∫(dx/(sinx+cosx))


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Question Number 141753 by mohammad17 last updated on 23/May/21

$\int \frac{d x}{s i n x + c o s x}$
Commented by mohammad17 last updated on 23/May/21

$h e l p m e s i r$
	Answered by rs4089 last updated on 23/May/21

	$\frac{1}{\sqrt{2}} \int \frac{d x}{s i n (x + \frac{π}{4})} = \frac{1}{\sqrt{2}} \int c o s e c (x + \frac{π}{4})$ $\frac{1}{\sqrt{2}} {l o g}_{e} [c o s e c (x + \frac{π}{4}) - c o t (x + \frac{π}{4})] + C$
	Commented by mohammad17 last updated on 23/May/21

	Commented by mohammad17 last updated on 23/May/21

	$s i r t h i s a n s w e r i s t r u e o r f a l s e ?$
	Commented by mohammad17 last updated on 23/May/21

	$i t h i n k o u t t h e i n t e g r a l i s \frac{1}{\sqrt{2}} n o t \sqrt{2}$
	Commented by rs4089 last updated on 23/May/21

	$y e s s i r, y o u a r e r i g h t,$ $t h a n k y o u v e r y m u c h$
	Answered by mathmax by abdo last updated on 23/May/21

	$I = \int \frac{dx}{sinx + cosx} we do the changement \tan (\frac{x}{2}) = y \Rightarrow$ $I = \int \frac{2 dy}{(1 + y^{2}) (\frac{2 y}{1 + y^{2}} + \frac{1 - y^{2}}{1 + y^{2}})} = \int \frac{2 dy}{2 y + 1 - y^{2}}$ $= - 2 \int \frac{dy}{y^{2} - 2 y - 1} = - 2 \int \frac{dy}{{(y - 1)}^{2} - 2}$ $=_{y - 1 = \sqrt{2} t} - 2 \int \frac{\sqrt{2} dt}{2 (t^{2} - 1)} = - \sqrt{2} \int \frac{dt}{t^{2} - 1} = - \frac{1}{\sqrt{2}} \int (\frac{1}{t - 1} - \frac{1}{t + 1}) dt$ $= \frac{1}{\sqrt{2}} \log ∣ \frac{t + 1}{t - 1} ∣ + C = \frac{1}{\sqrt{2}} \log ∣ \frac{\frac{y - 1}{\sqrt{2}} + 1}{\frac{y - 1}{\sqrt{2}} - 1} ∣ + C \Rightarrow$ $I = \frac{1}{\sqrt{2}} \log ∣ \frac{\tan (\frac{x}{2}) - 1 + \sqrt{2}}{\tan (\frac{x}{2}) - 1 - \sqrt{2}} ∣ + C$
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