Question

Evaluate the integral of \[\int\limits_{\text{0}}^{{{400\pi }}} {\sqrt {{\text{1 - cos2x}}} } {\text{ dx}}\]
a) $200\sqrt 2 $
b) $400\sqrt 2 $
c) $800\sqrt 2 $
d) None

Answer 1

Hint: In this type of problem you need to find the value of a given expression by using integration and differentiation methods. The key point in these questions is to find the integral of a given question by using integration and differentiation methods.
In this question, we have to evaluate the question to find the value of the required integral term. For that, we are going to solve that given integral term by using the integration and differentiation method. And also we are going to substitute the value of the limits to find the value of the integral term. Here given below there is a complete step-by-step solution.

Formula used: ${\text{log x = }}\dfrac{{\text{1}}}{{\text{x}}}$ where ${\text{x}}$ is a constant.
$\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{uv = u }}\dfrac{{{\text{dv}}}}{{{\text{dx}}}}{\text{ + v }}\dfrac{{{\text{du}}}}{{{\text{dx}}}}$
${\text{u dv = uv - }}\int {{\text{v du}}} $

Complete step-by-step solution:
Here it is given that integral of \[\int\limits_{\text{0}}^{{{400\pi }}} {\sqrt {{\text{1 - cos2x}}} } {\text{ dx}}\]. We have to find the integral of the given expression.
By using integration and differentiation methods we are going to find the value of given expression.
Now, consider given as
${\text{I = }}$\[\int\limits_{\text{0}}^{{{400\pi }}} {\sqrt {{\text{1 - cos2x}}} } {\text{ dx}}\]
Here limit value of $0$ to $\pi$
$\Rightarrow$${\text{I = }}$\[400\int\limits_{\text{0}}^{{\pi }} {\sqrt {{\text{1 - cos2x}}} } {\text{ dx}}\]
Next substitute the trigonometric values,
$\Rightarrow$${\text{I = }}$\[400\int\limits_{\text{0}}^{{\pi }} {\sqrt {2{{\sin }^2}{\text{x}}} } {\text{ dx}}\]
Now separate the square root,
$\Rightarrow$${\text{I = }}$\[400\int\limits_{\text{0}}^{{\pi }} {\sqrt 2 \sqrt {{{\sin }^2}{\text{x}}} } {\text{ dx}}\]
Here square of the term and square root be cancelled,
$\Rightarrow$${\text{I = }}$\[400 \times \sqrt 2 \int\limits_{\text{0}}^{{\pi }} {\sqrt {{{\sin }^2}{\text{x}}} } {\text{ dx}}\]
Hence we get,
$\Rightarrow$${\text{I = }}$\[400 \times \sqrt 2 \int\limits_0^{{\pi }} {\sin {\text{x}}} {\text{ dx}}\]
Integral of term is,
$\Rightarrow$${\text{I = }}$\[400 \times \sqrt 2 \left[ { - \cos {\text{x}}} \right]_0^{{\pi }}{\text{ dx}}\]
Substitute the value of limit,
$\Rightarrow$${\text{I = }}$\[ - 400 \times \sqrt 2 \left[ {\cos \pi - \cos 0} \right]{\text{ dx}}\]
Here substitute the value of terms,
$\Rightarrow$${\text{I = }}$\[ - 400 \times \sqrt 2 \left[ { - 1 - 1} \right]{\text{ dx}}\]
Subtracting the terms we get,
$\Rightarrow$${\text{I = }}$\[ - 400 \times \sqrt 2 \left[ { - 2} \right]{\text{ dx}}\]
Hence the final solution is,
$\Rightarrow$${\text{I = }}$\[800\sqrt 2 {\text{ dx}}\]

$\therefore $ The option (C) is the correct answer.

Note: A function $f$ is said to be periodic if, for some nonzero constant $P$, it is the case that $f\left( {x + P} \right) = f\left( x \right)$ for the values of $x$ in the domain. A nonzero constant $P$ for which this is the case is called a period of the function. If there exists a least positive constant $P$ with this property, it is called the fundamental period (also primitive period, basic period or prime period). Often the period of a function is used to mean its fundamental period. A function with period $P$ will repeat on intervals of length $P$, and these intervals are sometimes also referred to as periods of the function.