let’s use u = sinx and dv = sinx dx.

Differentiating u, we get du = cosx dx, and integrating dv, we get v = -cosx.

Applying the integration by parts formula, ∫u dv = uv – ∫v du, we have:

∫sin^2x dx = ∫(1 – cos2x)/2 dx

= (1/2)∫dx – (1/2)∫cos2x dx

= (1/2)x – (1/4)sin2x + C

where C is the constant of integration.

**Integral of cos^2x:**

The integral of cos^2x represents the indefinite integral of the function cos^2x with respect to x. It can be denoted as ∫cos^2x dx.

Using the identity cos^2x = (1 + cos2x)/2, the integral becomes:

∫(1 + cos2x)/2 dx

Splitting the integral and integrating each term separately, we get:

(1/2)∫dx + (1/2)∫cos2x dx

The first integral evaluates to (1/2)x + C1, where C1 is the constant of integration.

For the second integral, we can use the periodicity of the cosine function to simplify the evaluation:

(1/2)∫cos2x dx = (1/4)∫cos(4x) dx

Integrating, we get:

(1/4)(1/4)sin(4x) + C2 = (1/16)sin(4x) + C2

where C2 is the constant of integration.

Therefore, the integral of cos^2x is equal to (1/2)x + (1/16)sin(4x) + C, where C is the constant of integration.

**Integrate sin 2x dx from 0 to π/4:**

To evaluate the definite integral of sin 2x dx from 0 to π/4, we can use the formula sin 2x = 2sin x cos x.

The integral becomes:

∫[0, π/4]2sin x cos x dx

Splitting the integral and integrating each term separately, we get:

2∫[0, π/4]sin x cos x dx

Using the trigonometric identity for the product of sines, we can rewrite it as:

∫[0, π/4]sin 2x/2 dx

Now, integrating sin 2x/2, we get:

(1/2)(-cos 2x)|[0, π/4] = (1/2)(-cos(2π/4) + cos(0)) = (1/2)(-cos(π/2) +1)

Since cos(π/2) = 0, the integral simplifies to:

(1/2)(-0 + 1) = 1/2

Therefore, the definite integral of sin 2x dx from 0 to π/4 is equal to 1/2.

**FAQs**

**What is the cos2x formula?**

The relationship between an angle and its sides is explained by the trigonometric ratios of an angle in a right triangle. So what does cos2x mean? In addition to the cosine 2x or cos 2x formula, there is also the double angle formula or cos 2x formula. The formula has a double angle in it, which is why it’s called a double angle formula. It is driven by these trigonometric functions of the sum and difference of two numbers (angles) and related expressions. Now that we understand what the cos 2x formula is, we can move on to learn more about trigonometry and the formula of cos2x.

Now if you are wondering what the formula of cos2x is, let me tell you that we have 5 cos x formula.

- The trigonometric formula of $$cos2x = Cos²x – Sin²x$$
- The trigonometric formula of $$cos2x = 1 – 2Sin²x$$
- The trigonometric formula of $$cos2x = 2Cos²x – 1$$
- The trigonometric formula of $$cos2x = (1−tan²x)/(1+tan²x)$$
- The trigonometric formula of $$cos2x = (Cos²x−Sin²x)/(Cos²x+Sin²x)$$

**How To Find Integral Of Sin^2X?**

To find the integral of sin^2x, you can use the trigonometric identity sin^2x = (1 – cos2x)/2. This identity allows you to rewrite sin^2x in terms of cos2x, which can be integrated more easily.

Using the identity, the integral becomes:

∫sin^2x dx = ∫(1 – cos2x)/2 dx

Splitting the integral into two parts, we have:

(1/2)∫dx – (1/2)∫cos2x dx

Integrating each term separately, we get:

(1/2)x – (1/4)sin2x + C

where C is the constant of integration.

**How To Find Definite Integral Of Sin^2X?**

To find the definite integral of sin^2x over a specific interval [a, b], you can apply the same approach as finding the indefinite integral. First, find the indefinite integral of sin^2x using the method described above. Then, evaluate the expression at the upper limit (b) and subtract the value at the lower limit (a).

For example, let’s find the definite integral of sin^2x from 0 to π:

∫[0, π]sin^2x dx = [(1/2)x – (1/4)sin2x]_[0, π]

Plugging in the upper limit (π), we get:

(1/2)π – (1/4)sin(2π)

Plugging in the lower limit (0), we get:

(1/2)0 – (1/4)sin(0)

Simplifying, we have:

(1/2)π – 0 = π/2

Therefore, the definite integral of sin^2x from 0 to π is equal to π/2.

**How To Solve Integral Of Sin^2X?**

To solve the integral of sin^2x, you can use the trigonometric identity sin^2x = (1 – cos2x)/2. By rewriting sin^2x in terms of cos2x, the integral becomes easier to evaluate.

Using the identity, the integral can be expressed as:

∫sin^2x dx = ∫(1 – cos2x)/2 dx

Splitting the integral into two parts, we have:

(1/2)∫dx – (1/2)∫cos2x dx

Integrating each term separately, we get:

(1/2)x – (1/4)sin2x + C

where C is the constant of integration.

**What Is The Integral Of Sin 2X?**

The integral of sin 2x represents the indefinite integral of the function sin 2x with respect to x. It can be denoted as ∫sin 2x dx.

To solve this integral, you can use the trigonometric identity sin 2x = 2sin x cos x. By using this identity, you can rewrite the integral as:

∫2sin x cos x dx

Using the product-to-sum identity for sine, sin x cos x can be expressed as (1/2)sin(2x). The integral becomes:

∫2(1/2)sin(2x) dx

Simplifying, we get:

∫sin(2x) dx

Using the substitution u = 2x, du = 2 dx, we can rewrite the integral as:

(1/2)∫sin(u) du

Integrating sin(u), we get:

-(1/2

)cos(u) + C

Substituting back u = 2x, we have:

-(1/2)cos(2x) + C

Therefore, the integral of sin 2x is equal to -(1/2)cos(2x) + C, where C is the constant of integration.

**Why Can’t We Integrate Sin 2X?**

We can indeed integrate sin 2x, as shown in the previous explanation. The integral of sin 2x is -(1/2)cos(2x) + C, where C is the constant of integration.

It is a common misconception that sin 2x cannot be integrated. However, by applying appropriate trigonometric identities or using substitution, we can find the integral of sin 2x.

**What Is The Integration Of 2X?**

The integration of 2x represents the indefinite integral of the function 2x with respect to x. It can be denoted as ∫2x dx.

To solve this integral, we apply the power rule of integration. The power rule states that the integral of x^n with respect to x is (1/(n+1))x^(n+1), where n is any real number except -1.

Applying the power rule to the integral of 2x, we have:

∫2x dx = (2/2)x^2 + C

Simplifying, we get:

∫2x dx = x^2 + C

Therefore, the integral of 2x is x^2 + C, where C is the constant of integration.