Question

Find the integral \[\int\limits_{-{}^{\pi }/{}_{2}}^{{}^{\pi }/{}_{2}}{\sqrt{\dfrac{1}{2}\left( 1-\cos 2x \right)}}dx=\]
A. \[0\]
B. \[2\]
C. \[\dfrac{1}{2}\]
D. None of these

Answer 1

Hint: In this question, we are to find the given integral. The given integral is in the interval $[-1,1]$. So, we can go for an even or an odd function integral. By applying $x=-x$ in the given function, we can find that the function is an even or odd function. According to the type of function, we can evaluate the integral.



Formula Used:Definite integral:
Consider a function $f(x)$ is defined on $[a,b]$. If the integral of this function, $\int{f(x)dx=F(x)}$, then $F(b)-F(a)$ is called the definite integral of the function $f(x)$ over $[a,b]$.
I.e.,$\int\limits_{a}^{b}{f(x)dx}=\left. F(x) \right|_{a}^{b}=F(b)-F(a)$
Here $a$ (lower limit) and $b$(upper limit).
\[\int\limits_{a}^{b}{f(x)dx}=\int\limits_{a}^{b}{f(t)dt}\]
Some of the properties of the definite integrals are:
1) Interchanging the limits: \[\int\limits_{a}^{b}{f(x)dx=-}\int\limits_{b}^{a}{f(x)dx}\]
2) If $a3) $\int\limits_{0}^{a}{f(x)dx}=\int\limits_{0}^{a}{f(a-x)dx}$
4) $\int\limits_{-a}^{a}{f(x)dx}=2\int\limits_{0}^{a}{f(x)dx}$ if $f(x)$ is an even function
$\int\limits_{-a}^{a}{f(x)dx}=0$ if $f(x)$ is an odd function
5) $\begin{align}
  & \int\limits_{0}^{2a}{f(x)dx}=2\int\limits_{0}^{a}{f(x)dx};\text{if }f(2a-x)=f(x) \\
 & \text{ }=0\text{ if }f(2a-x)=-f(x) \\
\end{align}$
6) $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{f(\sin x)}{f(\sin x)+f(\cos x)}dx=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{f(\cos x)}{f(\sin x)+f(\cos x)}dx=\dfrac{\pi }{4}}}$
7) $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{f(\tan x)}{f(\tan x)+f(\cot x)}dx=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{f(\cot x)}{f(\tan x)+f(\cot x)}dx=\dfrac{\pi }{4}}}$



Complete step by step solution:Given integral is
\[I=\int\limits_{-{}^{\pi }/{}_{2}}^{{}^{\pi }/{}_{2}}{\sqrt{\dfrac{1}{2}\left( 1-\cos 2x \right)}}dx\]
Consider the function in the given interval as
\[f(x)=\sqrt{\dfrac{1}{2}\left( 1-\cos 2x \right)}\]
On substituting \[x=-x\] in the function $f(x)$, we get
\[\begin{align}
  & f(-x)=\sqrt{\dfrac{1}{2}\left( 1-\cos 2(-x) \right)} \\
 & \text{ }=\sqrt{\dfrac{1}{2}\left( 1-\cos 2x \right)} \\
 & \text{ }=f(x) \\
\end{align}\]
Since $f(-x)=f(x)$, the function in the given integral is an even function.
Thus, we can use the formula,
$\int\limits_{-a}^{a}{f(x)dx}=2\int\limits_{0}^{a}{f(x)dx}$
Then,
\[\begin{align}
  & I=\int\limits_{-{}^{\pi }/{}_{2}}^{{}^{\pi }/{}_{2}}{\sqrt{\dfrac{1}{2}\left( 1-\cos 2x \right)}}dx \\
 & \text{ }=2\int\limits_{0}^{{}^{\pi }/{}_{2}}{\sqrt{\dfrac{1}{2}\left( 1-\cos 2x \right)}}dx \\
\end{align}\]
Since we know that $(1-\cos 2x)=2{{\sin }^{2}}x$, the integral become
$\begin{align}
  & I=2\int\limits_{0}^{{}^{\pi }/{}_{2}}{\sqrt{\dfrac{1}{2}\left( 2{{\sin }^{2}}x \right)}}dx \\
 & \text{ }=2\int\limits_{0}^{{}^{\pi }/{}_{2}}{\sqrt{{{\sin }^{2}}x}}dx \\
 & \text{ }=2\int\limits_{0}^{{}^{\pi }/{}_{2}}{\sin x}dx \\
 & \text{ }=2\left[ -\cos x \right]_{0}^{{}^{\pi }/{}_{2}} \\
\end{align}$
On applying the limits, we get
\[\begin{align}
  & I=2\left[ -\cos \dfrac{\pi }{2}+\cos 0 \right] \\
 & \text{ }=2[0+1] \\
 & \text{ }=2 \\
\end{align}\]
Thus, \[I=\int\limits_{-{}^{\pi }/{}_{2}}^{{}^{\pi }/{}_{2}}{\sqrt{\dfrac{1}{2}\left( 1-\cos 2x \right)}}dx=2\]



Option ‘B’ is correct


Note: In this question, we have a trigonometric function in the integral. So, it is a little difficult to solve with a normal ILET integration method. Since the given integral has limits in the interval of $[-a, a]$ type, we can use the predefined formulae, where the type of the function plays the role. So, in order to know the type of the function, we need to substitute $x=-x$ in the function $f(x)$. Then, according to the type of the function i.e., either it is an even or odd function, we can evaluate the integral of the given function within the interval.