Integration of Sin2x: An introduction
In Mathematics, integration means the summation of given discrete data. Integration is done to find the function, given its derivative. The other use of integration is to calculate the area under the graph of a function. Integration is done to find the areas of the 2D region and the volumes of 3D objects. Integration is the opposite of differentiation. In the indefinite integration, we add the constant of integration. It is added to represent the constant term of the original function because we cannot obtain it through the anti-derivative process. The integral of sin2x and sin2x both is an indefinite integration because the limits are not defined here.
What is the Integration of sin2x?
The integral of sin2x is represented as \[\int {\sin 2xdx} \]. The given integral is calculated using the substitution method.
I = \[\int {\sin 2xdx} \]
Let u = 2x
Now, differentiating both sides, we get
du = 2dx \[ \Rightarrow dx = \dfrac{{du}}{2}\]
Substituting the value of u and du in the integral, we get,
\[\begin{array}{c}I = \int {\sin u{\rm{ }}\dfrac{{du}}{2}} \\ = \dfrac{1}{2}\int {\sin u{\rm{ }}} du\\ = \dfrac{1}{2}\left( { - \cos u} \right) + C\end{array}\]
Now convert the integral obtained in the terms of x,
\[I = \dfrac{{ - \cos 2x}}{2} + C\]
Thus, the integral of sin2x is \[\dfrac{{ - \cos 2x}}{2} + C\].
Integration of sin2x
The integral of sin2x is represented as \[\int {{{\sin }^2}xdx} \]. The given integral is calculated by first applying the formula of cos2x and then using the substitution method, the final answer is calculated.
I = \[\int {{{\sin }^2}xdx} \]
As we know, \[\cos 2x = 1 - 2{\sin ^2}x\].
Now find the value of sin2x from the above equation.
\[{\sin ^2}x = \dfrac{1}{2}\left( {1 - \cos 2x} \right)\]
Now, putting the value of sin2x in the integral, we get,
\[\begin{array}{c}I = \int {{{\sin }^2}xdx} \\ = \int {\dfrac{1}{2}\left( {1 - \cos 2x} \right)dx} \\ = \dfrac{1}{2}\int {\left( {1 - \cos 2x} \right)dx} \end{array}\]
The above integral can be broken into two integrals.
\[\begin{array}{c}I = \dfrac{1}{2}\int {\left( {1 - \cos 2x} \right)dx} \\ = \dfrac{1}{2}\int {dx} - \dfrac{1}{2}\int {\cos 2x{\rm{ }}dx} \\ = {I_1} - {I_2}\end{array}\]
Now integrating both the terms separately, we get
\[\begin{array}{c}{I_1} = \dfrac{1}{2}\int {dx} \\ = \dfrac{x}{2} + {C_1}\end{array}\]
\[{I_2} = \dfrac{1}{2}\int {\cos 2x{\rm{ }}dx} \]
Let u = 2x
Now, differentiating both sides, we get
du = 2dx \[ \Rightarrow dx = \dfrac{{du}}{2}\]
Substituting the value of u and du in the integral, we get
\[\begin{array}{c}{I_2} = \dfrac{1}{2}\int {\cos u{\rm{ }}\dfrac{{du}}{2}} \\ = \dfrac{1}{4}\int {\cos u{\rm{ du}}} \\ = \dfrac{1}{4}\left( {\sin u} \right) + {C_2}\end{array}\]
Now, convert the integral obtained in the terms of x.
\[{I_2} = \dfrac{{\sin 2x}}{4} + {C_2}\]
Now put the value of I1 and I2 in the original integral and find the value of I.
\[\begin{array}{c}I = {I_1} - {I_2}\\ = \dfrac{x}{2} + {C_1} - \dfrac{{\sin 2x}}{4} - {C_2}\\ = \dfrac{x}{2} - \dfrac{{\sin 2x}}{4} + C\end{array}\]
Thus, the integral of sin2x is \[\dfrac{x}{2} - \dfrac{{\sin 2x}}{4} + C\].
Solved Questions
1. How do we use substitution in integration?
Ans: We use a substitution method to change the independent variable from x to t. We assume that x = g(t). And substitute x = g(t) in the original function and then after differentiation, we substitute dx = \[g'(t)\].
