Question

The value of the integral $\int{\sqrt{\left( {{\sin }^{2}}x \right)}dx}$ is
(A) $-\cos x+c$
(B) $\cos x+c$
(C) $-\cos x\times \text{sgn} \left( \sin x \right)+c$
(D) None of These

Answer 1

Hint: We solve this question by first solving the function inside the integral, $\sqrt{\left( {{\sin }^{2}}x \right)}$, using the formula, $\sqrt{{{x}^{2}}}=\left| x \right|$. Then we consider the integral and solve in the integral when $\sin x>0$ and $\sin x<0$ using the formula $\int{\sin xdx}=-\cos x+c$. Then we multiply and divide the obtained value with $\left| \sin x \right|$. Then we simplify it and use the function $\text{sgn} x=\dfrac{\left| x \right|}{x}$ and simplify it to get the result and then combine the both results in the both cases if they are the same.

Complete step by step answer:
We are given the integral $\int{\sqrt{\left( {{\sin }^{2}}x \right)}dx}$.
First let us consider the function inside the integral, that is $\sqrt{\left( {{\sin }^{2}}x \right)}$.
Now let us consider the property of square roots.
$\sqrt{{{x}^{2}}}=\left\{ \begin{matrix}
   x\ \ \ \ if\ x>0 \\
   -x\ \ \ \ if\ x<0 \\
\end{matrix} \right.$
We can also write it as
$\sqrt{{{x}^{2}}}=\left| x \right|$
So, using the above property we can write our given function as,
$\sqrt{\left( {{\sin }^{2}}x \right)}=\left| \sin x \right|$
As we need to find the value of the integral $\int{\sqrt{\left( {{\sin }^{2}}x \right)}dx}$, let us substitute the above value in the integral. Then we get,
$\Rightarrow \int{\sqrt{\left( {{\sin }^{2}}x \right)}dx}=\int{\left| \sin x \right|dx}$
Now let us consider the cases when $\sin x>0$ and when $\sin x<0$. Then we can write it as,
$\begin{align}
  & \Rightarrow \int{\sqrt{\left( {{\sin }^{2}}x \right)}dx}=\left\{ \begin{matrix}
   \int{\sin xdx}\ \ \ \ if\ \ \sin x>0 \\
   \int{-\sin xdx}\ \ \ \ if\ \ \sin x<0 \\
\end{matrix} \right. \\
 & \Rightarrow \int{\sqrt{\left( {{\sin }^{2}}x \right)}dx}=\left\{ \begin{matrix}
   \int{\sin xdx}\ \ \ \ if\ \ \sin x>0 \\
   -\int{\sin xdx}\ \ \ \ if\ \ \sin x<0 \\
\end{matrix} \right. \\
\end{align}$
Now let us consider the formula for integration,
$\Rightarrow \int{\sin xdx}=-\cos x+c$
Using this formula, we can write our integral as,
$\begin{align}
  & \Rightarrow \int{\sqrt{\left( {{\sin }^{2}}x \right)}dx}=\left\{ \begin{matrix}
   -\cos x+c\ \ \ \ if\ \ \sin x>0 \\
   -\left( -\cos x+c \right)\ \ \ \ if\ \ \sin x<0 \\
\end{matrix} \right. \\
 & \Rightarrow \int{\sqrt{\left( {{\sin }^{2}}x \right)}dx}=\left\{ \begin{matrix}
   -\cos x+c\ \ \ \ if\ \ \sin x>0 \\
   \cos x+c\ \ \ \ if\ \ \sin x<0 \\
\end{matrix} \right. \\
\end{align}$
Here the sign of c is not changed because it is a constant so it can be positive or negative both so we can write it in the same way.
Now let us divide and multiply with $\left| \sin x \right|$. Then we get,
$\Rightarrow \int{\sqrt{\left( {{\sin }^{2}}x \right)}dx}=\left\{ \begin{matrix}
   -\cos x\times \dfrac{\left| \sin x \right|}{\left| \sin x \right|}+c\ \ \ \ if\ \ \sin x>0 \\
   \cos x\times \dfrac{\left| \sin x \right|}{\left| \sin x \right|}+c\ \ \ \ if\ \ \sin x<0 \\
\end{matrix} \right.$
Now let us use the formula of $\left| \sin x \right|$ discussed above. Then we get,
$\begin{align}
  & \Rightarrow \int{\sqrt{\left( {{\sin }^{2}}x \right)}dx}=\left\{ \begin{matrix}
   -\cos x\times \dfrac{\left| \sin x \right|}{\sin x}+c\ \ \ \ if\ \ \sin x>0 \\
   \cos x\times \dfrac{\left| \sin x \right|}{-\sin x}+c\ \ \ \ if\ \ \sin x<0 \\
\end{matrix} \right. \\
 & \Rightarrow \int{\sqrt{\left( {{\sin }^{2}}x \right)}dx}=\left\{ \begin{matrix}
   -\cos x\times \dfrac{\left| \sin x \right|}{\sin x}+c\ \ \ \ if\ \ \sin x>0 \\
   -\cos x\times \dfrac{\left| \sin x \right|}{\sin x}+c\ \ \ \ if\ \ \sin x<0 \\
\end{matrix} \right. \\
\end{align}$
As both values are the same now in both the cases.
$\Rightarrow \int{\sqrt{\left( {{\sin }^{2}}x \right)}dx}=-\cos x\times \dfrac{\left| \sin x \right|}{\sin x}+c$
Now let us consider the formula,
$\text{sgn} x=\dfrac{\left| x \right|}{x}$
Using this we can write our value above as,
$\Rightarrow \int{\sqrt{\left( {{\sin }^{2}}x \right)}dx}=-\cos x\times \text{sgn} \left( \sin x \right)+c$

So, the correct answer is “Option C”.

Note: There is a possibility of one making a mistake by taking the function inside the integral as $\sqrt{\left( {{\sin }^{2}}x \right)}=\sin x$. Then the integral becomes,
$\Rightarrow \int{\sqrt{\left( {{\sin }^{2}}x \right)}dx}=\int{\sin xdx}$
Now let us consider the formula,
$\int{\sin xdx}=-\cos x+c$
Using this formula, we get
$\Rightarrow \int{\sqrt{\left( {{\sin }^{2}}x \right)}dx}=-\cos x+c$
So, one might take the answer as Option A.
But here we need to remember that square root of a number is always positive and take the function as $\sqrt{\left( {{\sin }^{2}}x \right)}=\left| \sin x \right|$.