Question

What is the anti – derivative of ${{\left( \sin x \right)}^{2}}$?

Answer 1

The anti – derivative is nothing but the integration with respect to the variable of the given function. So, the required answer is the integration of a given function with respect to $'x'$.
Convert the given function into a standard form of trigonometric ratio by removing the square.
Use the standard formula of multiple angle of cosine ratio that is,
$\cos 2x=1-2{{\sin }^{2}}x$
Use the standard formulas of integration as,
$\int{\cos ax.dx}=\dfrac{1}{a}\sin ax+C$
$\int{dx}=x+C$

Complete step-by-step solution:
Let us assume that the given function as,
$\begin{aligned} & \Rightarrow f\left( x \right)={{\left( \sin x \right)}^{2}} \\\ & \Rightarrow f\left( x \right)={{\sin }^{2}}x \\\ \end{aligned}$
We are asked to find the anti – derivative of given function.
We know that the anti – derivative of a function is integration of that function.
So, let us assume that the integral of given function as,
$\Rightarrow I=\int{f\left( x \right).dx}............equation(i)$
We know that the integral of ${{\sin }^{2}}x$ is not defined directly.
So, let us convert the given equation in terms of standard trigonometric ratio.
We know that the standard formula of multiple angle of cosine ratio that is,
$\cos 2x=1-2{{\sin }^{2}}x$
By using the above formula we get,
$\begin{aligned} & \Rightarrow \cos 2x=1-2{{\sin }^{2}}x \\\ & \Rightarrow 2{{\sin }^{2}}x=1-\cos 2x \\\ & \Rightarrow {{\sin }^{2}}x=\dfrac{1-\cos 2x}{2} \\\ \end{aligned}$
Now, we can write the given function as
$\Rightarrow f\left( x \right)=\dfrac{1-\cos 2x}{2}$
Now, let us substitute the above function in equation (i) then we get,
$\Rightarrow I=\int{\left( \dfrac{1-\cos 2x}{2} \right).dx}$
Now, let us separate the above integral to its respective terms then we get,
$\Rightarrow I=\dfrac{1}{2}\int{dx}-\dfrac{1}{2}\int{\cos 2x.dx}$
We know that the standard formulas of integration as,
$\int{\cos ax.dx}=\dfrac{1}{a}\sin ax+C$
$\int{dx}=x+C$
By using the above formulas in the integral then we get,
$\Rightarrow I=\dfrac{1}{2}\left( x+{{C}_{1}} \right)-\dfrac{1}{2}\left( \dfrac{\sin 2x}{2}+{{C}_{2}} \right)$
By adding the terms in above equation using the LCM form we get the integral value as,
$\begin{aligned} & \Rightarrow I=\dfrac{2x-\sin 2x}{4}+\dfrac{{{C}_{1}}-{{C}_{2}}}{2} \\\ & \Rightarrow I=\dfrac{2x-\sin 2x}{4}+C \\\ \end{aligned}$
Where, $C=\dfrac{{{C}_{1}}-{{C}_{2}}}{2}$
Therefore, we can conclude that the anti – derivative of ${{\left( \sin x \right)}^{2}}$ as,
$\therefore \int{{{\left( \sin x \right)}^{2}}.dx}=\dfrac{2x-\sin 2x}{4}+C$

Note: The main mistake can be done in this problem is applying the integration directly without converting to standard ratio.
Some students may solve the given problem directly applying chain rule of integration as,
$\begin{aligned} & \Rightarrow I=\int{{{\left( \sin x \right)}^{2}}.dx} \\\ & \Rightarrow I=\dfrac{{{\left( \sin x \right)}^{2+1}}}{2+1}\times \dfrac{1}{\dfrac{d}{dx}\left( \sin x \right)} \\\ & \Rightarrow I=\dfrac{{{\sin }^{3}}x}{3\cos x} \\\ \end{aligned}$
This is a completely wrong solution because there is no such thing as chain rule as we have in differentiation.
So, first convert the given function into standard trigonometric ratio and then apply the integration.