Question Number 201516 by sonukgindia last updated on 08/Dec/23 Answered by Calculusboy last updated on 08/Dec/23 $$\boldsymbol{{Solution}}:\:\boldsymbol{{let}}\:\boldsymbol{{y}}=\frac{\boldsymbol{\pi}}{\mathrm{2}}−\boldsymbol{{x}}\:\:\:\boldsymbol{{dy}}=−\boldsymbol{{dx}} \\ $$$${when}\:\boldsymbol{{x}}=\frac{\boldsymbol{\pi}}{\mathrm{2}}\:\:\:\boldsymbol{{y}}=\mathrm{0}\:\:\boldsymbol{{and}}\:\boldsymbol{{when}}\:\boldsymbol{{x}}=\mathrm{0}\:\:\boldsymbol{{y}}=\frac{\boldsymbol{\pi}}{\mathrm{2}} \\ $$$$\boldsymbol{{I}}=\int_{\frac{\boldsymbol{\pi}}{\mathrm{2}}} ^{\mathrm{0}} \:\frac{\mathrm{1}}{\mathrm{1}+\left[\boldsymbol{{tan}}\left(\frac{\boldsymbol{\pi}}{\mathrm{2}}−\boldsymbol{{y}}\right)\right]^{\boldsymbol{{n}}} }\left(−\boldsymbol{{dy}}\right)\:\:\:\Leftrightarrow\:\:\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}}…
Question Number 201517 by sonukgindia last updated on 08/Dec/23 Answered by Calculusboy last updated on 08/Dec/23 $$\boldsymbol{{Solution}}:\:\boldsymbol{{let}}\:\boldsymbol{\theta}=\frac{\boldsymbol{\pi}}{\mathrm{2}}−\boldsymbol{{x}}\:\:\boldsymbol{{d}\theta}=−\boldsymbol{{dx}} \\ $$$$\boldsymbol{{when}}\:\boldsymbol{{x}}=\frac{\boldsymbol{\pi}}{\mathrm{2}}\:\:\:\boldsymbol{\theta}=\mathrm{0}\:\:\boldsymbol{{and}}\:\boldsymbol{{when}}\:\boldsymbol{{x}}=\mathrm{0}\:\:\boldsymbol{\theta}=\frac{\boldsymbol{\pi}}{\mathrm{2}} \\ $$$$\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \frac{\mathrm{1}}{\mathrm{1}+\left(\frac{\mathrm{1}}{\boldsymbol{{tanx}}}\right)^{\boldsymbol{{n}}} }\boldsymbol{{dx}}\:\:\Leftrightarrow\:\:\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}}…
Question Number 201519 by vahid last updated on 08/Dec/23 Answered by cortano12 last updated on 08/Dec/23 $$\left(\mathrm{1}\right)\:\int\:\frac{\mathrm{sin}\:^{\mathrm{3}} \mathrm{x}}{\mathrm{cos}\:^{\mathrm{6}} \mathrm{x}}\:\mathrm{dx}\:=\:−\int\:\frac{\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}}{\mathrm{cos}\:^{\mathrm{6}} \mathrm{x}}\:\mathrm{d}\left(\mathrm{cos}\:\mathrm{x}\right) \\ $$ Answered by…
Question Number 201544 by sonukgindia last updated on 08/Dec/23 Answered by aleks041103 last updated on 09/Dec/23 $${J}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{4}{sin}^{\mathrm{2}} \left({ln}\left({x}\right)\right)}{{ln}\left(\mathrm{1}/{x}\right)}{dx}=−\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{4}{sin}^{\mathrm{2}} \left(\mathrm{1}.{ln}\left({x}\right)\right)}{{ln}\left({x}\right)}{dx} \\ $$$${I}\left({s}\right)=−\int_{\mathrm{0}}…
Question Number 201545 by Calculusboy last updated on 08/Dec/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 201546 by Calculusboy last updated on 08/Dec/23 $$\:\:\:\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}? \\ $$ Commented by Calculusboy…
Question Number 201514 by sonukgindia last updated on 08/Dec/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 201547 by Calculusboy last updated on 08/Dec/23 Answered by mr W last updated on 08/Dec/23 $${let}\:{t}=\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}+{x} \\ $$$$\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}−{x}\right){t}=\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}−{x}\:\right)\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}+{x}\right)=\mathrm{2} \\…
Question Number 201515 by sonukgindia last updated on 08/Dec/23 Answered by Calculusboy last updated on 08/Dec/23 $$\boldsymbol{{Solution}}:\:\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\frac{\mathrm{1}}{\mathrm{1}+\left(\frac{\mathrm{1}}{\boldsymbol{{tanx}}}\right)^{\sqrt{\mathrm{2}}} }\boldsymbol{{dx}}\:\:\Leftrightarrow\:\:\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\frac{\mathrm{1}}{\frac{\left(\boldsymbol{{tanx}}\right)^{\sqrt{\mathrm{2}}} +\mathrm{1}}{\left(\boldsymbol{{tanx}}\right)^{\sqrt{\mathrm{2}}} }}\boldsymbol{{dx}} \\…
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…