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Question Number 79634 by TawaTawa last updated on 26/Jan/20

Commented by mathmax by abdo last updated on 27/Jan/20
$Ω =∫_(π/5) ^((3π)/(10)) (x/(sin(2x)))dx changement x=(π/2)−t givet=(π/2)−x Ω=∫_((3π)/(10)) ^(π/5) (((π/2)−t)/(sin(2t)))(−dt) =(π/2)∫_(π/5) ^((3π)/(10)) (dt/(sin(2t))) −Ω ⇒ 2Ω =(π/2) ∫_(π/5) ^((3π)/(10)) (dt/(sin(2t))) changement tant =u give ∫_(π/5) ^((3π)/(10)) (dt/(sin(2t))) =∫_(tan((π/5))) ^(tan(((3π)/(10)))) (1/((2u)/(1+u^2 )))×(du/(1+u^2 )) =(1/2)[ln∣u∣]_(tan((π/5))) ^(tan(((3π)/(10)))) =(1/2){ln(tan(((3π)/(10)))−ln(tan((π/5)))} =(1/2){ ln(tan((π/2)−(π/5)))−ln(tan((π/5)))} =(1/2){ln((1/(tan((π/5)))))−ln(tan((π/5)))}=−ln(tan((π/5))) ⇒ Ω=−(π/4)ln(tan((π/5)))$
$Ω = \int_{\frac{π}{5}}^{\frac{3 π}{10}} \frac{x}{s i n (2 x)} d x c h a n g e m e n t x = \frac{π}{2} - t g i v e t = \frac{π}{2} - x$ $Ω = \int_{\frac{3 π}{10}}^{\frac{π}{5}} \frac{\frac{π}{2} - t}{s i n (2 t)} (- d t) = \frac{π}{2} \int_{\frac{π}{5}}^{\frac{3 π}{10}} \frac{d t}{s i n (2 t)} - Ω \Rightarrow$ $2 Ω = \frac{π}{2} \int_{\frac{π}{5}}^{\frac{3 π}{10}} \frac{d t}{s i n (2 t)} c h a n g e m e n t t a n t = u g i v e$ $\int_{\frac{π}{5}}^{\frac{3 π}{10}} \frac{d t}{s i n (2 t)} = \int_{t a n (\frac{π}{5})}^{t a n (\frac{3 π}{10})} \frac{1}{\frac{2 u}{1 + u^{2}}} \times \frac{d u}{1 + u^{2}} = \frac{1}{2} {[l n ∣ u ∣]}_{t a n (\frac{π}{5})}^{t a n (\frac{3 π}{10})}$ $= \frac{1}{2} {l n (t a n (\frac{3 π}{10}) - l n (t a n (\frac{π}{5}))}$ $= \frac{1}{2} {l n (t a n (\frac{π}{2} - \frac{π}{5})) - l n (t a n (\frac{π}{5}))}$ $= \frac{1}{2} {l n (\frac{1}{t a n (\frac{π}{5})}) - l n (t a n (\frac{π}{5}))} = - l n (t a n (\frac{π}{5})) \Rightarrow$ $Ω = - \frac{π}{4} l n (t a n (\frac{π}{5}))$
Commented by TawaTawa last updated on 27/Jan/20

$God bless you sir$
Commented by mathmax by abdo last updated on 29/Jan/20

$y o u a r e w e l c o m e$
	Answered by mind is power last updated on 27/Jan/20

	$Ω = \int_{\frac{π}{5}}^{\frac{3 π}{10}} \frac{x}{s i n (2 x)} d x$ $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x \Rightarrow$ $\int_{\frac{π}{5}}^{\frac{3 π}{10}} \frac{\frac{π}{2} - x}{s i n (2 (\frac{π}{2} - x))} d x = \int_{\frac{π}{5}}^{\frac{3 π}{10}} \frac{\frac{π}{2} - x}{s i n (π - 2 x)} d x = - \int_{\frac{π}{5}}^{\frac{3 π}{10}} \frac{x}{s i n (2 x)} +$ $\frac{π}{2} \int_{\frac{π}{5}}^{\frac{3 π}{10}} \frac{d x}{s i n (2 x)} \Rightarrow$ $\Rightarrow 2 \int_{\frac{π}{5}}^{\frac{3 π}{10}} \frac{x d x}{s i n (2 x)} = \frac{π}{2} \int_{\frac{π}{5}}^{\frac{3 π}{10}} \frac{d x}{s i n (2 x)} = \frac{π}{4} \int_{\frac{π}{5}}^{\frac{3 π}{10}} \frac{d x}{s i n (x) c o s (x)}$ $= \frac{π}{4} \int_{\frac{π}{5}}^{\frac{3 π}{10}} \frac{c o s (x)}{s i n (x) . {c o s}^{2} (x)} d x = \frac{π}{4} \int_{\frac{π}{5}}^{\frac{3 π}{10}} c o t (x) . (1 + {t g}^{2} (x)) d x$ $= \frac{π}{4} \int_{\frac{π}{5}}^{\frac{3 π}{10}} t g (\frac{π}{2} - x) d (t g (x))$ $= \frac{π}{8} {[- {t g}^{2} (\frac{π}{2} - x)]}_{\frac{π}{5}}^{\frac{3 π}{10}} = - \frac{π}{8} ({t g}^{2} (\frac{π}{5}) - {t g}^{2} (\frac{3 π}{10}))$ $w e c a n f i n d e x a c t t g (\frac{π}{5}), a n d t g (\frac{3 π}{10})$
	Commented by TawaTawa last updated on 27/Jan/20

	$God bless you sir$
	Commented by mind is power last updated on 27/Jan/20

	$t h a k y o u, w i t h e p l e a s u r$
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