Question

The value of $\int\limits_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} {\dfrac{1}{{1 - \sin x}}} dx$ is equal to-
A) $1$
B) $ - 2$
C) $2$
D) $ - 1$

Answer 1

Hint:In the above question we are given a definite integral and we have to find its value. Since in case of integration there are only few functions whose integration is known to us. And here we do not directly know the integration of $\dfrac{1}{{1 - \sin x}}$, so we have to reduce the given function to some other function whose integration is known to us.So, we can use rationalization and further use trigonometric formulas to solve the above problem.

Complete step-by-step answer:
In the above question we have to find the value of definite integral-
$\int\limits_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} {\dfrac{1}{{1 - \sin x}}} dx ----(1)$
Now we know to calculate integration we need to reduce the function to another function whose integration is known to us.
Here the integration of a given function, $\dfrac{1}{{1 - \sin x}}$ is not known to us.
So, we will try to reduce this function into another function whose integration is not known.
Now consider (1)-
$\int\limits_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} {\dfrac{1}{{1 - \sin x}}} dx$
Now multiplying and dividing $1 - \sin x$ with the function $\dfrac{1}{{1 - \sin x}}$, we get
\[ = \int\limits_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} {\left( {\dfrac{1}{{1 - \sin x}}} \right)} \left( {\dfrac{{1 + \sin x}}{{1 + \sin x}}} \right)dx\]
Now using the algebraic identity $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$ for denominator, we get
\[ = \int\limits_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} {\left( {\dfrac{{1 + \sin x}}{{{{\left( 1 \right)}^2} - {{\left( {\sin x} \right)}^2}}}} \right)} dx\]
Now we know that
 $
  {\sin ^2}x + {\cos ^2}x = 1 \\
   \Rightarrow {\cos ^2}x = 1 - {\sin ^2}x \\
 $
And substituting value of $1 - {\sin ^2}x$ in denominator, we get
\[ = \int\limits_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} {\left( {\dfrac{{1 + \sin x}}{{{{\cos }^2}x}}} \right)} dx\]
Now solving further, we get
\[ = \int\limits_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} {\left( {\dfrac{1}{{{{\cos }^2}x}} + \dfrac{{\sin x}}{{{{\cos }^2}x}}} \right)} dx\]
Now using the properties of integral, we know $\int\limits_a^b {\left( {f(x) + g(x)} \right)} dx = \int\limits_a^b {f(x)} dx + \int\limits_a^b {g(x)} dx$
So, using this property, we get
\[ = \int\limits_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} {\dfrac{1}{{{{\cos }^2}x}}dx + \int\limits_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} {\dfrac{{\sin x}}{{{{\cos }^2}x}}} } dx\]
Now we can also write as
\[ = \int\limits_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} {\dfrac{1}{{{{\cos }^2}x}}dx + \int\limits_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} {\dfrac{1}{{\left( {\cos x} \right)}}\dfrac{{\sin x}}{{\left( {\cos x} \right)}}} } dx\]
Now by trigonometry we know that $\dfrac{1}{{\cos x}} = \sec x$ and $\dfrac{{sinx}}{{\cos x}} = \tan x$
So, we get
\[ = \int\limits_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} {{{\sec }^2}xdx + \int\limits_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} {\sec x \times \tan x} } dx ---(2)\]
Now we know the integration of \[{\sec ^2}x\] and \[\sec x \times \tan x\], that is
$\int {{{\sec }^2}xdx = \tan x + {c_1},{\text{ }}{c_1}{\text{ is constant}}}$
$\int \sec x dx$ = $\sec x \times \tan x$+ ${c_2},{\text{ }}{c_2}{\text{ is constant}}$
 Now substituting values of these integrals in (2), we get
$ = \left. {\tan x} \right|_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} + \left. {\sec x} \right|_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}}$
Now we know $\left. {f(x)} \right|_b^a = f(a) - f(b)$, so we get,
$ \Rightarrow = \tan \left( {\dfrac{\pi }{4}} \right) - \tan \left( { - \dfrac{\pi }{4}} \right) + \sec \left( {\dfrac{\pi }{4}} \right) - \sec \left( { - \dfrac{\pi }{4}} \right)$
now using trigonometry, we know $\tan \left( { - x} \right) = - \tan x$ and $\sec \left( { - x} \right) = \sec \left( x \right)$, so we get
$
   \Rightarrow = \tan \left( {\dfrac{\pi }{4}} \right) + \tan \left( {\dfrac{\pi }{4}} \right) + \sec \left( {\dfrac{\pi }{4}} \right) - \sec \left( {\dfrac{\pi }{4}} \right) \\
   \Rightarrow = 2\tan \left( {\dfrac{\pi }{4}} \right) \\
 $
Now using trigonometry, substituting $\tan \dfrac{\pi }{4} = 1$, we get
$ \Rightarrow \int\limits_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} {\dfrac{1}{{1 - \sin x}}} dx = 2$
So, the value of $\int\limits_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} {\dfrac{1}{{1 - \sin x}}} dx$ is equal to $2$

So, the correct answer is “Option C”.

Note:We can also solve the question $\int\limits_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} {\dfrac{1}{{1 - \sin x}}} dx$, By substituting $1$ in denominator by $1 = {\cos ^2}\dfrac{x}{2} + {\sin ^2}\dfrac{x}{2}$ and $\sin x$ by $\sin x = 2\sin \dfrac{x}{2}\cos \dfrac{x}{2}$ and further simplifying the numerator and denominator by using basic algebraic identities we get the same answer But using this way the question becomes very lengthy and sometimes confusing also.