Integration is the reverse of differentiation. When we talk about the integration of sin²x, we are trying to find the area under the curve of sin²x using calculus.

Sin²x simply means the square of the sine function, which is written as sin x × sin x.

Sine (sin) is one of the basic trigonometric functions. It is used to relate the angles and sides in a right-angled triangle. The sine of an angle is the ratio of the length of the side opposite the angle to the length of the hypotenuse (the longest side of the triangle).

There are six main trigonometric functions: sin, cos, tan, cot, sec, and cosec. These functions are widely used in geometry, trigonometry, and calculus to solve problems involving angles and distances.

The integration of sin²x is useful in many topics of mathematics, especially in solving calculus problems that involve trigonometric expressions.

This Math article will cover topics like integration of sin square x,\( Sin^2 x) function , Integration of Sin square x,formula, How to find the Integration of Sin square x , solved Examples, and FAQs. etc.

Sin^2 x Function 

The domain of a function means all the possible input values (x-values) you can use in it. For the function sin²x, you can plug in any real number, so its domain is from -infinity to infinity.

The range of a function means all the possible output values (y-values) it can give. Since sin x always gives values between -1 and 1, squaring those values makes them positive. So, sin²x will always be between 0 and 1.

That means the range of sin²x is from 0 to 1.

The following is graph of the function\( y= Sin^2 x \),

Definite Integral of sin 2x (∫ sin 2x dx)

A definite integral means we calculate the area under a curve between two specific points (called the lower and upper limits). To solve it, we first find the indefinite integral, then plug in the upper and lower values, and subtract.

1. Integral of sin 2x from 0 to π/2

We know:
∫ sin 2x dx = -½ cos 2x

Now plug in the values:
= -½ [cos(2 × π/2) − cos(2 × 0)]
= -½ [cos(π) − cos(0)]
= -½ [−1 − 1]
= -½ × (−2)
= 1

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

2. Integral of sin 2x from 0 to π

Using the same method:
∫ sin 2x dx from 0 to π
= -½ [cos(2π) − cos(0)]
= -½ [1 − 1]
= -½ × 0
= 0

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

Integration of Sin Square x 

The integral of sin square x is found by various integration methods like integration by using trigonometric identities and by integration parts method.

The formula for the integration of sin square x is represented as:

\(\int sin^{2}x dx = \frac{x}{2}-\frac{sin2x}{4}+c\)

where c is known as the constant integration.

The area under a curve between two points is found by doing a definite integral between the two points. To find the area under the curve y = f(x)=sin^{2}x between x = a & x = b, we integrate f(x)= sin^{2}x between the limits of a and b.

Formula

 The formula for the integral of \( sin^{2}x \) can be represented as 

• The integral of\( sin^{2}x \) is \(\frac{x}{2}-\frac{sin2x}{4}+c \) where C is the constant of integration.

How to Find the Integration of Sin Square x 

The Integration of Sin square x by using trigonometric identities.

\( \int sin^{2}x dx\) .

 Applying the identity\( cos2x=1-2sin^{2}x\) .

\(\int sin^{2}x dx\) 

=\(\int \frac{1-cos2x}{2} dx\) 

=\(\frac{1}{2}\int (1-cos2x) dx\) 

=\(\frac{1}{2}(x-\frac{sin2x}{2})+c\) 

=\(\frac{x}{2}-\frac{sin2x}{4}+c\) 

Therefore,the integration of Sin square x is \(\frac{x}{2}-\frac{sin2x}{4}+c\) .

The Integration of Sin square x by integration parts method

The integral of \( sin^{2}x\) is found by use of the integration by parts formula which is represented as :

\(\int p. q dx= p\int q \ dx - \int (\frac{d}{dx}p\int q\ dx) dx\) —- formula no.01.

The integral of the two functions is taken, considering the left term the first function, and the second term the second function. This method is called the Ilate rule. The first function is chosen in such a way that the derivative of the function can be easily integrated.

Therefore to find the integral of\( sin^{2}x\) we write \(sin^{2}x\) as \(sin x.sin x\). Assume sin x is the first function and sin x is the second function. Therefore, \(p= sin x \) and\( q= sin x\)

Therefore applying the integration by parts formula no.01, we get.

 \(\int sin^{2}x dx= \int (\sin x.\sin x) dx= \sin x\int \sin x dx - \int (\frac{d}{dx}(\ sin x)\int sin x dx \)

 \(\int sin^{2}x dx=\sin x\int \sin x dx - \int (\frac{d}{dx}(\ sin x)\int sin x dx\) 

=\(\sin x(-\cos x)-\int (\cos x(-\cos x))dx \)

=\(\sin x(-\cos x)+\int (\cos x^{2})dx\)

Applying identity \(\sin x^{2}+\cos x^{2}=1\)

=\(-\sin x \cos x+\int (1-\sin x^{2})dx\)

\(\int sin^{2}x dx=-\sin x \cos x+\int 1-\int \sin x^{2})dx\)

\(\int sin^{2}x dx=-\sin x \cosx+x -\int \sin x^{2})dx\)

 \(2\int \sin x^{2})dx=x-\sin x \cosx +c\)

\(\int \sin x^{2})dx=\frac{x}{2} - \frac{(\sin x \cosx)}{2} +c\)

=\(\frac{x}{2} - \frac{(2\sin x \cosx)}{4} +c\)

Applying identity\( sin 2x=2sin x cos x\)

=\(\frac{x}{2} - \frac{sin 2x}{4} +c\)

Therefore, the integration of Sin square x by integration by parts method is \(\frac{x}{2}-\frac{sin2x}{4}+c\).

