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Question Number 74123 by liki last updated on 19/Nov/19

Commented by liki last updated on 19/Nov/19

$. . . i n e e d h e l p r o m a n (i i i) p l z . .$
Commented by liki last updated on 19/Nov/19

$. . . (i i i) m r m i n d i s p o w e r p l z a s s i s t m e o r a n y o n e$
	Answered by ajfour last updated on 19/Nov/19

	$I = \int \tan^{- 1} (\sqrt{} (\frac{1 - \sin x}{1 + \sin x})) d x$ $= \int \tan^{- 1} ∣ \tan (\frac{π}{4} - \frac{x}{2}) ∣ d x$ $= \int (\frac{π}{4} - \frac{x}{2} + n π) d x$ $n s u c h t h a t 0 < \frac{π}{4} - \frac{x}{2} + n π < \frac{π}{2}$ $I = (\frac{π}{4} - \frac{x}{2} + n π) x + c .$
	Commented by ajfour last updated on 19/Nov/19

	$I t s h o u l d h a v e b e e n o n l y t h e r a d i c a l$ $s i g n i n s t e a d o f \int s i g n, b e c a u s e$ $t w o i n t e g r a l s s h o u l d t h e n$ $d e m a n d f o r t w o^{'} {d x}^{'} . A s t h e r e$ $i s o n l y s i n g l e d x t h e r e s h o u l d$ $b e s i n g l e \int .$ $I f i n i t i a l l y q u e s t i o n w a s g i v e n$ $b y t e a c h e r h a n d w r i t t e n o n p a p e r,$ $t h e t y p i s t c a n t a k e t h e r a d i c a l$ $t o b e t h e i n t e g r a l s i g n, i f t h e$ $h a n d w r i t i n g {i s n}^{'} t q u i t e c l e a r .$ $p l u s t h e q u e s t i o n i s o f 02 m a r k s .$
	Commented by liki last updated on 19/Nov/19

	$. . . t h a n k s s i r, b u t w h y d i d y o u p u t t h a t r a d i c a l s i g n$ $w h i l e i n o u r q u e s t i o n n o r a d i c a l . .$
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