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Question Number 78621 by mathmax by abdo last updated on 19/Jan/20

$$calculate{lim}_{x\to 1}{\int }_x^{x^3}\frac{sh\left({xt}^2\right)}{sin\left(xt\right)}dt$$

Commented by mathmax by abdo last updated on 22/Jan/20

$$letf\left(x\right)={\int }_x^{x^3}\frac{sh\left({xt}^2\right)}{sin\left(xt\right)}dt\Rightarrow f\left(x\right){=}_{xt=u}{\int }_{x^2}^{x^4}\frac{sh\left(x\frac{u^2}{x^2}\right)}{sin\left(u\right)}\frac{du}{x}$$$$=\frac{1}{x}{\int }_{x^2}^{x^4}\frac{sh\left(\frac{1}{x}{u}^2\right)}{sinu}du\mathrm{\exists }{c}_x\in ]x^2,x^4[/f\left(x\right)=\frac{1}{x}sh\left(\frac{c_x^2}{x}\right){\int }_{x^2}^{x^4}\frac{du}{sinu}$$$$changementtan\left(\frac{u}{2}\right)=zgive$$$${\int }_{x^2}^{x^4}\frac{du}{sinu}={\int }_{tan\left(\frac{x^2}{2}\right)}^{tan\left(\frac{x^4}{2}\right)}\frac{1}{\frac{2z}{1+z^2}}\frac{2dz}{1+z^2}={\int }_{tan\left(\frac{x^2}{2}\right)}^{tan\left(\frac{x^4}{2}\right)}\frac{dz}{z}=ln\mid tan\left(\frac{x^4}{2}\right)\mid -ln\mid tan\left(\frac{x^2}{2}\right)\mid$$$$\Rightarrow {lim}_{x\to 1}f\left(x\right)={lim}_{x\to 1}\frac{1}{x}sh\left(\frac{c_x^2}{x}\right)ln\mid \frac{tan\left(\frac{x^4}{2}\right)}{tan\left(\frac{x^2}{2}\right)}\mid$$$$=sh\left(1\right)\times ln\left(1\right)=0$$

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