Question

How do you find the integral of \[{\sin ^2}x\]

Answer 1

Hint: Applying the integration directly to the function it may be complicated. So, to solve this question we are simplifying the trigonometry ratio by using the properties and identities and then we are going to simplify and then we are applying the integration to the function.

Complete step by step solution:
The integral means we have to apply the integration to the given function. In the integral we have two kinds of definite and indefinite integral. This question is of the form indefinite integral where the limits points are not given. The function here is a trigonometric function. Here in this question we have found the integral of \[{\sin ^2}x\] . Let us consider
 \[I = \int {{{\sin }^2}x} dx\] - (1)
We can’t apply integration to \[{\sin ^2}x\] , it may turn out to be complicated. So first we simplify the \[{\sin ^2}x\] by trigonometry identity.
As we known that \[\cos 2x = 1 - 2{\sin ^2}x\]
Take cos2x to RHS and \[{\sin ^2}x\] to LHS, so we have
 \[ \Rightarrow 2{\sin ^2}x = 1 - \cos 2x\]
Divide the above equation by 2 we have
 \[ \Rightarrow {\sin ^2}x = \dfrac{{1 - \cos 2x}}{2}\] - (2)
Substitute the equation (2) in the equation (1) we have
 \[ \Rightarrow I = \int {\left( {\dfrac{{1 - \cos 2x}}{2}} \right)} dx\]
The above integral is written as
 \[ \Rightarrow I = \int {\left( {\dfrac{1}{2} - \dfrac{{\cos 2x}}{2}} \right)} dx\]
Apply the integral to each term we have
 \[ \Rightarrow I = \int {\dfrac{1}{2}} dx - \int {\dfrac{{\cos 2x}}{2}dx} \]
Let we take the constant terms outside the integral and it is written as
 \[ \Rightarrow I = \dfrac{1}{2}\int {dx} - \dfrac{1}{2}\int {\cos 2xdx} \]
Applying the integration, we have
 \[ \Rightarrow I = \dfrac{1}{2}x - \dfrac{1}{2}\dfrac{{\sin 2x}}{2} + c\]
On simplification we have
 \[ \Rightarrow I = \dfrac{x}{2} - \dfrac{{\sin 2x}}{4} + c\]
Hence we have found the integral.
So, the correct answer is “$\dfrac{x}{2} - \dfrac{{\sin 2x}}{4} + c$”.

Note: We have two types of integrals one is definite integral and another is indefinite integral. The definite integral where the limit points of the integration are mentioned and whereas in the indefinite integral the limits points of integration is not mentioned. If the function is a trigonometry ratio then we can use trigonometry identities and properties we can solve these types of questions.