2. Integration of an integrable function f(x) gives a family of curves differing by a constant value. True or false.
Ans: True, the integration of an integrable function gives F(x) + C which represents a family of curves as we change the value of constant C.
Practice Questions
1. How to find the definite integral of sin2x from 0 to \[\pi \]?
Ans: 0
2. How to find the definite integral of sin2x from 0 to \[\pi \]?
Ans: \[\dfrac{\pi }{2}\]
Interesting Facts
Integrals and derivatives are a part of a branch of Mathematics known as calculus.
Primitively while integration we used to divide the whole graph into rectangles of small width and added them together. This is known as the Riemann sum.
Integration is known as the anti-derivative.
Many rules are defined in substitution like the power rule, sum and difference rules, exponential rule, reciprocal rule, etc.
The fundamental theorem of calculus relates integration to differentiation.
Key Features
We add an arbitrary constant after integration to represent the constant term of the function that cannot be calculated by the antiderivative process.
The integral of sin2x is \[\dfrac{{ - \cos 2x}}{2} + C\].
The integral of \[sin^2x\] is \[\dfrac{x}{2} - \dfrac{{\sin 2x}}{4} + C\].
FAQs on Integration of sin 2x
1. What is the integral of sin(2x) with respect to x?
The integral of sin(2x) with respect to x is -cos(2x)/2 + C. This result is found using the method of substitution, a core technique in the CBSE Class 12 syllabus. By setting u = 2x, the integral simplifies. The constant 'C' is added because the derivative of any constant is zero, so the antiderivative is a family of functions.
2. How do you find the integral of sin²x?
To integrate sin²x, you cannot use a simple substitution. Instead, you must first apply the trigonometric power-reduction identity sin²x = (1 - cos(2x))/2. This transforms the integrand into a form that is easy to integrate. The final answer after integrating term-by-term is x/2 - sin(2x)/4 + C.
3. What is the fundamental difference between integrating sin(2x) and sin²x?
The key difference lies in the integration technique required:
- For sin(2x), the argument of the sine function is linear (2x), making it ideal for a direct u-substitution.
- For sin²x, the entire function is squared. This requires a trigonometric identity to reduce the power and express it in terms of functions we know how to integrate, like cos(2x).
This highlights a crucial concept in integration: simplifying the integrand is often the first and most important step.
4. Why is a constant of integration, 'C', always added to indefinite integrals?
The constant of integration, 'C', represents all possible constant values that could have been part of the original function. Since the process of differentiation eliminates any constant term (the derivative of x² + 5 is the same as x² + 99, which is 2x), the reverse process of integration must account for this lost information. Therefore, ∫2x dx = x² + C represents an entire family of curves, not just a single one.
5. How does the integration of cos²x compare to that of sin²x?
The method for integrating cos²x is nearly identical to that for sin²x, as both rely on a power-reduction identity. For cos²x, you use the identity cos²x = (1 + cos(2x))/2. Notice the only change is the sign from minus to plus. This leads to a final answer of x/2 + sin(2x)/4 + C, differing only in the sign connecting the two terms.
6. How does a definite integral of sin(2x), for example from 0 to π/4, differ from its indefinite integral?
An indefinite integral, ∫sin(2x)dx, provides a general function or a family of functions: -cos(2x)/2 + C. In contrast, a definite integral, such as ∫₀^(π/4) sin(2x)dx, calculates a specific numerical value. This value represents the net area under the function's curve between the specified limits. For this example, the definite integral evaluates to exactly 1/2.
7. Is it efficient to use integration by parts for ∫sin²x dx?
While technically possible, using integration by parts for ∫sin²x dx is not the recommended or efficient method as per the NCERT curriculum. It involves a more complex process that often leads back to the original integral, requiring further algebraic steps. The standard and most direct approach is to use the trigonometric identity sin²x = (1-cos(2x))/2, which simplifies the problem instantly.
8. What are some real-world examples where integration is applied?
Integration is a fundamental concept with wide-ranging applications in science and engineering. For example:
- In Physics, it is used to calculate the total distance travelled from a velocity function or the work done by a variable force.
- In Electrical Engineering, it helps determine the total charge accumulated over time from a current.
- In Medicine, it can be used to model the concentration of a drug in the bloodstream over time.
