Question Number 201817 by Calculusboy last updated on 13/Dec/23 | ||
![]() | ||
Answered by witcher3 last updated on 13/Dec/23 | ||
![]() | ||
$$=\int\left(\mathrm{sin}\left(\mathrm{ln}\left(\mathrm{x}+\mathrm{1}\right)\right)\left(\sqrt{\mathrm{x}+\mathrm{1}}−\mathrm{1}\right)\right)^{\mathrm{2}} \mathrm{dx} \\ $$$$=\int\mathrm{sin}^{\mathrm{2}} \left(\mathrm{ln}\left(\mathrm{x}+\mathrm{1}\right)\right)\left(\mathrm{x}+\mathrm{2}−\mathrm{2}\sqrt{\mathrm{x}+\mathrm{1}}\right) \\ $$$$\mathrm{ln}\left(\sqrt{\mathrm{x}+\mathrm{1}}\right)=\mathrm{t} \\ $$$$\mathrm{x}+\mathrm{1}=\mathrm{e}^{\mathrm{2t}} \\ $$$$\Leftrightarrow\int\mathrm{sin}^{\mathrm{2}} \left(\mathrm{2t}\right)\left(\mathrm{e}^{\mathrm{2t}} +\mathrm{1}−\mathrm{2e}^{\mathrm{t}} \right).\mathrm{2e}^{\mathrm{2t}} \mathrm{dt} \\ $$$$\mathrm{basic}\:\mathrm{calculus} \\ $$ | ||
Commented by justenspi last updated on 15/Dec/23 | ||
![]() | ||
$${sir}\:{Can}\:{you}\:{help}\:{me}\:{with}\:{this} \\ $$$$ \\ $$ | ||
Commented by justenspi last updated on 15/Dec/23 | ||
![]() | ||
Commented by Calculusboy last updated on 15/Dec/23 | ||
![]() | ||
$$\boldsymbol{{nice}}\:\boldsymbol{{sir}},\boldsymbol{{i}}\:\boldsymbol{{don}}\:\boldsymbol{{get}}\:\boldsymbol{{am}} \\ $$ | ||