∫(e^x /(cosx))dx


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Question Number 142389 by alcohol last updated on 31/May/21

$\int \frac{e^{x}}{c o s x} d x$
	Answered by ArielVyny last updated on 31/May/21

	$= [e^{x} \times \frac{1}{c o s x}] - \int e^{x} \times - \frac{s i n x}{{c o s}^{2} x} d x$ Nous considérons U'= $\int e^{x} \times \frac{s i n x}{{c o s}^{2} x} d x = [t g x \times e^{x} s i n x] - \int t g x (e^{x} s i n x + e^{x} c o s x)$ $I = [\frac{e^{x}}{c o s x} + t g x \times e^{x} s i n x] - \int e^{x} \times \frac{{s i n}^{2} x}{c o s x} d x - \int e^{x} s i n x d x$ $l i n t e g r a l e \int e^{x} s i n x d x e t a n t s i m p l e$ $c h e r c h o n s \int e^{x} \times \frac{{s i n}^{2} x}{c o s x} d x$ $\int e^{x} \times \frac{1 - {c o s}^{2} x}{c o s x} d x$ $\int \frac{e^{x}}{c o s x} - \int e^{x} c o s x d x$ $o n o b t i e n t$ $I = [e^{x} (\frac{1}{c o s x} + t g x \times s i n x)] - \int e^{x} s i n x d x - (\int \frac{e^{x}}{c o s x} - \int e^{x} c o s x)$ $2 I = [e^{x} (\frac{1}{c o s x} + \frac{{s i n}^{2} x}{c o s x})] + \int e^{x} c o s x d x - \int e^{x} s i n x d x$ $\int e^{x} c o s x d x = [e^{x} c o s x] + \int e^{x} s i n x d x$ $2 I = [e^{x} (\frac{1 + {s i n}^{2} x}{c o s x}) + c o s x]$ $I = \frac{1}{2} [e^{x} (\frac{1 + {s i n}^{2} x}{c o s x}) + c o s x] + c t e$
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