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Hint- Rationalize the given function which needs to be integrated.

Let the given integral be $I = \int_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} {\left( {\dfrac{1}{{1 + \sin x}}} \right)dx} $

Now rationalizing the function on the RHS of the above equation, we get

$

I = \int_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} {\left[ {\left( {\dfrac{1}{{1 + \sin x}}} \right) \times \left( {\dfrac{{1 - \sin x}}{{1 - \sin x}}} \right)} \right]dx} = \int_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} {\left[ {\dfrac{{1 - \sin x}}{{\left( {1 + \sin x} \right)\left( {1 - \sin x} \right)}}} \right]dx} \\

\Rightarrow I = \int_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} {\left[ {\dfrac{{1 - \sin x}}{{1 - {{\left( {\sin x} \right)}^2}}}} \right]dx} \\

$

Using the identity ${\left( {\sin x} \right)^2} + {\left( {\cos x} \right)^2} = 1 \Rightarrow 1 - {\left( {\sin x} \right)^2} = {\left( {\cos x} \right)^2}$ , we have

\[ \Rightarrow I = \int_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} {\left[ {\dfrac{{1 - \sin x}}{{{{\left( {\cos x} \right)}^2}}}} \right]dx = \int_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} {\left[ {\dfrac{1}{{{{\left( {\cos x} \right)}^2}}} - \dfrac{{\sin x}}{{\left( {\cos x} \right)\left( {\cos x} \right)}}} \right]dx} } \]

Since, $\sec x = \dfrac{1}{{\cos x}}$ and $\tan x = \dfrac{{\sin x}}{{\cos x}}$

\[

\therefore I = \int_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} {\left[ {{{\left( {\sec x} \right)}^2} - \left( {\tan x} \right)\left( {\sec x} \right)} \right]dx} = \int_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} {{{\left( {\sec x} \right)}^2}dx - \int_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} {\left( {\tan x} \right)\left( {\sec x} \right)} dx} \\

\Rightarrow I = \left[ {\tan x} \right]_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} - \left[ {\sec x} \right]_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} = \left[ {\tan \left( {\dfrac{\pi }{4}} \right) - \tan \left( { - \dfrac{\pi }{4}} \right)} \right] - \left[ {\sec \left( {\dfrac{\pi }{4}} \right) - \sec \left( { - \dfrac{\pi }{4}} \right)} \right] \\

\Rightarrow I = \left[ {\tan \left( {\dfrac{\pi }{4}} \right) - \tan \left( { - \dfrac{\pi }{4}} \right)} \right] - \left[ {\sec \left( {\dfrac{\pi }{4}} \right) - \sec \left( { - \dfrac{\pi }{4}} \right)} \right] \\

\]

Since, we know that $\tan \left( { - \theta } \right) = \tan \theta $ and $\sec \left( { - \theta } \right) = \sec \theta $

$\therefore I = \left[ {\tan \left( {\dfrac{\pi }{4}} \right) + \tan \left( {\dfrac{\pi }{4}} \right)} \right] - \left[ {\sec \left( {\dfrac{\pi }{4}} \right) - \sec \left( {\dfrac{\pi }{4}} \right)} \right] = 2\tan \left( {\dfrac{\pi }{4}} \right)$

Also, $\tan \left( {\dfrac{\pi }{4}} \right) = 1$

$ \Rightarrow I = 2 \times 1 = 2$.

Note- These types of problems can be solved by rationalizing the function which needs to be integrated in order to get a function for which the formula of integration is known.

Let the given integral be $I = \int_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} {\left( {\dfrac{1}{{1 + \sin x}}} \right)dx} $

Now rationalizing the function on the RHS of the above equation, we get

$

I = \int_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} {\left[ {\left( {\dfrac{1}{{1 + \sin x}}} \right) \times \left( {\dfrac{{1 - \sin x}}{{1 - \sin x}}} \right)} \right]dx} = \int_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} {\left[ {\dfrac{{1 - \sin x}}{{\left( {1 + \sin x} \right)\left( {1 - \sin x} \right)}}} \right]dx} \\

\Rightarrow I = \int_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} {\left[ {\dfrac{{1 - \sin x}}{{1 - {{\left( {\sin x} \right)}^2}}}} \right]dx} \\

$

Using the identity ${\left( {\sin x} \right)^2} + {\left( {\cos x} \right)^2} = 1 \Rightarrow 1 - {\left( {\sin x} \right)^2} = {\left( {\cos x} \right)^2}$ , we have

\[ \Rightarrow I = \int_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} {\left[ {\dfrac{{1 - \sin x}}{{{{\left( {\cos x} \right)}^2}}}} \right]dx = \int_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} {\left[ {\dfrac{1}{{{{\left( {\cos x} \right)}^2}}} - \dfrac{{\sin x}}{{\left( {\cos x} \right)\left( {\cos x} \right)}}} \right]dx} } \]

Since, $\sec x = \dfrac{1}{{\cos x}}$ and $\tan x = \dfrac{{\sin x}}{{\cos x}}$

\[

\therefore I = \int_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} {\left[ {{{\left( {\sec x} \right)}^2} - \left( {\tan x} \right)\left( {\sec x} \right)} \right]dx} = \int_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} {{{\left( {\sec x} \right)}^2}dx - \int_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} {\left( {\tan x} \right)\left( {\sec x} \right)} dx} \\

\Rightarrow I = \left[ {\tan x} \right]_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} - \left[ {\sec x} \right]_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} = \left[ {\tan \left( {\dfrac{\pi }{4}} \right) - \tan \left( { - \dfrac{\pi }{4}} \right)} \right] - \left[ {\sec \left( {\dfrac{\pi }{4}} \right) - \sec \left( { - \dfrac{\pi }{4}} \right)} \right] \\

\Rightarrow I = \left[ {\tan \left( {\dfrac{\pi }{4}} \right) - \tan \left( { - \dfrac{\pi }{4}} \right)} \right] - \left[ {\sec \left( {\dfrac{\pi }{4}} \right) - \sec \left( { - \dfrac{\pi }{4}} \right)} \right] \\

\]

Since, we know that $\tan \left( { - \theta } \right) = \tan \theta $ and $\sec \left( { - \theta } \right) = \sec \theta $

$\therefore I = \left[ {\tan \left( {\dfrac{\pi }{4}} \right) + \tan \left( {\dfrac{\pi }{4}} \right)} \right] - \left[ {\sec \left( {\dfrac{\pi }{4}} \right) - \sec \left( {\dfrac{\pi }{4}} \right)} \right] = 2\tan \left( {\dfrac{\pi }{4}} \right)$

Also, $\tan \left( {\dfrac{\pi }{4}} \right) = 1$

$ \Rightarrow I = 2 \times 1 = 2$.

Note- These types of problems can be solved by rationalizing the function which needs to be integrated in order to get a function for which the formula of integration is known.

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