∫ (dx/(sec x + csc x)) = ?


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Question Number 82185 by jagoll last updated on 19/Feb/20

$\int \frac{d x}{\sec x + c s c x} = ?$
Commented by john santu last updated on 19/Feb/20

$\sec x + c s c x = \frac{1}{\cos x} + \frac{1}{\sin x}$ $\frac{\sin x + \cos x}{\sin x \cos x} .$ $\int \frac{\sin x \cos x}{\sin x + \cos x} d x =$ $l e t \tan (\frac{x}{2}) = t$
Commented by mathmax by abdo last updated on 19/Feb/20

$l e t I = \int \frac{d x}{\frac{1}{c o s x} + \frac{1}{s i n x}} \Rightarrow I = \int \frac{c o s x . s i n x}{c o s x + s i n x} d x c h a n g e m e n t$ $t a n (\frac{x}{2}) = t g i v e I = \int \frac{\frac{1 - t^{2}}{1 + t^{2}} \times \frac{2 t}{1 + t^{2}}}{\frac{1 - t^{2}}{1 + t^{2}} + \frac{2 t}{1 + t^{2}}} \times \frac{2 d t}{1 + t^{2}} = \int \frac{2 t (1 - t^{2})}{{(1 + t^{2})}^{2} (- t^{2} + 2 t + 1)} d t$ $= \int \frac{2 t (t^{2} - 1)}{{(t^{2} + 1)}^{2} (t^{2} - 2 t - 1)} d t = \int \frac{2 t^{3} - 2 t}{{(t^{2} + 1)}^{2} (t^{2} - 2 t - 1)} d t$ $t^{2} - 2 t - 1 = 0 \to Δ^{^{'}} = 1 + 1 = 2 \Rightarrow t_{1} = 1 + \sqrt{2} a n d t_{2} = 1 - \sqrt{2} l e t d e c o m p o s e$ $F (t) = \frac{2 t^{3} - 2 t}{{(t^{2} + 1)}^{2} (t - t_{1}) (t - t_{2})} \Rightarrow F (t) = \frac{a}{t - t_{1}} + \frac{b}{t - t_{2}} + \frac{c t + d}{t^{2} + 1} + \frac{e t + f}{{(t^{2} + 1)}^{2}}$ $\Rightarrow \int F (t) d t = a l n ∣ t - t_{1} ∣ + b l n ∣ t - t_{2} ∣ + \frac{c}{2} l n (t^{2} + 1) + d a r c t a n (t)$ $+ \int \frac{e t + f}{{(t^{2} + 1)}^{2}} d t r e s t c a l c u l u s o f c o e f f i c i e n t s . . . b e c o n t i n u e d . . .$
	Answered by TANMAY PANACEA last updated on 20/Feb/20

	$\int \frac{s i n x + c o s x}{s i n x c o s x} d x$ $2 \int \frac{s i n x + c o s x}{1 - (1 - 2 s i n x c o s x)} d x$ $2 \int \frac{d (s i n x - c o s x)}{1 - {(s i n x - c o s x)}^{2}} [f o r m u l a \int \frac{d a}{1 - a^{2}}]$ $l n (\frac{s i n x - c o s x + 1}{s i n x - c o s x - 1}) + c$
	Commented by jagoll last updated on 20/Feb/20

	$\frac{1}{c s c x + \sec x} = \frac{1}{\frac{1}{\sin x} + \frac{1}{\cos x}}$ $= \frac{\sin x . \cos x}{\sin x + \cos x} \neq \frac{\sin x + \cos x}{\sin x \cos x}$ $s o r r y s i r . y o u r a n s w e r n o t c o r r e c t$
	Commented by TANMAY PANACEA last updated on 20/Feb/20

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