Integration Questions

Question Number 13364 by tawa tawa last updated on 19/May/17

$$\int_{\:\:\:\mathrm{0}} ^{\:\:\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}} \:\:\:\frac{\mathrm{x}^{\mathrm{3}} }{\sqrt{\mathrm{4x}^{\mathrm{2}} \:−\:\mathrm{9}}}\:\:\mathrm{dx} \\$$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 19/May/17

$${you}\:{should}\:{change}\:{the}\:{limits}\:{of}\:{integral}. \\$$$${function}\:{have}\:{not}\:{real}\:{valves}\:{in}\:{interval}:\left[\mathrm{0},\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}\right]. \\$$

Answered by ajfour last updated on 19/May/17

$${considering}\:\:\:{I}=\int\frac{{x}^{\mathrm{3}} }{\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{9}}}\:{dx} \\$$$${let}\:\boldsymbol{{t}}=\mathrm{4}{x}^{\mathrm{2}} −\mathrm{9} \\$$$${dt}=\mathrm{8}{xdx} \\$$$${tdt}=\mathrm{32}{x}^{\mathrm{3}} {dx}−\mathrm{9}×\mathrm{8}{xdx} \\$$$$\:\:\:\:\:\mathrm{32}{x}^{\mathrm{3}} {dx}\:=\left({t}+\mathrm{9}\right){dt} \\$$$${so},\:\: \\$$$${I}=\frac{\mathrm{1}}{\mathrm{32}}\int\:\frac{\left({t}+\mathrm{9}\right){dt}}{\sqrt{{t}}} \\$$$$\:=\frac{\mathrm{1}}{\mathrm{32}}\left(\frac{\mathrm{3}}{\mathrm{2}}{t}^{\mathrm{3}/\mathrm{2}} +\mathrm{18}\sqrt{{t}}\right)+{C} \\$$$$\:=\frac{\mathrm{3}}{\mathrm{64}}\left(\mathrm{4}{x}^{\mathrm{2}} −\mathrm{9}\right)^{\mathrm{3}/\mathrm{2}} +\frac{\mathrm{9}}{\mathrm{16}}\left(\mathrm{4}{x}^{\mathrm{2}} −\mathrm{9}\right)+{C} \\$$$${if}\:{the}\:{limits}\:{were}\:{appropriate} \\$$$${i}\:{would}\:{have}\:{tried}\:{to}\:{get}\:{an}\:{answer}. \\$$

