Question

How do you evaluate the integral $\int{\dfrac{1}{1-\sin x}dx}$ from $0$ to $\dfrac{\pi }{2}$?

Answer 1

Hint: In this problem we need to calculate the definite integral value of the given function in the given range or limits. For this we need to first calculate the indefinite integral value of the given function. In the given function we can observe that the trigonometric ratio $\sin x$ in the denominator. So, we will use a substitution method to solve the indefinite integral value. In this method we are going to use the substitution $u=\tan \left( \dfrac{x}{2} \right)$ and calculate the differentiation of the substitution. Now we will convert the given function into simplified form by using the substitution value and some trigonometric formulas. Now we will simplify the equation and apply the integration formulas to get the value of indefinite integral value. After that we will apply the limits for the calculated value to get the required result.

Complete step by step solution:
Given that, $\int{\dfrac{1}{1-\sin x}dx}$.
Take the substitution $u=\tan \left( \dfrac{x}{2} \right)$.
Differentiating the above value with respect to $x$, then we will get
$\Rightarrow du=\dfrac{1}{2}{{\sec }^{2}}\left( \dfrac{x}{2} \right)dx$
From the trigonometric identity ${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1$, we can write the above value as
$\begin{align}
  & \Rightarrow du=\dfrac{1}{2}\left( 1+{{\tan }^{2}}\dfrac{x}{2} \right)dx \\
 & \Rightarrow du=\dfrac{1}{2}\left( 1+{{u}^{2}} \right)dx \\
\end{align}$
In the trigonometry we have the formula $\sin \theta =\dfrac{2\tan \theta }{1+{{\tan }^{2}}\theta }$. From this formula we can write
$\Rightarrow \sin x=\dfrac{2u}{1+{{u}^{2}}}$
Substituting all the values we have in the given function, then we will get
$\Rightarrow \int{\dfrac{1}{1-\sin x}dx}=\int{\dfrac{1}{1-\dfrac{2u}{1+{{u}^{2}}}}\left( \dfrac{2du}{1+{{u}^{2}}} \right)}$
Simplifying the above equation, then we will get
$\begin{align}
  & \Rightarrow \int{\dfrac{1}{1-\sin x}dx}=\int{\dfrac{1+{{u}^{2}}}{1+{{u}^{2}}-2u}\times \dfrac{2}{1+{{u}^{2}}}du} \\
 & \Rightarrow \int{\dfrac{1}{1-\sin x}dx}=\int{\dfrac{2}{{{\left( u-1 \right)}^{2}}}du} \\
\end{align}$
Using the integration formula $\int{\dfrac{1}{{{x}^{2}}}dx}=-\dfrac{1}{x}+C$ in the above equation, then we will get
$\begin{align}
  & \Rightarrow \int{\dfrac{1}{1-\sin x}dx}=-\dfrac{2}{u-1}+C \\
 & \Rightarrow \int{\dfrac{1}{1-\sin x}dx}=-\dfrac{2}{\tan \left( \dfrac{x}{2} \right)-1}+C \\
\end{align}$
The definite integral for the above equation will be
$\begin{align}
  & \Rightarrow \int\limits_{0}^{y}{\dfrac{1}{1-\sin x}dx}=\left[ -\dfrac{2}{\tan \left( \dfrac{x}{2} \right)-1} \right]_{0}^{y} \\
 & \Rightarrow \int\limits_{0}^{y}{\dfrac{1}{1-\sin x}dx}=\left( -\dfrac{2}{\tan \left( \dfrac{y}{2} \right)-1} \right)-\left( -\dfrac{2}{0-1} \right) \\
 & \Rightarrow \int\limits_{0}^{y}{\dfrac{1}{1-\sin x}dx}=-2-\dfrac{2}{\tan \left( \dfrac{y}{2} \right)-1} \\
\end{align}$
Let us check the value of above integral when $y=\dfrac{\pi }{2}$, by substituting $y=\dfrac{\pi }{2}$ in the above equation, then we will get
$\Rightarrow -2-\dfrac{2}{\tan \left( \dfrac{\dfrac{\pi }{2}}{2} \right)-1}=-2-\dfrac{2}{\tan \left( \dfrac{\pi }{4} \right)-1}$
We know that the value of $\tan \left( \dfrac{\pi }{4} \right)=1$, then the above equation is modified as
$\begin{align}
  & \Rightarrow -2-\dfrac{2}{\tan \left( \dfrac{\pi }{4} \right)-1}=-2-\dfrac{2}{1-1} \\
 & \Rightarrow -2-\dfrac{2}{\tan \left( \dfrac{\pi }{4} \right)-1}=-2-\dfrac{2}{0} \\
\end{align}$
Here we have the infinite form which is $\dfrac{2}{0}$. We can’t divide the any number with zero. So we can write that
$\Rightarrow \displaystyle \lim_{y \to \dfrac{\pi }{2}}\left( -2-\dfrac{2}{\tan \left( \dfrac{y}{2} \right)-1} \right)=-\infty $
Hence the above integral is divergent. So, we can’t find the integration of the given function.


Note: For this problem we have got the divergent integral, so we have terminated the process. But if we don’t get the divergent integral then we need to continue the process by applying the integral to the calculated value and simplifying the equation.