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Question Number 117511 by Canovas last updated on 12/Oct/20

Answered by bemath last updated on 12/Oct/20

∫ (dx/(cos^2 x (4−9tan^2 x))) = ∫ ((sec^2 x)/(4−9tan^2 x)) dx = ∫ ((d(tan x))/(4−9tan^2 x)) = ∫ (dϕ/(4−9ϕ^2 )) =∫ (dϕ/((2+3ϕ)(2−3ϕ))) =(1/4) ∫ (1/(2−3ϕ))+(1/(2+3ϕ)) dϕ = (1/4){−(1/3)ln(2−3ϕ)+(1/3)ln (2+3ϕ)}+c = (1/(12)) ln ∣((2+3ϕ)/(2−3ϕ))∣+ c = (1/(12)) ln ∣((2+3tan x)/(2−3tan x)) ∣+ c

$$\int\:\frac{\mathrm{dx}}{\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\:\left(\mathrm{4}−\mathrm{9tan}\:^{\mathrm{2}} \mathrm{x}\right)}\:=\:\int\:\frac{\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}}{\mathrm{4}−\mathrm{9tan}\:^{\mathrm{2}} \mathrm{x}}\:\mathrm{dx} \\ $$$$=\:\int\:\frac{\mathrm{d}\left(\mathrm{tan}\:\mathrm{x}\right)}{\mathrm{4}−\mathrm{9tan}\:^{\mathrm{2}} \mathrm{x}}\:=\:\int\:\frac{\mathrm{d}\varphi}{\mathrm{4}−\mathrm{9}\varphi^{\mathrm{2}} } \\ $$$$=\int\:\frac{\mathrm{d}\varphi}{\left(\mathrm{2}+\mathrm{3}\varphi\right)\left(\mathrm{2}−\mathrm{3}\varphi\right)}\:=\frac{\mathrm{1}}{\mathrm{4}}\:\int\:\frac{\mathrm{1}}{\mathrm{2}−\mathrm{3}\varphi}+\frac{\mathrm{1}}{\mathrm{2}+\mathrm{3}\varphi}\:\mathrm{d}\varphi \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{4}}\left\{−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}\left(\mathrm{2}−\mathrm{3}\varphi\right)+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}\:\left(\mathrm{2}+\mathrm{3}\varphi\right)\right\}+\mathrm{c}\: \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{12}}\:\mathrm{ln}\:\mid\frac{\mathrm{2}+\mathrm{3}\varphi}{\mathrm{2}−\mathrm{3}\varphi}\mid+\:\mathrm{c} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{12}}\:\mathrm{ln}\:\mid\frac{\mathrm{2}+\mathrm{3tan}\:\mathrm{x}}{\mathrm{2}−\mathrm{3tan}\:\mathrm{x}}\:\mid+\:\mathrm{c} \\ $$

Commented by Canovas last updated on 12/Oct/20

Really I haven′t understood sir

$${Really}\:{I}\:{haven}'{t}\:{understood}\:{sir} \\ $$$$ \\ $$

Commented by bobhans last updated on 12/Oct/20

it is by substitution sir. ϕ = tan x so dϕ = sec^2 x dx

$$\mathrm{it}\:\mathrm{is}\:\mathrm{by}\:\mathrm{substitution}\:\mathrm{sir}. \\ $$$$\varphi\:=\:\mathrm{tan}\:\mathrm{x}\:\mathrm{so}\:\mathrm{d}\varphi\:=\:\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}\:\mathrm{dx} \\ $$