Commented by tawa tawa last updated on 19/May/17

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\$$

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 19/May/17

$$\frac{\mathrm{2}{x}}{\mathrm{3}}={i}.{tgt}\Rightarrow{dx}=\frac{\mathrm{3}}{\mathrm{2}}{i}\left(\mathrm{1}+{tg}^{\mathrm{2}} {t}\right){dt}\Rightarrow \\$$$${cost}=\frac{\mathrm{1}}{\sqrt{\mathrm{1}−\frac{\mathrm{4}{x}^{\mathrm{2}} }{\mathrm{9}}}}=\frac{\mathrm{3}}{\sqrt{\mathrm{9}−\mathrm{4}{x}^{\mathrm{2}} }}=\frac{−\mathrm{3}{i}}{\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{9}}} \\$$$$\mathrm{4}{x}^{\mathrm{2}} −\mathrm{9}=−\mathrm{4}.\frac{\mathrm{9}}{\mathrm{4}}{tg}^{\mathrm{2}} {t}−\mathrm{9}=−\mathrm{9}\left({tg}^{\mathrm{2}} {t}+\mathrm{1}\right)= \\$$$$=\frac{−\mathrm{9}}{{cos}^{\mathrm{2}} {t}}\:\:\:\:\:\:\left(\boldsymbol{{i}}^{\mathrm{2}} =−\mathrm{1}\right) \\$$$${I}=\int\frac{\frac{\mathrm{27}}{\mathrm{8}}\boldsymbol{{i}}^{\mathrm{3}} .{tg}^{\mathrm{3}} {t}.\frac{\mathrm{3}}{\mathrm{2}}\boldsymbol{{i}}.\left(\mathrm{1}+{tg}^{\mathrm{2}} {t}\right){dt}}{\mathrm{3}\boldsymbol{{i}}/\boldsymbol{{cost}}}= \\$$$$=−\frac{\mathrm{27}}{\mathrm{16}}\boldsymbol{{i}}\int\frac{\boldsymbol{{cost}}.\boldsymbol{{tg}}^{\mathrm{3}} \boldsymbol{{t}}}{\boldsymbol{{cos}}^{\mathrm{2}} \boldsymbol{{t}}}\boldsymbol{{dt}}=−\frac{\mathrm{27}\boldsymbol{{i}}}{\mathrm{16}}\int\frac{{sin}^{\mathrm{3}} {t}}{{cos}^{\mathrm{4}} {t}}{dt}= \\$$$$=−\frac{\mathrm{27}\boldsymbol{{i}}}{\mathrm{16}}\int\frac{{sint}\left(\mathrm{1}−{cos}^{\mathrm{2}} {t}\right){dt}}{{cos}^{\mathrm{4}} {t}}= \\$$$$=\frac{−\mathrm{27}\boldsymbol{{i}}}{\mathrm{16}}\int\left[\frac{{sint}}{{cos}^{\mathrm{4}} {t}}−\frac{{sint}}{{cos}^{\mathrm{2}} {t}}\right]{dt}= \\$$$$=\frac{−\mathrm{27}\boldsymbol{{i}}}{\mathrm{16}}\left(−\frac{\mathrm{1}}{\mathrm{5}\boldsymbol{{cos}}^{\mathrm{5}} \boldsymbol{{t}}}+\frac{\mathrm{1}}{\boldsymbol{{cos}}^{\mathrm{3}} \boldsymbol{{t}}}\right)+\boldsymbol{{C}}.= \\$$$$=\frac{−\mathrm{27}\boldsymbol{{i}}}{\mathrm{16}}\left(−\frac{\mathrm{1}}{\mathrm{5}}\frac{\left(\mathrm{4}\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{9}\right)^{\mathrm{2}} \sqrt{\mathrm{4}\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{9}}}{−\mathrm{3}^{\mathrm{5}} \boldsymbol{{i}}}+\frac{\left(\mathrm{4}\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{9}\right)\sqrt{\mathrm{4}\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{9}}}{−\mathrm{27}\boldsymbol{{i}}}\right)+\boldsymbol{{C}}= \\$$$$=\frac{−\left(\mathrm{4}{x}^{\mathrm{2}} −\mathrm{9}\right)\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{9}}}{\mathrm{16}×\mathrm{45}}\left(\mathrm{4}{x}^{\mathrm{2}} −\mathrm{54}\right)+\boldsymbol{{C}}=\boldsymbol{{F}}\left({x}\right) \\$$$${F}\left(\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)=\frac{−\mathrm{18}\sqrt{\mathrm{18}}}{\mathrm{16}×\mathrm{45}}\left(\mathrm{27}−\mathrm{54}\right)=+\frac{\mathrm{81}}{\mathrm{40}}\sqrt{\mathrm{2}} \\$$$${F}\left(\mathrm{0}\right)=\frac{\mathrm{27}\boldsymbol{{i}}}{\mathrm{16}×\mathrm{45}}\left(−\mathrm{54}\right)=−\frac{\mathrm{81}}{\mathrm{40}}\boldsymbol{{i}} \\$$$$\Rightarrow\boldsymbol{{I}}=\frac{\mathrm{81}}{\mathrm{40}}\left(\sqrt{\mathrm{2}}−\boldsymbol{{i}}\right)\:\:\:\:.\blacksquare \\$$