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Question Number 128349 by mathocean1 last updated on 06/Jan/21
function g is defined in [0;(π/4)] by g(x)=((sinx)/(cos^3 x)). 1. Determinate a;b ∈R such that g′(x)=(a/(cos^4 x)) + (b/(cos^2 x)). 2.Deduct the primitive G of the function t(x)=(1/(cos^4 x)) such that t((π/4))=1.
$${function}\:{g}\:{is}\:{defined}\:{in}\:\left[\mathrm{0};\frac{\pi}{\mathrm{4}}\right] \\ $$$${by}\:{g}\left({x}\right)=\frac{{sinx}}{{cos}^{\mathrm{3}} {x}}. \\ $$$$\mathrm{1}.\:{Determinate}\:{a};{b}\:\in\mathbb{R}\:{such}\:{that} \\ $$$${g}'\left({x}\right)=\frac{{a}}{{cos}^{\mathrm{4}} {x}}\:+\:\frac{{b}}{{cos}^{\mathrm{2}} {x}}. \\ $$$$\mathrm{2}.{Deduct}\:{the}\:{primitive}\:{G}\:{of}\:{the}\: \\ $$$${function}\:{t}\left({x}\right)=\frac{\mathrm{1}}{{cos}^{\mathrm{4}} {x}}\:\:{such}\:{that} \\ $$$${t}\left(\frac{\pi}{\mathrm{4}}\right)=\mathrm{1}. \\ $$
Answered by mathmax by abdo last updated on 06/Jan/21
g(x)=((sinx)/(cos^3 x)) ⇒g^′ (x)=((cos^4 x−3cosx^2 (−sinx)sinx)/(cos^6 x)) =((cos^4 x+3cos^2 x sin^2 x)/(cos^6 x)) =(1/(cos^2 x)) +((3sin^2 x)/(cos^4 x)) =(1/(cos^2 x))+((3(1−cos^2 x))/(cos^4 x)) =(1/(cos^2 x))−(3/(cos^2 x))+(3/(cos^4 x))=((−2)/(cos^2 x)) +(3/(cos^4 x)) ⇒ ∫g^′ (x)dx =−2 ∫(dx/(cos^2 x)) +3∫ (dx/(cos^4 x)) ⇒ g(x)=−2∫ (dx/(cos^2 x)) +3∫ (dx/(cos^4 x)) ⇒ 3∫ (dx/(cos^4 x)) =((sinx)/(cos^3 x)) +2∫ (dx/(cos^2 x)) we have ∫ (dx/(cos^2 x)) =∫ ((2dx)/(1+cos(2x))) =_(tanx=t) 2 ∫ (dt/((1+t^2 )(1+((1−t^2 )/(1+t^2 ))))) =2 ∫ (dt/(1+t^2 +1−t^2 )) =t +c =tanx +c ⇒ ∫ (dx/(cos^4 x)) =((sinx)/(3cos^3 x)) +(2/3)tanx +C
$$\mathrm{g}\left(\mathrm{x}\right)=\frac{\mathrm{sinx}}{\mathrm{cos}^{\mathrm{3}} \mathrm{x}}\:\Rightarrow\mathrm{g}^{'} \left(\mathrm{x}\right)=\frac{\mathrm{cos}^{\mathrm{4}} \mathrm{x}−\mathrm{3cosx}^{\mathrm{2}} \left(−\mathrm{sinx}\right)\mathrm{sinx}}{\mathrm{cos}^{\mathrm{6}} \mathrm{x}} \\ $$$$=\frac{\mathrm{cos}^{\mathrm{4}} \mathrm{x}+\mathrm{3cos}^{\mathrm{2}} \mathrm{x}\:\mathrm{sin}^{\mathrm{2}} \mathrm{x}}{\mathrm{cos}^{\mathrm{6}} \mathrm{x}}\:=\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \mathrm{x}}\:+\frac{\mathrm{3sin}^{\mathrm{2}} \mathrm{x}}{\mathrm{cos}^{\mathrm{4}} \mathrm{x}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \mathrm{x}}+\frac{\mathrm{3}\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \mathrm{x}\right)}{\mathrm{cos}^{\mathrm{4}} \mathrm{x}}\:=\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \mathrm{x}}−\frac{\mathrm{3}}{\mathrm{cos}^{\mathrm{2}} \mathrm{x}}+\frac{\mathrm{3}}{\mathrm{cos}^{\mathrm{4}} \mathrm{x}}=\frac{−\mathrm{2}}{\mathrm{cos}^{\mathrm{2}} \mathrm{x}}\:+\frac{\mathrm{3}}{\mathrm{cos}^{\mathrm{4}} \mathrm{x}}\:\Rightarrow \\ $$$$\int\mathrm{g}^{'} \left(\mathrm{x}\right)\mathrm{dx}\:=−\mathrm{2}\:\int\frac{\mathrm{dx}}{\mathrm{cos}^{\mathrm{2}} \mathrm{x}}\:+\mathrm{3}\int\:\frac{\mathrm{dx}}{\mathrm{cos}^{\mathrm{4}} \mathrm{x}}\:\Rightarrow \\ $$$$\mathrm{g}\left(\mathrm{x}\right)=−\mathrm{2}\int\:\frac{\mathrm{dx}}{\mathrm{cos}^{\mathrm{2}} \mathrm{x}}\:+\mathrm{3}\int\:\frac{\mathrm{dx}}{\mathrm{cos}^{\mathrm{4}} \mathrm{x}}\:\Rightarrow \\ $$$$\mathrm{3}\int\:\frac{\mathrm{dx}}{\mathrm{cos}^{\mathrm{4}} \mathrm{x}}\:=\frac{\mathrm{sinx}}{\mathrm{cos}^{\mathrm{3}} \mathrm{x}}\:+\mathrm{2}\int\:\:\frac{\mathrm{dx}}{\mathrm{cos}^{\mathrm{2}} \mathrm{x}}\:\:\mathrm{we}\:\mathrm{have} \\ $$$$\int\:\:\frac{\mathrm{dx}}{\mathrm{cos}^{\mathrm{2}} \mathrm{x}}\:=\int\:\frac{\mathrm{2dx}}{\mathrm{1}+\mathrm{cos}\left(\mathrm{2x}\right)}\:=_{\mathrm{tanx}=\mathrm{t}} \:\mathrm{2}\:\:\int\:\:\frac{\mathrm{dt}}{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\left(\mathrm{1}+\frac{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\right)} \\ $$$$=\mathrm{2}\:\int\:\frac{\mathrm{dt}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} +\mathrm{1}−\mathrm{t}^{\mathrm{2}} }\:=\mathrm{t}\:+\mathrm{c}\:=\mathrm{tanx}\:+\mathrm{c}\:\Rightarrow \\ $$$$\int\:\:\frac{\mathrm{dx}}{\mathrm{cos}^{\mathrm{4}} \mathrm{x}}\:=\frac{\mathrm{sinx}}{\mathrm{3cos}^{\mathrm{3}} \mathrm{x}}\:+\frac{\mathrm{2}}{\mathrm{3}}\mathrm{tanx}\:+\mathrm{C} \\ $$

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