Answered by 1549442205PVT last updated on 12/Oct/20

F=∫(dx/(4cos^2 x−9sin^2 x))=∫((1/(cos^2 x))/(4−9tan^2 x))dx =−∫((1+tan^2 x)/(9tan^2 x−4))dx.Put tanx=t ⇒dt=(1+t^2 )dx F=−∫(((1+t^2 )dt)/((1+t^2 )(9t^2 −4)))=+∫(dt/(9t^2 −4)) =−(1/9)∫(dt/(t^2 −(4/9)))=−(1/9)×(3/4)∫((1/(t−(2/3)))−(1/(t+(2/3))))dt =−(1/(12))ln∣((3t−2)/(3t+2))∣+C=(1/(12))ln∣((3tanx+2)/(3tanx−2))∣+C

$$\mathrm{F}=\int\frac{\mathrm{dx}}{\mathrm{4cos}^{\mathrm{2}} \mathrm{x}−\mathrm{9sin}^{\mathrm{2}} \mathrm{x}}=\int\frac{\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \mathrm{x}}}{\mathrm{4}−\mathrm{9tan}^{\mathrm{2}} \mathrm{x}}\mathrm{dx} \\ $$$$=−\int\frac{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{x}}{\mathrm{9tan}^{\mathrm{2}} \mathrm{x}−\mathrm{4}}\mathrm{dx}.\mathrm{Put}\:\mathrm{tanx}=\mathrm{t} \\ $$$$\Rightarrow\mathrm{dt}=\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\mathrm{dx} \\ $$$$\mathrm{F}=−\int\frac{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\mathrm{dt}}{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\left(\mathrm{9t}^{\mathrm{2}} −\mathrm{4}\right)}=+\int\frac{\mathrm{dt}}{\mathrm{9t}^{\mathrm{2}} −\mathrm{4}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{9}}\int\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} −\frac{\mathrm{4}}{\mathrm{9}}}=−\frac{\mathrm{1}}{\mathrm{9}}×\frac{\mathrm{3}}{\mathrm{4}}\int\left(\frac{\mathrm{1}}{\mathrm{t}−\frac{\mathrm{2}}{\mathrm{3}}}−\frac{\mathrm{1}}{\mathrm{t}+\frac{\mathrm{2}}{\mathrm{3}}}\right)\mathrm{dt} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{12}}\mathrm{ln}\mid\frac{\mathrm{3t}−\mathrm{2}}{\mathrm{3t}+\mathrm{2}}\mid+\mathrm{C}=\frac{\mathrm{1}}{\mathrm{12}}\mathrm{ln}\mid\frac{\mathrm{3tanx}+\mathrm{2}}{\mathrm{3tanx}−\mathrm{2}}\mid+\mathrm{C} \\ $$

Answered by mathmax by abdo last updated on 12/Oct/20

I =∫ (dx/(4cos^2 x−9sin^2 x)) ⇒ I =∫ (dx/(4((1+cos(2x))/2)−9((1−cos(2x))/2))) =∫ (dx/(2+2cos(2x)−(9/2)+(9/2)cos(2x))) =∫ ((2dx)/(4+4cos(2x)−9+9cos(2x))) =∫ ((2dx)/(13cos(2x)−5)) =_(2x=t) ∫ (dt/(13 cost−5)) =_(tan((t/2))=u) ∫ ((2du)/((1+u^2 )(13((1−u^2 )/(1+u^2 ))−5))) =∫ ((2du)/(13(1−u^2 )−5(1+u^2 ))) =∫ ((2du)/(13−13u^2 −5−5u^2 )) =∫ ((2du)/(8−18u^2 )) =∫ (du/(4−9u^2 )) =∫ (du/((2−3u)(2+3u))) =(1/4)∫((1/(2−3u))+(1/(2+3u)))du=(1/4)ln∣4−9u^2 ∣ +c =(1/4)ln∣4−9tan^2 (x)∣ +c