The integral of sin²x is written as ∫ sin²x dx, and its answer is:

x/2 − (sin 2x)/4 + C

Let’s see how we can get this using two easy methods.

Method 1: Using the Formula of cos 2x

We use a trigonometric identity:

cos 2x = 1 − 2 sin²x

From this, we can write:

sin²x = (1 − cos 2x)/2

Now plug this into the integral:

∫ sin²x dx = ∫ (1 − cos 2x)/2 dx
= (1/2) ∫ (1 − cos 2x) dx
= (1/2) ∫ 1 dx − (1/2) ∫ cos 2x dx
= (1/2)x − (1/2) × (sin 2x / 2) + C
= x/2 − (sin 2x)/4 + C

Method 2: Using Integration by Parts

We know sin²x is the same as sin x × sin x.
So, use integration by parts:

Let
u = sin x → du = cos x dx
dv = sin x dx → v = −cos x

Now apply the formula:
∫ u dv = uv − ∫ v du

So,
∫ sin x · sin x dx = sin x (−cos x) − ∫ (−cos x)(cos x) dx
= −sin x cos x + ∫ cos²x dx

We now use:
cos²x = 1 − sin²x

So,
∫ sin²x dx = −sin x cos x + ∫ (1 − sin²x) dx
= −sin x cos x + ∫ 1 dx − ∫ sin²x dx

Now bring ∫ sin²x dx to one side:

∫ sin²x dx + ∫ sin²x dx = −sin x cos x + x
2 ∫ sin²x dx = x − sin x cos x
So,
∫ sin²x dx = (x − sin x cos x) / 2

Now use 2 sin x cos x = sin 2x:
So sin x cos x = sin 2x / 2

Final answer:
∫ sin²x dx = x/2 − sin 2x / 4 + C

So, both methods give the same result. The integral of sin²x is:

x/2 − (sin 2x)/4 + C

Summary

• The inverse process of differentiation is known as integration
•  The formula for the integral of \( sin^{2}x \) can be represented as 
• The integral of\( sin^{2}x \) is \(\frac{x}{2}-\frac{sin2x}{4}+c \) where C is the constant of integration.
• \(Sin^2 x\) has a real number for every value of x, so the function \(Sin^2 x \)has the domain from -infinity to infinity.
• Since any number's square is always positive, its value will range from 0 to 1. Thus the function\( Sin^2 x\) has a range from 0 to 1.

Solved Example

Example 1: Evaluate the integral \(\int sin^{2}x cos x/).

Solution: Let \(=sinx/)

\(dt=cos x dx/)

\(dx=\frac{dt}{cos x}/)

\(\int sin^{2}x cos x dx/)

=\(\int t^{2} cos x .\frac{dt}{cos x}/)

=\(\int t^{2}{dt}/)

=\(\frac{t^{3}}{3}+c/)

=\(\frac{sin^{3}x}{3}+c/)

Answer: The integral of \( \int sin^{2}x cos x is \(\frac{sin^{3}x}{3}+c/).

Example 2:  Find the integral of ∫ sin(3x) · cos(3x) dx

Solution: We use the identity: 2 sin A cos A = sin 2A

So we can write:
sin(3x) · cos(3x) = (1/2) sin(6x)

∫ sin(3x) cos(3x) dx
= ∫ (1/2) sin(6x) dx
= (1/2) ∫ sin(6x) dx
= (1/2) [ −(1/6) cos(6x) ] + C
= −1/12 cos(6x) + C

Final answer: ∫ sin(3x) cos(3x) dx = −1/12 cos(6x) + C

Example 3: Evaluate the integral \(\int\frac{1}{sin^{2}x cos^{2}x}dx/)

Solution: Applying identity \(sin^{2}x+cos^{2}x =1/)

\(\int\frac{1}{sin^{2}x cos^{2}x}dx/)

=\(\int\frac{sin^{2}x+cos^{2}x}{sin^{2}x cos^{2}x}dx/)

=\(\int\frac{sin^{2}x}{sin^{2}x cos^{2}x}dx+\int\frac{cos^{2}x}{sin^{2}x cos^{2}x}dx/)

=\(\int sec^{2}x +\int cosec^{2}x/)

=\(tanx-cotx+c/)

Answer: The integral of \(\int\frac{1}{sin^{2}x cos^{2}x}dx /) is \(tanx-cotx+c/) .

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