Question Number 78621 by mathmax by abdo last updated on 19/Jan/20
$${calculate}\:{lim}_{{x}\rightarrow\mathrm{1}} \:\:\:\:\int_{{x}} ^{{x}^{\mathrm{3}} } \:\:\frac{{sh}\left({xt}^{\mathrm{2}} \right)}{{sin}\left({xt}\right)}{dt} \\ $$
Commented by mathmax by abdo last updated on 22/Jan/20
![let f(x)=∫_x ^x^3 ((sh(xt^2 ))/(sin(xt)))dt ⇒f(x)=_(xt=u) ∫_x^2 ^x^4 ((sh(x(u^2 /x^2 )))/(sin(u)))(du/x) =(1/x) ∫_x^2 ^x^4 ((sh((1/x)u^2 ))/(sinu))du ∃ c_x ∈]x^2 ,x^4 [ /f(x)=(1/x)sh((c_x ^2 /x))∫_x^2 ^x^4 (du/(sinu)) changement tan((u/2))=z give ∫_x^2 ^x^4 (du/(sinu)) =∫_(tan((x^2 /2))) ^(tan((x^4 /2))) (1/((2z)/(1+z^2 )))((2dz)/(1+z^2 )) =∫_(tan((x^2 /2))) ^(tan((x^4 /2))) (dz/z) =ln∣tan((x^4 /2))∣−ln∣tan((x^2 /2))∣ ⇒lim_(x→1) f(x) =lim_(x→1) (1/x)sh((c_x ^2 /x))ln∣((tan((x^4 /2)))/(tan((x^2 /2))))∣ =sh(1)×ln(1) =0](https://www.tinkutara.com/question/Q79048.png)
$${let}\:{f}\left({x}\right)=\int_{{x}} ^{{x}^{\mathrm{3}} } \:\frac{{sh}\left({xt}^{\mathrm{2}} \right)}{{sin}\left({xt}\right)}{dt}\:\Rightarrow{f}\left({x}\right)=_{{xt}={u}} \:\int_{{x}^{\mathrm{2}} } ^{{x}^{\mathrm{4}} } \:\:\frac{{sh}\left({x}\frac{{u}^{\mathrm{2}} }{{x}^{\mathrm{2}} }\right)}{{sin}\left({u}\right)}\frac{{du}}{{x}} \\ $$$$\left.=\frac{\mathrm{1}}{{x}}\:\int_{{x}^{\mathrm{2}} } ^{{x}^{\mathrm{4}} } \:\:\:\frac{{sh}\left(\frac{\mathrm{1}}{{x}}{u}^{\mathrm{2}} \right)}{{sinu}}{du}\:\:\:\exists\:{c}_{{x}} \:\in\right]{x}^{\mathrm{2}} ,{x}^{\mathrm{4}} \left[\:/{f}\left({x}\right)=\frac{\mathrm{1}}{{x}}{sh}\left(\frac{{c}_{{x}} ^{\mathrm{2}} }{{x}}\right)\int_{{x}^{\mathrm{2}} } ^{{x}^{\mathrm{4}} } \:\:\frac{{du}}{{sinu}}\right. \\ $$$${changement}\:{tan}\left(\frac{{u}}{\mathrm{2}}\right)={z}\:{give} \\ $$$$\int_{{x}^{\mathrm{2}} } ^{{x}^{\mathrm{4}} } \:\frac{{du}}{{sinu}}\:=\int_{{tan}\left(\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)} ^{{tan}\left(\frac{{x}^{\mathrm{4}} }{\mathrm{2}}\right)} \:\frac{\mathrm{1}}{\frac{\mathrm{2}{z}}{\mathrm{1}+{z}^{\mathrm{2}} }}\frac{\mathrm{2}{dz}}{\mathrm{1}+{z}^{\mathrm{2}} }\:=\int_{{tan}\left(\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)} ^{{tan}\left(\frac{{x}^{\mathrm{4}} }{\mathrm{2}}\right)} \:\frac{{dz}}{{z}}\:={ln}\mid{tan}\left(\frac{{x}^{\mathrm{4}} }{\mathrm{2}}\right)\mid−{ln}\mid{tan}\left(\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)\mid \\ $$$$\Rightarrow{lim}_{{x}\rightarrow\mathrm{1}} {f}\left({x}\right)\:={lim}_{{x}\rightarrow\mathrm{1}} \:\frac{\mathrm{1}}{{x}}{sh}\left(\frac{{c}_{{x}} ^{\mathrm{2}} }{{x}}\right){ln}\mid\frac{{tan}\left(\frac{{x}^{\mathrm{4}} }{\mathrm{2}}\right)}{{tan}\left(\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)}\mid \\ $$$$={sh}\left(\mathrm{1}\right)×{ln}\left(\mathrm{1}\right)\:=\mathrm{0} \\ $$