calculate lim_(x→π) ∫_(π/2) ^x (dx/(1+sinx−cosx))


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Question Number 64448 by mathmax by abdo last updated on 18/Jul/19

$c a l c u l a t e {l i m}_{x \to π} \int_{\frac{π}{2}}^{x} \frac{d x}{1 + s i n x - c o s x}$
Commented by mathmax by abdo last updated on 19/Jul/19
$let A(x) =∫_(π/2) ^x (dt/(1+sint−cost)) changement tan((t/2))=u vive A(x) =∫_1 ^(tan((x/2))) ((2du)/((1+u^2 )(1+((2u)/(1+u^2 ))−((1−u^2 )/(1+u^2 ))))) =∫_1 ^(tan((x/2))) ((2du)/(1+u^2 +2u−1+u^2 )) =∫_1 ^(tan((x/2))) ((2du)/(2u^2 +2u)) =∫_1 ^(tan((x/2))) (du/(u(u+1))) =∫_1 ^(tan((x/2))) {(1/u)−(1/(u+1))}du =[ln∣(u/(u+1))∣]_1 ^(tan((x/2))) =ln∣((tan((x/2)))/(tan((x/2))+1))∣−ln((1/2)) ⇒lim_(x→π) A(x) =0+ln(2) =ln(2).$
$l e t A (x) = \int_{\frac{π}{2}}^{x} \frac{d t}{1 + s i n t - c o s t} c h a n g e m e n t t a n (\frac{t}{2}) = u v i v e$ $A (x) = \int_{1}^{t a n (\frac{x}{2})} \frac{2 d u}{(1 + u^{2}) (1 + \frac{2 u}{1 + u^{2}} - \frac{1 - u^{2}}{1 + u^{2}})}$ $= \int_{1}^{t a n (\frac{x}{2})} \frac{2 d u}{1 + u^{2} + 2 u - 1 + u^{2}} = \int_{1}^{t a n (\frac{x}{2})} \frac{2 d u}{2 u^{2} + 2 u} = \int_{1}^{t a n (\frac{x}{2})} \frac{d u}{u (u + 1)}$ $= \int_{1}^{t a n (\frac{x}{2})} {\frac{1}{u} - \frac{1}{u + 1}} d u = {[l n ∣ \frac{u}{u + 1} ∣]}_{1}^{t a n (\frac{x}{2})} = l n ∣ \frac{t a n (\frac{x}{2})}{t a n (\frac{x}{2}) + 1} ∣ - l n (\frac{1}{2})$ $\Rightarrow {l i m}_{x \to π} A (x) = 0 + l n (2) = l n (2) .$
	Answered by Tanmay chaudhury last updated on 18/Jul/19

	$\int \frac{d x}{2 s i n \frac{x}{2} c o s \frac{x}{2} + 2 {s i n}^{2} \frac{x}{2}}$ $\int \frac{{s e c}^{2} \frac{x}{2} d x}{2 t a n \frac{x}{2} + 2 {t a n}^{2} \frac{x}{2}} d x$ $\int \frac{d (t a n \frac{x}{2})}{t a n \frac{x}{2} (1 + t a n \frac{x}{2})}$ $\int \frac{d (t a n \frac{x}{2})}{t a n \frac{x}{2}} - \int \frac{d (t a n \frac{x}{2})}{1 + t a n \frac{x}{2}}$ $∣ l n (\frac{t a n \frac{x}{2}}{1 + t a n \frac{x}{2}}) ∣_{\frac{π}{2}}^{π}$ $l n 1 - l n (\frac{1}{1 + 1})$ $= l n 2$
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