Question-201817

Question Number 201817 by Calculusboy last updated on 13/Dec/23

Answered by witcher3 last updated on 13/Dec/23

=∫(sin(ln(x+1))((√(x+1))−1))^2 dx =∫sin^2 (ln(x+1))(x+2−2(√(x+1))) ln((√(x+1)))=t x+1=e^(2t) ⇔∫sin^2 (2t)(e^(2t) +1−2e^t ).2e^(2t) dt basic calculus

$$=\int\left(\mathrm{sin}\left(\mathrm{ln}\left(\mathrm{x}+\mathrm{1}\right)\right)\left(\sqrt{\mathrm{x}+\mathrm{1}}−\mathrm{1}\right)\right)^{\mathrm{2}} \mathrm{dx} \\ $$$$=\int\mathrm{sin}^{\mathrm{2}} \left(\mathrm{ln}\left(\mathrm{x}+\mathrm{1}\right)\right)\left(\mathrm{x}+\mathrm{2}−\mathrm{2}\sqrt{\mathrm{x}+\mathrm{1}}\right) \\ $$$$\mathrm{ln}\left(\sqrt{\mathrm{x}+\mathrm{1}}\right)=\mathrm{t} \\ $$$$\mathrm{x}+\mathrm{1}=\mathrm{e}^{\mathrm{2t}} \\ $$$$\Leftrightarrow\int\mathrm{sin}^{\mathrm{2}} \left(\mathrm{2t}\right)\left(\mathrm{e}^{\mathrm{2t}} +\mathrm{1}−\mathrm{2e}^{\mathrm{t}} \right).\mathrm{2e}^{\mathrm{2t}} \mathrm{dt} \\ $$$$\mathrm{basic}\:\mathrm{calculus} \\ $$

Commented by justenspi last updated on 15/Dec/23