Question

What is the antiderivative of ${{\sin }^{2}}x$?

Answer 1

Hint: This question needs the knowledge of trigonometry and integration. The trigonometric identity should be known to solve the problem. The trigonometric identity used here to solve the question is $2{{\sin }^{2}}x=1-\cos 2x$. Further the problem is solved by integrating. So the formulas used in this problem are $\int{1dx}=x$ and $\int{\cos axdx=\dfrac{\sin ax}{a}}$.

Complete step by step solution:
To solve this question, the first step is to use the trigonometric identity. We would convert the square of the trigonometric function which is ${{\sin }^{2}}x$ in terms of trigonometric function, which has power $1$.
Here too the trigonometric function will change to such a function which has power $1$, so the trigonometric identity which will be used here is $2{{\sin }^{2}}x=1-\cos 2x$ , which on further solving becomes ${{\sin }^{2}}x=\dfrac{1-\cos 2x}{2}$ .
Antiderivative of ${{\sin }^{2}}x$ could be mathematically represented as,$\int{{{\sin }^{2}}x}dx$
We will have to substitute the ${{\sin }^{2}}x$, so on substituting the value of ${{\sin }^{2}}x$ with $\dfrac{1-\cos 2x}{2}$ , we get:
$\int{{{\sin }^{2}}xdx=\int{\left( \dfrac{1-\cos 2x}{2} \right)}}dx$
On solving the integration part wise , so as to make calculation much easier we get:
$\Rightarrow \int{\dfrac{1}{2}dx-\int{\dfrac{\cos 2x}{2}}}dx$
The constant is put outside the integral to make the calculation easier, On putting the constant out of the integral and thus integrating just the function, we get
$\Rightarrow \dfrac{1}{2}\int{dx-\dfrac{1}{2}\int{\cos 2x}}$
Firstly integrating the first function which is $1$ with respect to $dx$ , we use formula $\int{{{x}^{n}}=\dfrac{{{x}^{n+1}}}{n+1}}$ , applying the same in the given function, where $n=0$ , we get
$\Rightarrow \dfrac{1}{2}\int{dx=}\dfrac{1}{2}\dfrac{{{x}^{0+1}}}{0+1}$
$\Rightarrow \dfrac{x}{2}$
Now integrating the second function which is $\cos 2x$ , the formula used will be $\int{\cos ax=\dfrac{\sin ax}{a}}$ ,thus applying the same for the integration of $\cos 2x$, we get:
$\dfrac{1}{2}\int{\cos 2x=\dfrac{1}{2}\dfrac{\sin 2x}{2}}$
$\Rightarrow \dfrac{1}{2}\dfrac{\sin 2x}{2}$
Thus on adding the result of the two integration we get
$\int{\dfrac{1}{2}dx-\int{\dfrac{\cos 2x}{2}}}dx=\dfrac{x}{2}-\dfrac{\sin 2x}{2}+c$
$\therefore $ Antiderivative of ${{\sin }^{2}}x$ is $\dfrac{x}{2}-\dfrac{\sin 2x}{2}+c$.

Note: Whenever we are asked to solve a big integration question, we can solve it on breaking it part-wise, as done in the above question this makes the question easy to solve and also error free. Integration of a function should always give a certain constant as shown here, $\int{f\left( x \right)=F\left( x \right)+c}$. In definite integral the constant $c$ plays an important role.