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Question Number 32914 by mondodotto@gmail.com last updated on 06/Apr/18

Commented by mrW2 last updated on 06/Apr/18

(b) ∫((sin x+cos x)/(cos x−sin x)) dx =−∫((cos x+sin x)/(sin x−cos x)) dx =−∫(1/(sin x−cos x)) d(sin x−cos x) =−ln ∣sin x−cos x∣+C

$$\left({b}\right) \\ $$$$\int\frac{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}{\mathrm{cos}\:{x}−\mathrm{sin}\:{x}}\:{dx} \\ $$$$=−\int\frac{\mathrm{cos}\:{x}+\mathrm{sin}\:{x}}{\mathrm{sin}\:{x}−\mathrm{cos}\:{x}}\:{dx} \\ $$$$=−\int\frac{\mathrm{1}}{\mathrm{sin}\:{x}−\mathrm{cos}\:{x}}\:{d}\left(\mathrm{sin}\:{x}−\mathrm{cos}\:{x}\right) \\ $$$$=−\mathrm{ln}\:\mid\mathrm{sin}\:{x}−\mathrm{cos}\:{x}\mid+{C} \\ $$

Commented by prof Abdo imad last updated on 06/Apr/18

a) let use the changement x=5sht I= ∫5 cht .5cht dt = 25 ∫ ch^2 t dt =((25)/2)∫( 1+ch(2t))dt = ((25t)/2) + ((25)/2) ∫ ch(2t)dt =((25)/2)t + ((25)/4) sh(2t) +λ but t=argsh((x/5)) =ln((x/5) +(√(1+((x/5))^2 )) ) sh(2t) =2 sh(t)ch(t) =2(x/5)(√(1+((x/5))^2 )) ⇒ I = ((25)/2)ln((x/5) +(√(1+(x^2 /(25)))) ) +(5/2) x(√(1 +(x^2 /(25)))) +λ .

$$\left.{a}\right)\:{let}\:{use}\:{the}\:{changement}\:{x}=\mathrm{5}{sht} \\ $$$${I}=\:\int\mathrm{5}\:{cht}\:.\mathrm{5}{cht}\:{dt}\:=\:\mathrm{25}\:\int\:{ch}^{\mathrm{2}} {t}\:{dt} \\ $$$$=\frac{\mathrm{25}}{\mathrm{2}}\int\left(\:\mathrm{1}+{ch}\left(\mathrm{2}{t}\right)\right){dt}\:\:=\:\frac{\mathrm{25}{t}}{\mathrm{2}}\:+\:\frac{\mathrm{25}}{\mathrm{2}}\:\int\:{ch}\left(\mathrm{2}{t}\right){dt} \\ $$$$=\frac{\mathrm{25}}{\mathrm{2}}{t}\:\:+\:\frac{\mathrm{25}}{\mathrm{4}}\:{sh}\left(\mathrm{2}{t}\right)\:+\lambda\:\:{but}\:{t}={argsh}\left(\frac{{x}}{\mathrm{5}}\right) \\ $$$$={ln}\left(\frac{{x}}{\mathrm{5}}\:+\sqrt{\mathrm{1}+\left(\frac{{x}}{\mathrm{5}}\right)^{\mathrm{2}} }\:\right) \\ $$$${sh}\left(\mathrm{2}{t}\right)\:=\mathrm{2}\:{sh}\left({t}\right){ch}\left({t}\right) \\ $$$$=\mathrm{2}\frac{{x}}{\mathrm{5}}\sqrt{\mathrm{1}+\left(\frac{{x}}{\mathrm{5}}\right)^{\mathrm{2}} }\:\:\Rightarrow \\ $$$${I}\:=\:\frac{\mathrm{25}}{\mathrm{2}}{ln}\left(\frac{{x}}{\mathrm{5}}\:+\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{25}}}\:\right)\:+\frac{\mathrm{5}}{\mathrm{2}}\:{x}\sqrt{\mathrm{1}\:+\frac{{x}^{\mathrm{2}} }{\mathrm{25}}}\:\:+\lambda\:. \\ $$

Commented by mondodotto@gmail.com last updated on 06/Apr/18

thanx

$$\boldsymbol{\mathrm{thanx}} \\ $$

Commented by prof Abdo imad last updated on 07/Apr/18

c) let put I = ∫ (dx/(2x^2 +x−3)) roots of 2x^2 +x−3 ⇒Δ=1−4(2)(−3)=25 x_1 =((−1 +5)/4) =1 and x_2 =((−1−5)/2) =−3 so F(x)= (1/(2x^2 +x−3)) = (1/(2(x−1)(x+3))) = (a/(x−1)) +(b/(x+3)) a= lim_(x→1) (x−1)F(x)= (1/8) b=lim_(x→−3) (x+3)F(x) =−(1/8) ⇒ F(x)= (1/(8(x−1))) −(1/(8(x+3))) I = (1/8) ∫( (1/(x−1)) −(1/(x+3))) dx +λ I = (1/8)ln∣((x−1)/(x+3))∣ +λ .

$$\left.{c}\right)\:{let}\:{put}\:{I}\:=\:\int\:\:\:\frac{{dx}}{\mathrm{2}{x}^{\mathrm{2}} \:+{x}−\mathrm{3}} \\ $$$${roots}\:{of}\:\mathrm{2}{x}^{\mathrm{2}} \:+{x}−\mathrm{3}\:\Rightarrow\Delta=\mathrm{1}−\mathrm{4}\left(\mathrm{2}\right)\left(−\mathrm{3}\right)=\mathrm{25} \\ $$$${x}_{\mathrm{1}} \:=\frac{−\mathrm{1}\:+\mathrm{5}}{\mathrm{4}}\:=\mathrm{1}\:\:{and}\:\:{x}_{\mathrm{2}} =\frac{−\mathrm{1}−\mathrm{5}}{\mathrm{2}}\:=−\mathrm{3}\:{so} \\ $$$${F}\left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} +{x}−\mathrm{3}}\:=\:\frac{\mathrm{1}}{\mathrm{2}\left({x}−\mathrm{1}\right)\left({x}+\mathrm{3}\right)} \\ $$$$=\:\frac{{a}}{{x}−\mathrm{1}}\:+\frac{{b}}{{x}+\mathrm{3}} \\ $$$${a}=\:{lim}_{{x}\rightarrow\mathrm{1}} \left({x}−\mathrm{1}\right){F}\left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{8}} \\ $$$${b}={lim}_{{x}\rightarrow−\mathrm{3}} \left({x}+\mathrm{3}\right){F}\left({x}\right)\:=−\frac{\mathrm{1}}{\mathrm{8}}\:\Rightarrow \\ $$$${F}\left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{8}\left({x}−\mathrm{1}\right)}\:−\frac{\mathrm{1}}{\mathrm{8}\left({x}+\mathrm{3}\right)} \\ $$$${I}\:=\:\frac{\mathrm{1}}{\mathrm{8}}\:\int\left(\:\frac{\mathrm{1}}{{x}−\mathrm{1}}\:−\frac{\mathrm{1}}{{x}+\mathrm{3}}\right)\:{dx}\:+\lambda \\ $$$${I}\:=\:\frac{\mathrm{1}}{\mathrm{8}}{ln}\mid\frac{{x}−\mathrm{1}}{{x}+\mathrm{3}}\mid\:+\lambda\:. \\ $$

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