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Question Number 81591 by john santu last updated on 14/Feb/20

if g(−2)=−5 and   g′(x)= (x^2 /(cos^2 (x)+1))  find g(4)

$$\mathrm{if}\:\mathrm{g}\left(−\mathrm{2}\right)=−\mathrm{5}\:\mathrm{and}\: \\ $$$$\mathrm{g}'\left(\mathrm{x}\right)=\:\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{x}\right)+\mathrm{1}} \\ $$$$\mathrm{find}\:\mathrm{g}\left(\mathrm{4}\right)\: \\ $$

Commented by mr W last updated on 14/Feb/20

g(4)=g(−2)+∫_(−2) ^4 (x^2 /(cos^2  x+1))dx  ≈−5+15.650787=10.650787    have you got an exact solution?

$${g}\left(\mathrm{4}\right)={g}\left(−\mathrm{2}\right)+\int_{−\mathrm{2}} ^{\mathrm{4}} \frac{{x}^{\mathrm{2}} }{\mathrm{cos}^{\mathrm{2}} \:{x}+\mathrm{1}}{dx} \\ $$$$\approx−\mathrm{5}+\mathrm{15}.\mathrm{650787}=\mathrm{10}.\mathrm{650787} \\ $$$$ \\ $$$${have}\:{you}\:{got}\:{an}\:{exact}\:{solution}? \\ $$

Commented by john santu last updated on 14/Feb/20

how get 15.650787 ?  i have not got the answer .

$$\mathrm{how}\:\mathrm{get}\:\mathrm{15}.\mathrm{650787}\:? \\ $$$$\mathrm{i}\:\mathrm{have}\:\mathrm{not}\:\mathrm{got}\:\mathrm{the}\:\mathrm{answer}\:. \\ $$

Commented by mr W last updated on 14/Feb/20

Commented by john santu last updated on 14/Feb/20

solving by wolframalpha

$$\mathrm{solving}\:\mathrm{by}\:\mathrm{wolframalpha} \\ $$

Commented by mr W last updated on 14/Feb/20

yes. for numerical integral there are  many tools. nobody calculates by hand.

$${yes}.\:{for}\:{numerical}\:{integral}\:{there}\:{are} \\ $$$${many}\:{tools}.\:{nobody}\:{calculates}\:{by}\:{hand}. \\ $$

Commented by mathmax by abdo last updated on 14/Feb/20

g^′ (x)=(x^2 /(1+cos^2 x)) ⇒g(x)=∫_(−2) ^x  (t^2 /(1+cos^2 t))dt +g(−2)  =∫_(−2) ^x  (t^2 /(1+cos^2 t))dt −5 ⇒g(4)=∫_(−2) ^4  (t^2 /(1+cos^2 t))dt −5  ∫_(−2) ^4  (t^2 /(1+cos^2 t))dt =∫_(−2) ^4  (t^2 /(1+((1+cos(2t))/2)))dt  =2 ∫_(−2) ^4  (t^2 /(3+cos(2t)))dt =_(2t=u)   2 ∫_(−4) ^8  (u^2 /(4(3+cosu)))×(du/2)  =(1/4) ∫_(−4) ^8  (u^2 /(3+cosu))du  =(1/(12)) ∫_(−4) ^8   (u^2 /(1+((1/3)cosu)))du =(1/(12))∫_(−4) ^8 u^2 (Σ_(n=0) ^∞ (−1)^n ((1/3))^n  cos^n (u)du  =(1/(12))Σ_(n=0) ^∞ (−(1/3))^n  ∫_(−4) ^8  u^2  cos^n udu   ....be continued...

$${g}^{'} \left({x}\right)=\frac{{x}^{\mathrm{2}} }{\mathrm{1}+{cos}^{\mathrm{2}} {x}}\:\Rightarrow{g}\left({x}\right)=\int_{−\mathrm{2}} ^{{x}} \:\frac{{t}^{\mathrm{2}} }{\mathrm{1}+{cos}^{\mathrm{2}} {t}}{dt}\:+{g}\left(−\mathrm{2}\right) \\ $$$$=\int_{−\mathrm{2}} ^{{x}} \:\frac{{t}^{\mathrm{2}} }{\mathrm{1}+{cos}^{\mathrm{2}} {t}}{dt}\:−\mathrm{5}\:\Rightarrow{g}\left(\mathrm{4}\right)=\int_{−\mathrm{2}} ^{\mathrm{4}} \:\frac{{t}^{\mathrm{2}} }{\mathrm{1}+{cos}^{\mathrm{2}} {t}}{dt}\:−\mathrm{5} \\ $$$$\int_{−\mathrm{2}} ^{\mathrm{4}} \:\frac{{t}^{\mathrm{2}} }{\mathrm{1}+{cos}^{\mathrm{2}} {t}}{dt}\:=\int_{−\mathrm{2}} ^{\mathrm{4}} \:\frac{{t}^{\mathrm{2}} }{\mathrm{1}+\frac{\mathrm{1}+{cos}\left(\mathrm{2}{t}\right)}{\mathrm{2}}}{dt} \\ $$$$=\mathrm{2}\:\int_{−\mathrm{2}} ^{\mathrm{4}} \:\frac{{t}^{\mathrm{2}} }{\mathrm{3}+{cos}\left(\mathrm{2}{t}\right)}{dt}\:=_{\mathrm{2}{t}={u}} \:\:\mathrm{2}\:\int_{−\mathrm{4}} ^{\mathrm{8}} \:\frac{{u}^{\mathrm{2}} }{\mathrm{4}\left(\mathrm{3}+{cosu}\right)}×\frac{{du}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\:\int_{−\mathrm{4}} ^{\mathrm{8}} \:\frac{{u}^{\mathrm{2}} }{\mathrm{3}+{cosu}}{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{12}}\:\int_{−\mathrm{4}} ^{\mathrm{8}} \:\:\frac{{u}^{\mathrm{2}} }{\mathrm{1}+\left(\frac{\mathrm{1}}{\mathrm{3}}{cosu}\right)}{du}\:=\frac{\mathrm{1}}{\mathrm{12}}\int_{−\mathrm{4}} ^{\mathrm{8}} {u}^{\mathrm{2}} \left(\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{{n}} \:{cos}^{{n}} \left({u}\right){du}\right. \\ $$$$=\frac{\mathrm{1}}{\mathrm{12}}\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\frac{\mathrm{1}}{\mathrm{3}}\right)^{{n}} \:\int_{−\mathrm{4}} ^{\mathrm{8}} \:{u}^{\mathrm{2}} \:{cos}^{{n}} {udu}\:\:\:....{be}\:{continued}... \\ $$

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