Question Number 202418 by MathematicalUser2357 last updated on 26/Dec/23 $$\mathrm{Hard}\:\mathrm{integral} \\ $$$$\int\int\int\int\int\int\int\int\int\begin{vmatrix}{{a}}&{{b}}&{{c}}\\{{f}}&{{g}}&{{h}}\\{{j}}&{{k}}&{{l}}\end{vmatrix}{dl}\:{dk}\:{dj}\:{dh}\:{dg}\:{df}\:{dc}\:{db}\:{da}= \\ $$ Answered by Frix last updated on 26/Dec/23 $$\mathrm{Not}\:\mathrm{hard}\:\mathrm{at}\:\mathrm{all} \\ $$$$=\frac{{abcfghjkl}}{\mathrm{8}}\left({a}\left({gl}−{hk}\right)−{b}\left({fl}−{hj}\right)+{c}\left({fk}−{gj}\right)\right) \\…
Question Number 202415 by MathematicalUser2357 last updated on 26/Dec/23 $$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\int{g}'\left({x}\right){f}'\left({g}\left({x}\right)\right){dx}\:\mathrm{is}… \\ $$ Answered by cortano12 last updated on 26/Dec/23 $$\:\mathrm{let}\:\mathrm{u}=\mathrm{g}\left(\mathrm{x}\right)\Rightarrow\mathrm{du}=\:\mathrm{g}'\left(\mathrm{x}\right)\:\mathrm{dx} \\ $$$$\:\mathrm{I}=\:\int\:\mathrm{g}'\left(\mathrm{x}\right)\:\mathrm{f}\:'\left(\mathrm{g}\left(\mathrm{x}\right)\right)\:\mathrm{dx}\: \\ $$$$\:\:\:=\:\int\:\mathrm{f}\:'\left(\mathrm{u}\right)\:\mathrm{du}=\:\int\:\frac{\mathrm{df}\left(\mathrm{u}\right)}{\mathrm{du}}.\:\mathrm{du} \\…
Question Number 202406 by mou0113 last updated on 26/Dec/23 Answered by witcher3 last updated on 26/Dec/23 $$\mathrm{f}\left(\mathrm{s}\right)=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{t}^{\mathrm{s}} }{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\mathrm{dt}\Rightarrow\mathrm{f}'\left(\mathrm{0}\right)=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}\left(\mathrm{x}\right)}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}}…
Question Number 202388 by Calculusboy last updated on 25/Dec/23 $$\:\:\boldsymbol{{P}}\:\boldsymbol{{rove}}\:\boldsymbol{{that}}:\:\:\:\:\int\:\frac{\boldsymbol{{dx}}}{\boldsymbol{{b}}^{\mathrm{4}} +\mathrm{2}\boldsymbol{{ax}}^{\mathrm{2}} +\boldsymbol{{c}}}=\frac{\boldsymbol{{tan}}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{2}}\sqrt{\boldsymbol{{a}}}\boldsymbol{{x}}}{\:\sqrt{\boldsymbol{{c}}+\boldsymbol{{b}}^{\mathrm{4}} }}\right)}{\:\sqrt{\mathrm{2}}\sqrt{\boldsymbol{{a}}}\sqrt{\boldsymbol{{c}}+\boldsymbol{{b}}^{\mathrm{4}} }}+\boldsymbol{{C}} \\ $$$$\boldsymbol{{if}}\:\:\boldsymbol{{a}}\centerdot\left(\boldsymbol{{c}}+\boldsymbol{{b}}^{\mathrm{4}} \right)>\mathrm{0} \\ $$$$ \\ $$ Answered by witcher3…
Question Number 202167 by tri26112004 last updated on 22/Dec/23 $$\underset{\mathrm{0}} {\int}^{\mathrm{1}} \underset{{x}} {\int}^{\mathrm{1}} {sin}\left({y}^{\mathrm{2}} \right){dydx}\:=\:¿ \\ $$ Answered by mnjuly1970 last updated on 22/Dec/23 $$\:{answer}:=\:{sin}^{\mathrm{2}}…
Question Number 202212 by Calculusboy last updated on 22/Dec/23 Commented by BOYQOBILOV last updated on 23/Dec/23 $$ \\ $$ Answered by shunmisaki007 last updated on…
Question Number 202125 by Calculusboy last updated on 21/Dec/23 $$\int\:\frac{\boldsymbol{{sin}}\left(\mathrm{3}\boldsymbol{{x}}\right)}{\mathrm{1}+\boldsymbol{{sin}}^{\mathrm{3}} \boldsymbol{{x}}}\boldsymbol{{dx}} \\ $$ Answered by Frix last updated on 21/Dec/23 $$\mathrm{Let}\:{s}=\mathrm{sin}\:{x} \\ $$$$\frac{\mathrm{sin}\:\mathrm{3}{x}}{\mathrm{1}+\mathrm{sin}^{\mathrm{3}} \:{x}}=\frac{{s}\left(\mathrm{4}{s}^{\mathrm{2}} −\mathrm{3}\right)}{\left({s}+\mathrm{1}\right)\left({s}^{\mathrm{2}}…
Question Number 202127 by Calculusboy last updated on 21/Dec/23 Answered by qaz last updated on 21/Dec/23 $$\int_{\pi/\mathrm{6}} ^{\pi/\mathrm{3}} \frac{{dx}}{\mathrm{1}+\mathrm{tan}^{\pi} \:{x}}=\int_{\pi/\mathrm{6}} ^{\pi/\mathrm{3}} \frac{{dx}}{\mathrm{1}+\mathrm{cot}\:^{\pi} {x}}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\pi/\mathrm{6}} ^{\pi/\mathrm{3}} \left(\frac{\mathrm{1}}{\mathrm{1}+\mathrm{tan}^{\pi}…
Question Number 201980 by Calculusboy last updated on 17/Dec/23 Answered by Sutrisno last updated on 18/Dec/23 $${misal}\:{x}^{\mathrm{2}} ={t} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{dx}=\frac{{dt}}{\mathrm{2}{x}} \\ $$$$=\int{x}.{x}^{\mathrm{2}} {cot}\left({x}^{\mathrm{2}} \right)\frac{{dt}}{\mathrm{2}{x}} \\…
Question Number 201982 by Calculusboy last updated on 17/Dec/23 Terms of Service Privacy Policy Contact: info@tinkutara.com