Question-169164

Question Number 169164 by 0731619 last updated on 25/Apr/22

Commented by infinityaction last updated on 25/Apr/22

y = ln ((√(((1−sin x)/cos x)/((1+sin x)/cos x))) y = ln (√(((sec x−tan x)(sec x−tan x))/((sec x+tan x)(sec x−tan x)))) y = ln (sec x−tan x) (dy/dx) = ((sec x(tan x−sec x))/((sec x−tan x))) (dy/dx) = −sec x

$${y}\:=\:\mathrm{ln}\:\left(\sqrt{\frac{\left(\mathrm{1}−\mathrm{sin}\:{x}\right)/\mathrm{cos}\:{x}}{\left(\mathrm{1}+\mathrm{sin}\:{x}\right)/\mathrm{cos}\:{x}}}\right. \\ $$$${y}\:=\:\mathrm{ln}\:\sqrt{\frac{\left(\mathrm{sec}\:{x}−\mathrm{tan}\:{x}\right)\left(\mathrm{sec}\:{x}−\mathrm{tan}\:{x}\right)}{\left(\mathrm{sec}\:{x}+\mathrm{tan}\:{x}\right)\left(\mathrm{sec}\:{x}−\mathrm{tan}\:{x}\right)}} \\ $$$${y}\:=\:\mathrm{ln}\:\left(\mathrm{sec}\:{x}−\mathrm{tan}\:{x}\right) \\ $$$$\frac{{dy}}{{dx}}\:=\:\frac{\mathrm{sec}\:{x}\left(\mathrm{tan}\:{x}−\mathrm{sec}\:{x}\right)}{\left(\mathrm{sec}\:{x}−\mathrm{tan}\:{x}\right)} \\ $$$$\frac{{dy}}{{dx}}\:=\:−\mathrm{sec}\:{x} \\ $$$$ \\ $$

Commented by infinityaction last updated on 25/Apr/22

let p = ((x^2 +3x)/((x+1)(x^2 +1))) p = (((x+1)^2 +(x−1))/((x+1)(x^2 +1))) p = ((x+1)/(x^2 +1)) + ((x−1)/((x+1)(x^2 +1))) p = ((x+1)/(x^2 +1)) − (((x^2 +1)−x(x+1))/((x+1)(x^2 +1))) p = ((x+1)/(x^2 +1)) − (1/(x+1)) +(x/(x^2 +1)) p = ((2x)/(x^2 +1))+(1/(x^2 +1)) − (1/(x+1)) I = ∫_0 ^1 {((2x)/(x^2 +1))+(1/(x^2 +1))−(1/(x+1))}dx I = [log (x^2 +1) + tan^(−1) (x) + log (x+1)]_0 ^1 I = log 2 + (π/4) + log 2 I = (π/4) + log 4

$$\:\:{let}\: \\ $$$$\:\:\:\:{p}\:\:\:\:\:=\:\:\:\:\:\frac{{x}^{\mathrm{2}} +\mathrm{3}{x}}{\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$\:\:\:\:{p}\:\:\:\:\:=\:\:\:\:\frac{\left({x}+\mathrm{1}\right)^{\mathrm{2}} +\left({x}−\mathrm{1}\right)}{\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$\:\:\:\:{p}\:\:\:\:\:\:=\:\:\:\:\frac{{x}+\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}\:\:+\:\frac{{x}−\mathrm{1}}{\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$\:\:\:\:{p}\:\:\:\:\:\:=\:\:\:\:\:\frac{{x}+\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}\:−\:\frac{\left({x}^{\mathrm{2}} +\mathrm{1}\right)−{x}\left({x}+\mathrm{1}\right)}{\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$\:\:\:\:{p}\:\:\:\:\:\:\:=\:\:\:\:\frac{{x}+\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}\:−\:\frac{\mathrm{1}}{{x}+\mathrm{1}}\:+\frac{{x}}{{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\:\:\:\:{p}\:\:\:\:\:\:=\:\:\:\:\:\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}\:−\:\frac{\mathrm{1}}{{x}+\mathrm{1}} \\ $$$$\:\:\:\:\:{I}\:\:\:\:\:\:=\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \left\{\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}−\frac{\mathrm{1}}{{x}+\mathrm{1}}\right\}{dx} \\ $$$$\:\:\:\:\:{I}\:\:\:\:\:=\:\:\:\left[\mathrm{log}\:\left({x}^{\mathrm{2}} +\mathrm{1}\right)\:+\:\mathrm{tan}^{−\mathrm{1}} \left({x}\right)\:+\:\mathrm{log}\:\left({x}+\mathrm{1}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$\:\:\:\:{I}\:\:\:\:\:=\:\:\:\:\mathrm{log}\:\mathrm{2}\:\:+\:\frac{\pi}{\mathrm{4}}\:+\:\mathrm{log}\:\mathrm{2} \\ $$$$\:\:\:\:{I}\:\:\:\:=\:\:\frac{\pi}{\mathrm{4}}\:+\:\mathrm{log}\:\mathrm{4}\: \\ $$$$\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\: \\ $$

Answered by Mathspace last updated on 25/Apr/22

1)F(x)=ln((√((1−cos((π/2)−x))/(1+cos((π/2)−x))))) =ln(√((2sin^2 ((π/4)−(x/2)))/(2cos^2 ((π/4)−(x/2))))) =ln∣tan((π/4)−(x/2)) ⇒ F^′ (x)=((−(1/2)(1+tan^2 ((π/4)−(x/2))))/(tan((π/4)−(x/2))))

$$ \\ $$$$\left.\mathrm{1}\right){F}\left({x}\right)={ln}\left(\sqrt{\frac{\mathrm{1}−{cos}\left(\frac{\pi}{\mathrm{2}}−{x}\right)}{\mathrm{1}+{cos}\left(\frac{\pi}{\mathrm{2}}−{x}\right)}}\right) \\ $$$$={ln}\sqrt{\frac{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}−\frac{{x}}{\mathrm{2}}\right)}{\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}−\frac{{x}}{\mathrm{2}}\right)}} \\ $$$$={ln}\mid{tan}\left(\frac{\pi}{\mathrm{4}}−\frac{{x}}{\mathrm{2}}\right)\:\Rightarrow \\ $$$${F}^{'} \left({x}\right)=\frac{−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+{tan}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}−\frac{{x}}{\mathrm{2}}\right)\right)}{{tan}\left(\frac{\pi}{\mathrm{4}}−\frac{{x}}{\mathrm{2}}\right)} \\ $$

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