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Question Number 30390 by mondodotto@gmail.com last updated on 21/Feb/18

Commented by abdo imad last updated on 21/Feb/18

we have x^2 +4x +3=(x+2)^2 −1 tbe ch .x+2=cht give I= ∫ (√(ch^2 t −1)) shtdt = ∫ sh^2 t dt = ∫((1−ch(2t))/2)dt = (1/2)t −(1/2) ∫ch(2t)dt=(t/2) −(1/4)sh(2t) = (t/2) −(1/2) sh(t)ch(t)=(1/2)argch(x+2) −(1/2)(√((x+2)^2 −1)) (x+2) I=(1/2)ln((x+2)^2 +(√((x+2)^2 −1))) −(1/2)(x+2)(√((x+2)^2 −1)) +λ

$${we}\:{have}\:{x}^{\mathrm{2}} \:+\mathrm{4}{x}\:+\mathrm{3}=\left({x}+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{1}\:\:{tbe}\:{ch}\:.{x}+\mathrm{2}={cht}\:{give} \\ $$$${I}=\:\int\:\sqrt{{ch}^{\mathrm{2}} {t}\:−\mathrm{1}}\:{shtdt}\:=\:\int\:{sh}^{\mathrm{2}} {t}\:{dt}\:=\:\int\frac{\mathrm{1}−{ch}\left(\mathrm{2}{t}\right)}{\mathrm{2}}{dt} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}{t}\:\:−\frac{\mathrm{1}}{\mathrm{2}}\:\int{ch}\left(\mathrm{2}{t}\right){dt}=\frac{{t}}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{4}}{sh}\left(\mathrm{2}{t}\right) \\ $$$$=\:\frac{{t}}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{2}}\:{sh}\left({t}\right){ch}\left({t}\right)=\frac{\mathrm{1}}{\mathrm{2}}{argch}\left({x}+\mathrm{2}\right)\:−\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\left({x}+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{1}}\:\left({x}+\mathrm{2}\right) \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\left({x}+\mathrm{2}\right)^{\mathrm{2}} +\sqrt{\left({x}+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{1}}\right)\:−\frac{\mathrm{1}}{\mathrm{2}}\left({x}+\mathrm{2}\right)\sqrt{\left({x}+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{1}}\:\:+\lambda \\ $$

Commented by mondodotto@gmail.com last updated on 22/Feb/18

thAnx

$$\mathrm{thAnx} \\ $$

Commented by mondodotto@gmail.com last updated on 22/Feb/18

what does it mean? ch and sh????

$$\mathrm{what}\:\mathrm{does}\:\mathrm{it}\:\mathrm{mean}?\:\mathrm{ch}\:\mathrm{and}\:\mathrm{sh}???? \\ $$

Commented by prof Abdo imad last updated on 22/Feb/18

ch(x)= ((e^x +e^(−x) )/2) ,sh(x)= ((e^x −e^(−x) )/2) th(x)= ((shx)/(chx)) with x from R and the same definition on set C.

$${ch}\left({x}\right)=\:\frac{{e}^{{x}} \:+{e}^{−{x}} }{\mathrm{2}}\:\:,{sh}\left({x}\right)=\:\frac{{e}^{{x}} \:−{e}^{−{x}} }{\mathrm{2}} \\ $$$${th}\left({x}\right)=\:\frac{{shx}}{{chx}}\:\:\:\:\:{with}\:{x}\:{from}\:{R}\:{and}\:{the}\:{same} \\ $$$${definition}\:{on}\:{set}\:\mathbb{C}. \\ $$

Commented by mondodotto@gmail.com last updated on 22/Feb/18

now i undstnd!

$$\mathrm{now}\:\mathrm{i}\:\mathrm{undstnd}!\: \\ $$

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