For x ( - 2 , 2 ) , if y ( x ) = c o s e c x + sin x c o s e c x sec x + tan x sin 2 x d x , and lim x ( 2 ) - y ( x ) = 0 , then y is equal to [2024]

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Solution

Q.
$For x \in (- \frac{π}{2}, \frac{π}{2}), if y (x) = \int \frac{c o s e c x + \sin x}{c o s e c x \sec x + \tan x \sin^{2} x} d x,$ $and \lim_{x \to {(\frac{π}{2})}^{-}} y (x) = 0, then y (\frac{π}{4}) is equal to$ [2024]

1 $- \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{1}{\sqrt{2}})$

2 $\tan^{- 1} (\frac{1}{\sqrt{2}})$

3 $\frac{1}{2} \tan^{- 1} (\frac{1}{\sqrt{2}})$

4 $\frac{1}{\sqrt{2}} \tan^{- 1} (- \frac{1}{2})$

Ans.
(4)

Let $I = y (x) = \int \frac{c o s e c x + \sin x}{c o s e c x \sec x + \tan x \sin^{2} x} d x$

$= \int \frac{\frac{1}{\sin x} + \sin x}{\frac{1}{\sin x} \cdot \frac{1}{\cos x} + \frac{\sin x}{\cos x} \cdot \sin^{2} x} d x = \int \frac{(1 + \sin^{2} x) \cos x}{(1 + \sin^{4} x)} d x$

Let $\sin x = t \Rightarrow \cos x d x = d t$

$∴ I = \int \frac{1 + t^{2}}{1 + t^{4}} d t = \int \frac{(1 + \frac{1}{t^{2}})}{(t^{2} + \frac{1}{t^{2}})} d t$

Put $t - \frac{1}{t} = u \Rightarrow (1 + \frac{1}{t^{2}}) d t = d u and t^{2} + \frac{1}{t^{2}} = u^{2} + 2$

$∴ I = \int \frac{d u}{u^{2} + 2} = \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{u}{\sqrt{2}}) + C = \frac{1}{\sqrt{2}} \tan^{- 1} \frac{1}{\sqrt{2}} (t - \frac{1}{t}) + C$

$y (x) = \frac{1}{\sqrt{2}} \tan^{- 1} \frac{1}{\sqrt{2}} (\sin x - \cos e c x) + C$

$\lim_{x \to {(\frac{π}{2})}^{-}} y (x) = \lim_{h \to 0} [\frac{1}{\sqrt{2}} \tan^{- 1} \frac{1}{\sqrt{2}} {\sin (\frac{π}{2} - h) - c o s e c (\frac{π}{2} - h)} + c]$

$\Rightarrow \frac{1}{\sqrt{2}} \tan^{- 1} \frac{1}{\sqrt{2}} (0) + c = 0 \Rightarrow c = 0$

$∴ y (\frac{π}{4}) = \frac{1}{\sqrt{2}} \tan^{- 1} {\frac{1}{\sqrt{2}} (\sin \frac{π}{4} - c o s e c \frac{π}{4})}$

$= \frac{1}{\sqrt{2}} \tan^{- 1} {\frac{1}{\sqrt{2}} (\frac{1}{\sqrt{2}} - \sqrt{2})} = \frac{1}{\sqrt{2}} \tan^{- 1} (- \frac{1}{2})$

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