INTEGRAL (1)∫_0 ^( ln 2) x^(−2) .e^(−(1/x)) dx =? (2) ∫ ((sin x+cos x)/(sin^4 x+cos^4 x)) dx =?


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Question Number 134636 by bobhans last updated on 06/Mar/21

$INTEGRAL$ $(1) \int_{0}^{\ln 2} x^{- 2} . e^{- \frac{1}{x}} d x = ?$ $(2) \int \frac{\sin x + \cos x}{\sin^{4} x + \cos^{4} x} dx = ?$
	Answered by benjo_mathlover last updated on 06/Mar/21

	$(1) \int_{0}^{\ln 2} \frac{e^{- \frac{1}{x}}}{x^{2}} dx = \int_{0}^{\ln 2} e^{- \frac{1}{x}} d (- \frac{1}{x})$ $= \lim_{p \to 0} {[e^{- \frac{1}{x}}]}_{p}^{\ln 2} = e^{- \frac{1}{\ln 2}} - e^{- \infty}$ $= e^{- \frac{1}{\ln 2}} = \frac{1}{\sqrt[\ln 2]{e}}$
	Answered by rs4089 last updated on 06/Mar/21

	$\int \frac{s i n x + c o s x}{1 - 2 {s i n}^{2} {x c o s}^{2} x} d x$ $2 \int \frac{s i n x + c o s x}{2 - {s i n}^{2} 2 x} d x$ $l e t s i n x - c o s x = t \Rightarrow (s i n x + c o s x) d x = d t$ $a n d s i n 2 x = 1 - t^{2}$ $2 \int \frac{d t}{2 - (1 - t^{2})}$ $2 \int \frac{d t}{1 + t^{2}} = 2 {t a n}^{- 1} t + c$ $= 2 {t a n}^{- 1} (s i n x - c o s x) + C$
	Answered by john_santu last updated on 06/Mar/21

	$(2) \int \frac{\sin x + \cos x}{\sin^{4} x + \cos^{4} x} d x$ $= \int \frac{\sqrt{2} \sin (x + π / 4)}{1 - \frac{1}{2} \sin^{2} 2 x} d x$ $= 2 \sqrt{2} \int \frac{\sin (x + π / 4)}{2 - \sin^{2} 2 x} d x$ $s u b s t i t u t e h = x + π / 4 a n d 2 x = 2 h - π / 2$ $I = 2 \sqrt{2} \int \frac{\sin h}{2 - \cos^{2} 2 h} d h$ $l e t t i n g \sqrt{2} \cos h = u$ $I = - 2 \int \frac{d u}{1 + u^{2}} = - 2 \arctan (u) + c$ $= - 2 \arctan (\sqrt{2} \cos (x + π / 4)) + c$
	Answered by mathmax by abdo last updated on 27/Mar/21

	$1) we have \frac{d}{dx} (e^{- \frac{1}{x}}) = \frac{1}{x^{2}} e^{- \frac{1}{x}} \Rightarrow$ $\int_{0}^{\ln 2} x^{- 2} e^{- \frac{1}{x}} ex = {[e^{- \frac{1}{x}}]}_{0}^{\ln 2} = e^{- \frac{1}{\ln 2}} - 0 = e^{- \frac{1}{\ln 2}}$
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