Question

If \[\int\limits_{0}^{50\pi }{\sqrt{1-\cos 2x}dx}\], then the value of \[\lambda \]is
A.$50\sqrt{2}$
B. 100
C. 0
D. None of these

Answer 1

Hint: Replace \[\left( 1-\cos 2x \right)\] by using the identity $''\cos 2x=1-2{{\sin }^{2}}x''$ in I then we use the formula that $''\int\limits_{0}^{nT}{f\left( x \right)dx=n\times \int\limits_{0}^{T}{f\left( x \right)dx''}}$ for further integrating. Where ‘T’ is the period of $f\left( x \right)$. Draw the graph of $y=\left| \sin x \right|$ to get the fundamental period of $\left| \sin x \right|$.

Complete step-by-step answer:


$I=\int\limits_{0}^{50\pi }{\sqrt{1-\cos x}dx}$

We know,

$\begin{align}

  & \cos 2x=1-2{{\sin }^{2}}x \\

 & \Rightarrow 2{{\sin }^{2}}x=1-\cos 2x \\

\end{align}$

Replacing $\left( 1-\cos 2x \right)$ with $2{{\sin }^{2}}x$ in I, we will get,

$\begin{align}

  & I=\int\limits_{0}^{50\pi }{\sqrt{2{{\sin }^{2}}x}dx} \\

 & \Rightarrow I=\int\limits_{0}^{50\pi }{\sqrt{2}\left| \sin x \right|dx} \\

\end{align}$

We know that $\int{cf\left( x \right)dx=c\int{f\left( x \right)dx}}$ if ‘c’ is any constant.

Here$'\sqrt{2}'$ is constant, so,

$I=\sqrt{2}\int\limits_{0}^{50\pi }{\left| \sin x \right|dx}$

Graph of $y=\left| \sin x \right|$

From the graph above, we can see that $y=\left| \sin x \right|$ is a periodic function with period $\pi $.

We know that,

$\int\limits_{0}^{nT}{f\left( x \right)dx=n\times \int\limits_{0}^{T}{f\left( x \right)dx}}$

Where T is the fundamental period of $f\left( x \right)$.

$\Rightarrow \int\limits_{0}^{50\pi }{\left| \sin x \right|dx}=50\times \int\limits_{0}^{\pi }{\left| \sin x \right|dx}$

Graph of $y=\sin x$

In $x\in \left[ 0,\pi \right],\sin x$ is positive. So, $\left| \sin x \right|=\sin x$.

\[\begin{align}

  & \Rightarrow \int\limits_{0}^{50\pi }{\left| \sin x \right|dx}=50\times \int\limits_{0}^{\pi }{\left| \sin x \right|dx} \\

 & =50\times \int\limits_{0}^{\pi }{\left( \sin x \right)dx} \\

 & \int{\sin xdx=-\cos x} \\

 & \Rightarrow \int\limits_{0}^{50\pi }{\left| \sin x \right|dx}=50\times \left[ -\cos x \right]_{0}^{\pi } \\

 & =50\times \left[ -\cos \pi -\left( -\cos 0 \right) \right] \\

 & =50\times \left[ -\cos \pi +\cos 0 \right]\ \ \ \ \left[ As\ \cos \pi =-1\ and\ \cos 0=1 \right] \\

 & =50\times \left[ -\left( -1 \right)+1 \right] \\

 & =50\times \left( 2 \right) \\

 & =100 \\

 & I=\sqrt{2}\int\limits_{0}^{50\pi }{\left| \sin x \right|dx} \\

 & \Rightarrow I=\sqrt{2}\left( 100 \right) \\

 & \Rightarrow I=100\sqrt{2} \\

\end{align}\]

Hence, $I=\int\limits_{0}^{50\pi }{\sqrt{1-\cos x}dx}=100\sqrt{2}$ and option (D) is the correct answer.


Note: Students can make mistakes by writing $\sqrt{2{{\sin }^{2}}x}=\sqrt{2}\sin x$ and not considering the modulus sign. But $\sqrt{2{{\sin }^{2}}x}=\sqrt{2}\left| \sin x \right|$ and fundamental period of $\sin x$ is $2\pi $while fundamental period of $\left| \sin x \right|$ will be $\pi $and the answer will be different.