Question Number 201762 by Calculusboy last updated on 11/Dec/23 Answered by mr W last updated on 12/Dec/23 $$\int_{−\mathrm{2}} ^{\mathrm{2}} \left({x}^{\mathrm{3}} \mathrm{cos}\:\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }\:{dx} \\ $$$$=\int_{−\mathrm{2}} ^{\mathrm{2}}…
Question Number 201646 by Calculusboy last updated on 10/Dec/23 Answered by aleks041103 last updated on 10/Dec/23 $$\sqrt[{{ln}\left({x}\right)}]{{x}}={x}^{\frac{\mathrm{1}}{{ln}\left({x}\right)}} =\left({e}^{{ln}\left({x}\right)} \right)^{\frac{\mathrm{1}}{{ln}\left({x}\right)}} ={e}={const} \\ $$$$\Rightarrow\int\sqrt[{{ln}\left({x}\right)}]{{x}}\:{dx}\:=\:{ex}\:+\:{C} \\ $$ Terms…
Question Number 201644 by Calculusboy last updated on 10/Dec/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 201553 by Calculusboy last updated on 08/Dec/23 Answered by som(math1967) last updated on 09/Dec/23 $$\mathrm{1}.\:\int\frac{\mathrm{1}+{logx}−\mathrm{1}}{\left(\mathrm{1}+{logx}\right)^{\mathrm{2}} }{dx} \\ $$$$=\int\frac{{dx}}{\left(\mathrm{1}+{logx}\right)}\:−\int\frac{{dx}}{\left(\mathrm{1}+{logx}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}+{logx}}\int{dx}−\int\left\{\frac{{d}}{{dx}}×\frac{\mathrm{1}}{\mathrm{1}+{logx}}\int{dx}\right\}{dx} \\ $$$$\:\:\:\:\:\:−\int\frac{{dx}}{\left(\mathrm{1}+{logx}\right)^{\mathrm{2}}…
Question Number 201548 by Calculusboy last updated on 08/Dec/23 Answered by MathematicalUser2357 last updated on 20/Jan/24 $$\left[−\left({x}−\mathrm{1}\right)\left\{\mathrm{2}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }+\sqrt{\mathrm{1}−{x}}{x}−\mathrm{2}{x}+\sqrt{\mathrm{1}−{x}}−\mathrm{2}\sqrt{{x}+\mathrm{1}}\mathrm{log}\left(\sqrt{\mathrm{1}−{x}}+\mathrm{1}\right)−\sqrt{{x}+\mathrm{1}}\mathrm{log}\:{x}+\mathrm{2}\sqrt{{x}+\mathrm{1}}\mathrm{log}\left(\sqrt{{x}+\mathrm{1}}+\mathrm{1}\right)−\mathrm{2}\right\}−\sqrt{\mathrm{1}−{x}}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\mathrm{tanh}^{−\mathrm{1}} \sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right]+{C} \\ $$ Terms of…
Question Number 201546 by Calculusboy last updated on 08/Dec/23 $$\:\:\:\int\boldsymbol{{Sin}}\left(\boldsymbol{{Inx}}\right)\boldsymbol{{dx}} \\ $$ Commented by aleks041103 last updated on 09/Dec/23 $${is}\:{this}\:{sine}\:{of}\:{natural}\:{log}\:{of}\:{x}\:{or}\:{sth}\:{else}? \\ $$ Commented by Calculusboy…
Question Number 201510 by Calculusboy last updated on 08/Dec/23 Answered by MathematicalUser2357 last updated on 04/Jan/24 $$=−\mathrm{2tan}^{−\mathrm{1}} \left[\mathrm{cos}\left\{\frac{\mathrm{1}}{\mathrm{2}}\left({t}−\mathrm{sin}\:{t}\right)\right\}\mathrm{cosec}\left\{\frac{\mathrm{1}}{\mathrm{2}}\left({t}−\mathrm{sin}\:{t}\right)\right\}\right]+{C} \\ $$ Terms of Service Privacy Policy…
Question Number 201509 by Calculusboy last updated on 07/Dec/23 Answered by witcher3 last updated on 09/Dec/23 $$\mathrm{tack}\:\mathrm{principal}\:\:\mathrm{definition}\:\mathrm{of}\:\mathrm{Log}\left(\mathrm{z}\right)=\mathrm{ln}\mid\mathrm{z}\mid+\mathrm{iarg}\left(\mathrm{z}\right) \\ $$$$\left.\mathrm{z}\in\right]−\frac{\pi}{\mathrm{2}},\frac{\mathrm{3}\pi}{\mathrm{2}}\left[\:\right. \\ $$$$\mathrm{ln}\left(\mathrm{ix}+\mathrm{ln}\left(\mathrm{cos}\left(\mathrm{x}\right)\right)=\mathrm{ln}\mid\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{ln}^{\mathrm{2}} \left(\mathrm{cos}\left(\mathrm{x}\right)\right.}\mid+\mathrm{i}\:\mathrm{arg}\left(\mathrm{ln}\left(\mathrm{cos}\left(\mathrm{x}\right)+\mathrm{ix}\right)\right.\right. \\ $$$$\mathrm{arg}\left(\mathrm{ln}\left(\mathrm{cos}\left(\mathrm{x}\right)+\mathrm{ix}\right)\in\left[\frac{\pi}{\mathrm{2}},\pi\left[\:\right.\right.\right.…
Question Number 201456 by Calculusboy last updated on 06/Dec/23 Answered by MathematicalUser2357 last updated on 04/Jan/24 $$? \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 201452 by tri26112004 last updated on 06/Dec/23 Answered by Calculusboy last updated on 06/Dec/23 $$\int\boldsymbol{{x}}^{−\mathrm{2}} \boldsymbol{{e}}^{−\mathrm{4}\boldsymbol{{x}}} \boldsymbol{{dx}} \\ $$$$\boldsymbol{{Solution}}:\:\:\boldsymbol{{by}}\:\boldsymbol{{using}}\:\boldsymbol{{IBP}} \\ $$$$\boldsymbol{{let}}\:\boldsymbol{{u}}=\boldsymbol{{e}}^{−\mathrm{4}\boldsymbol{{x}}} \:\:\:\boldsymbol{{du}}=−\mathrm{4}\boldsymbol{{e}}^{−\mathrm{4}\boldsymbol{{x}}} \boldsymbol{{dx}}\:\:\boldsymbol{{dv}}=\boldsymbol{{x}}^{−\mathrm{2}}…