Question Number 179107 by Acem last updated on 25/Oct/22
$${I}=\:\int\:\frac{{x}^{\mathrm{2}} }{\mathrm{sin}\:\left(\mathrm{2arctan}\:\left({e}^{{x}} \right)\right)}\:{dx}\:\:,\:{Find}\:{I} \\ $$
Answered by MJS_new last updated on 25/Oct/22
![∫(x^2 /(sun 2arctan e^x ))dx=∫x^2 cosh x dx= [by parts] =x^2 sinh x −2∫xsinh x dx= [by parts] =x^2 sinh x −2xcosh x +2∫sinh x dx = =x^2 sinh x −2xcosh x +2sinh x = =(x^2 +2)sinh x −2xcosh x +C](https://www.tinkutara.com/question/Q179115.png)
$$\int\frac{{x}^{\mathrm{2}} }{\mathrm{sun}\:\mathrm{2arctan}\:\mathrm{e}^{{x}} }{dx}=\int{x}^{\mathrm{2}} \mathrm{cosh}\:{x}\:{dx}= \\ $$$$\:\:\:\:\:\left[\mathrm{by}\:\mathrm{parts}\right] \\ $$$$={x}^{\mathrm{2}} \mathrm{sinh}\:{x}\:−\mathrm{2}\int{x}\mathrm{sinh}\:{x}\:{dx}= \\ $$$$\:\:\:\:\:\left[\mathrm{by}\:\mathrm{parts}\right] \\ $$$$={x}^{\mathrm{2}} \mathrm{sinh}\:{x}\:−\mathrm{2}{x}\mathrm{cosh}\:{x}\:+\mathrm{2}\int\mathrm{sinh}\:{x}\:{dx}\:= \\ $$$$={x}^{\mathrm{2}} \mathrm{sinh}\:{x}\:−\mathrm{2}{x}\mathrm{cosh}\:{x}\:+\mathrm{2sinh}\:{x}\:= \\ $$$$=\left({x}^{\mathrm{2}} +\mathrm{2}\right)\mathrm{sinh}\:{x}\:−\mathrm{2}{x}\mathrm{cosh}\:{x}\:+{C} \\ $$
Commented by Acem last updated on 25/Oct/22
$${Exactly}!\:{good}\:{Sir}! \\ $$