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# How to integrate $\int {\left( {\dfrac{{\sqrt {1 - \cos x} }}{2}} \right)} dx$ by using u-substitution?

Last updated date: 20th Jun 2024
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Hint:
Here, we will first rationalize the given numerator and try to simplify it further so that we can apply the u-substitution method. Solving this further and using the formula of integration of ${x^n}$, we will be able to find the required answer. Integration is defined as the summation of all the discrete data.

Formula Used:
We will use the following formulas:
1) $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$
2) ${\sin ^2}x + {\cos ^2}x = 1$
3) $\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} + C$

Complete step by step solution:
The given integral function is $\int {\left( {\dfrac{{\sqrt {1 - \cos x} }}{2}} \right)} dx$.
This can also be written as: $\dfrac{1}{2}\int {\sqrt {1 - \cos x} } dx$
Now, we will do the rationalization by multiplying and dividing by $\sqrt {1 + \cos x}$
Hence, we get,
$\dfrac{1}{2}\int {\sqrt {1 - \cos x} } dx = \dfrac{1}{2}\int {\dfrac{{\sqrt {1 - \cos x} \times \sqrt {1 + \cos x} }}{{\sqrt {1 + \cos x} }}} dx$
Using the identity $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$ in the numerator, we get,
$\Rightarrow \dfrac{1}{2}\int {\sqrt {1 - \cos x} } dx = \dfrac{1}{2}\int {\dfrac{{\sqrt {1 - {{\cos }^2}x} }}{{\sqrt {1 + \cos x} }}} dx$
Also, ${\sin ^2}x + {\cos ^2}x = 1$
Or $1 - {\cos ^2}x = {\sin ^2}x$
Thus, substituting this value in the numerator,
$\Rightarrow \dfrac{1}{2}\int {\sqrt {1 - \cos x} } dx = \dfrac{1}{2}\int {\dfrac{{\sqrt {{{\sin }^2}x} }}{{\sqrt {1 + \cos x} }}} dx = \dfrac{1}{2}\int {\dfrac{{\sin x}}{{\sqrt {1 + \cos x} }}} dx$…………………..$\left( 1 \right)$
Now, by u-substitution, let $1 + \cos x = t$
Differentiating both sides, with respect to $x$, we get,
$- \sin xdx = dt$
Or $\sin xdx = - dt$
Substituting this in $\left( 1 \right)$, we get,
$\Rightarrow \dfrac{1}{2}\int {\sqrt {1 - \cos x} } dx = - \dfrac{1}{2}\int {\dfrac{1}{{\sqrt t }}} dt$
This can also be written as:
$\Rightarrow \dfrac{1}{2}\int {\sqrt {1 - \cos x} } dx = - \dfrac{1}{2}\int {{t^{ - \dfrac{1}{2}}}} dt$
We know that, $\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} + C$
Hence, we get,
$\Rightarrow \dfrac{1}{2}\int {\sqrt {1 - \cos x} } dx = - \dfrac{1}{2}\left( {\dfrac{{{t^{ - \dfrac{1}{2} + 1}}}}{{ - \dfrac{1}{2} + 1}}} \right) + C = - \dfrac{1}{2}\left( {\dfrac{{{t^{\dfrac{1}{2}}}}}{{\dfrac{1}{2}}}} \right) + C$
$\Rightarrow \dfrac{1}{2}\int {\sqrt {1 - \cos x} } dx = - 2 \times \dfrac{1}{2}\left( {{t^{\dfrac{1}{2}}}} \right) + C = - \sqrt t + C$
But we know that, $1 + \cos x = t$
Hence,

$\int {\left( {\dfrac{{\sqrt {1 - \cos x} }}{2}} \right)} dx = - \sqrt {1 + \cos x} + C$
Therefore, this is the required answer.

Note:
In calculus, integration by substitution, also known as u-substitution or change of variables, is a method which is used for evaluating integrals or anti-derivatives. It is the counterpart to the chain rule for differentiation, in fact, it can also be considered as doing the chain rule "backwards". Also, there is another method which is known as integration by parts which is a process of finding the integral of a product of functions in terms of the integral of the product of their derivative and antiderivative. But, clearly, in this question, we didn’t have any product of functions thus, we cannot use the by parts method.