$$\mathrm{I}\:=\int\:\:\frac{\mathrm{dx}}{\mathrm{4cos}^{\mathrm{2}} \mathrm{x}−\mathrm{9sin}^{\mathrm{2}} \mathrm{x}}\:\Rightarrow\:\mathrm{I}\:=\int\:\:\frac{\mathrm{dx}}{\mathrm{4}\frac{\mathrm{1}+\mathrm{cos}\left(\mathrm{2x}\right)}{\mathrm{2}}−\mathrm{9}\frac{\mathrm{1}−\mathrm{cos}\left(\mathrm{2x}\right)}{\mathrm{2}}} \\ $$$$=\int\:\:\frac{\mathrm{dx}}{\mathrm{2}+\mathrm{2cos}\left(\mathrm{2x}\right)−\frac{\mathrm{9}}{\mathrm{2}}+\frac{\mathrm{9}}{\mathrm{2}}\mathrm{cos}\left(\mathrm{2x}\right)}\:=\int\:\frac{\mathrm{2dx}}{\mathrm{4}+\mathrm{4cos}\left(\mathrm{2x}\right)−\mathrm{9}+\mathrm{9cos}\left(\mathrm{2x}\right)} \\ $$$$=\int\:\frac{\mathrm{2dx}}{\mathrm{13cos}\left(\mathrm{2x}\right)−\mathrm{5}}\:=_{\mathrm{2x}=\mathrm{t}} \:\:\:\:\int\:\frac{\mathrm{dt}}{\mathrm{13}\:\mathrm{cost}−\mathrm{5}}\:=_{\mathrm{tan}\left(\frac{\mathrm{t}}{\mathrm{2}}\right)=\mathrm{u}} \\ $$$$\int\:\:\:\frac{\mathrm{2du}}{\left(\mathrm{1}+\mathrm{u}^{\mathrm{2}} \right)\left(\mathrm{13}\frac{\mathrm{1}−\mathrm{u}^{\mathrm{2}} }{\mathrm{1}+\mathrm{u}^{\mathrm{2}} }−\mathrm{5}\right)}\:=\int\:\frac{\mathrm{2du}}{\mathrm{13}\left(\mathrm{1}−\mathrm{u}^{\mathrm{2}} \right)−\mathrm{5}\left(\mathrm{1}+\mathrm{u}^{\mathrm{2}} \right)} \\ $$$$=\int\:\frac{\mathrm{2du}}{\mathrm{13}−\mathrm{13u}^{\mathrm{2}} −\mathrm{5}−\mathrm{5u}^{\mathrm{2}} }\:=\int\:\frac{\mathrm{2du}}{\mathrm{8}−\mathrm{18u}^{\mathrm{2}} }\:=\int\:\:\frac{\mathrm{du}}{\mathrm{4}−\mathrm{9u}^{\mathrm{2}} } \\ $$$$=\int\:\:\frac{\mathrm{du}}{\left(\mathrm{2}−\mathrm{3u}\right)\left(\mathrm{2}+\mathrm{3u}\right)}\:=\frac{\mathrm{1}}{\mathrm{4}}\int\left(\frac{\mathrm{1}}{\mathrm{2}−\mathrm{3u}}+\frac{\mathrm{1}}{\mathrm{2}+\mathrm{3u}}\right)\mathrm{du}=\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\mid\mathrm{4}−\mathrm{9u}^{\mathrm{2}} \mid\:+\mathrm{c} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\mid\mathrm{4}−\mathrm{9tan}^{\mathrm{2}} \left(\mathrm{x}\right)\mid\:+\mathrm{c} \\ $$

Commented by bobhans last updated on 13/Oct/20

how do you get (1/4)∫ ((1/(2−3u)) +(1/(2+3u))) du = (1/4)ln ∣4−9u^2 ∣ ? it not correct

$$\mathrm{how}\:\mathrm{do}\:\mathrm{you}\:\mathrm{get}\:\frac{\mathrm{1}}{\mathrm{4}}\int\:\left(\frac{\mathrm{1}}{\mathrm{2}−\mathrm{3u}}\:+\frac{\mathrm{1}}{\mathrm{2}+\mathrm{3u}}\right)\:\mathrm{du} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\mid\mathrm{4}−\mathrm{9u}^{\mathrm{2}} \mid\:?\:\mathrm{it}\:\mathrm{not}\:\mathrm{correct} \\ $$

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