Question

Evaluate $\int\limits_0^{400\pi } {\sqrt {1 - \cos 2x} dx} $
(A) $200\sqrt 2 $
(B) $400\sqrt 2 $
(C) $800\sqrt 2 $
(D) None

Answer 1

Hint: Simplify the integrand using the half angle formula. Check whether the function obtained is periodic or not. If the function obtained is periodic, then apply the integration properties and simply the result. Finally apply the limits and find the value of the given integrand.

Formula used: Half angle formula for cosine function:
 $\cos 2x = 1 - 2{\sin ^2}x$
A function $f\left( x \right)$ is said to be periodic, if there exists a positive real number $T$such that $f\left( {x + T} \right) = f\left( x \right)$
Property of sine function: $\sin \left( {\pi - x} \right) = - \sin x$
Property of definite integral:
If $f\left( x \right)$ is a periodic function with period value $T$, then we have$\int\limits_0^{nT} {f\left( x \right)dx} = n\int\limits_0^T {f\left( x \right)dx} $
Integration of sine function is given by $\int {\sin xdx} = - \cos x$
Evaluation of definite integral on a continuous function $f\left( x \right)$ defined on $\left[ {a,b} \right]$
$\int\limits_a^b {f\left( x \right)dx} = \left[ {\phi \left( x \right)} \right]_a^b = \phi \left( b \right) - \phi \left( a \right)$
Value of cosine function:
$\cos \pi = - 1$ and $\cos 0 = 1$

Complete step-by-step solution:
It is given that integral we have,$\int\limits_0^{400\pi } {\sqrt {1 - \cos 2x} dx} ....\left( 1 \right)$
Using the half angle formula, $\cos 2x = 1 - 2{\sin ^2}x$ and we can write it as,
Now, the given integral changes to
$ \Rightarrow \int\limits_0^{400\pi } {\sqrt {1 - \left( {1 - 2{{\sin }^2}x} \right)} dx} $
On splitting the bracket term and we get
$ \Rightarrow \int\limits_0^{400\pi } {\sqrt {1 - 1 + 2{{\sin }^2}x} dx} $
On subtract the term and we get,
$ \Rightarrow \int\limits_0^{400\pi } {\sqrt {2{{\sin }^2}x} dx} $
Taking the square term out we get,
\[ \Rightarrow \int\limits_0^{400\pi } {\sqrt 2 \left| {\sin x} \right|dx} \]
Now$\sqrt 2 $, being a constant can be taken out of the integral sign.
\[ \Rightarrow \sqrt 2 \int\limits_0^{400\pi } {\left| {\sin x} \right|dx} ....\left( 2 \right)\]
Now we have to check the given function $\left| {\sin x} \right|$ is a periodic function or it is not a periodic function.
We know that if $f\left( x \right)$ is said to be periodic, if there exists a positive real number $T$ such that $f\left( {x + T} \right) = f\left( x \right)$
Here, we can write it as, $f\left( x \right) = \left| {\sin x} \right|$
Since sine function has the property,$\sin \left( {\pi - x} \right) = - \sin x$
So,$\left| {\sin \left( {\pi - x} \right)} \right| = \left| {\sin x} \right|$
Thus, $\left| {\sin x} \right|$ is a periodic function with period $\pi $.
Now using the following property of definite integrals on a given integral.
If $f\left( x \right)$ is a periodic function with period value $T$, then we have
$ \Rightarrow \int\limits_0^{nT} {f\left( x \right)dx} = n\int\limits_0^T {f\left( x \right)dx} $
Here $f\left( x \right) = \left| {\sin x} \right|$ is a periodic function with period $\pi $.
So, the integral \[\left( 2 \right)\] becomes
\[ \Rightarrow \sqrt 2 \times 400\int\limits_0^\pi {\left| {\sin x} \right|dx} \]
On rewritten as
\[ \Rightarrow 400\sqrt 2 \int\limits_0^\pi {\sin xdx} \]
Use the formula $\int {\sin xdx} = - \cos x$ in the above equation and apply the limits.
\[ \Rightarrow 400\sqrt 2 \left[ { - \cos x} \right]_0^\pi \]
Again we use the property $\int\limits_a^b {f\left( x \right)dx} = \left[ {\phi \left( x \right)} \right]_a^b = \phi \left( b \right) - \phi \left( a \right)$ in above equation where$f\left( x \right) = \cos x$, $a = 0$ and $b = \pi $.
The integral will become
\[ \Rightarrow - 400\sqrt 2 \left[ {\cos \pi - \cos 0} \right]\]
Put the value of $\cos \pi = - 1$ and $\cos 0 = 1$ in the above integral and find the value of integrand.
\[ \Rightarrow - 400\sqrt 2 \left[ { - 1 - 1} \right]\]
On adding the bracket term, we get
\[ \Rightarrow - 400\sqrt 2 \left[ { - 2} \right]\]
Let us multiply the term and we get,
\[ \Rightarrow 800\sqrt 2 \]
Thus, $\int\limits_0^{400\pi } {\sqrt {1 - \cos 2x} dx} = 800\sqrt 2 $

Hence the correct option is (C).

Note: In the property $\int\limits_a^b {f\left( x \right)dx} = \left[ {\phi \left( x \right)} \right]_a^b = \phi \left( b \right) - \phi \left( a \right)$, it does not matter which anti-derivative is used to evaluate the definite integral, because if $\int {f\left( x \right)dx} = \phi \left( x \right) + C$, then $\int\limits_a^b {f\left( x \right)dx} = \left[ {\phi \left( x \right) + C} \right]_a^b = \left[ {\phi \left( b \right) + C} \right] - \left[ {\phi \left( a \right) + C} \right] = \phi \left( b \right) - \phi \left( a \right)$
In other words, we have to evaluate the definite integral there is no need to keep the constant value of integration.