Question Number 211381 by Spillover last updated on 07/Sep/24
Commented by MathematicalUser2357 last updated on 10/Sep/24
$$\mathrm{Is}\:\mathrm{it}\:\int_{\mathrm{0}} ^{\mathrm{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \mathrm{sinh}^{\mathrm{3}} {x}\:\mathrm{cosh}^{\mathrm{11}} {x}\:{dx} \\ $$
Answered by Frix last updated on 07/Sep/24
$$\mathrm{Simply}\:\mathrm{use}\:{t}=\mathrm{e}^{{x}} \:\mathrm{to}\:\mathrm{get}\:\int=\frac{\mathrm{107}}{\mathrm{28}} \\ $$
Answered by mehdee1342 last updated on 07/Sep/24
![∫sih^3 x×coh^(11) xdx =∫(cosh^2 x−1)sinhx×cosh^(11) xdx =∫cosh^(13) xsinhxdx−∫cosh^(11) xsinhxdx =(1/(14))cosh^(14) x−(1/(12))cosh^(12) x+c ⇒I=(1/(14))[(((e^(ln(1+(√2))) +e^(−ln(1+(√2))) )/2))^(14) −1]− (1/(12))[(((e^(ln(1+(√2))) +e^(−(1+(√2))) )/2))^(12) −1] =(1/(14))[(((1+(√2)+(√2)−1)/2) )^(14) −1] −(1/(12))[(((1+(√2)+(√2)−1)/2))^(12) −1] =(1/(14))(127)−(1/(12))(63) =((107)/(28)) ✓](https://www.tinkutara.com/question/Q211390.png)
$$\int{sih}^{\mathrm{3}} {x}×{coh}^{\mathrm{11}} {xdx} \\ $$$$=\int\left({cosh}^{\mathrm{2}} {x}−\mathrm{1}\right){sinhx}×{cosh}^{\mathrm{11}} {xdx} \\ $$$$=\int{cosh}^{\mathrm{13}} {xsinhxdx}−\int{cosh}^{\mathrm{11}} {xsinhxdx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{14}}{cosh}^{\mathrm{14}} {x}−\frac{\mathrm{1}}{\mathrm{12}}{cosh}^{\mathrm{12}} {x}+{c} \\ $$$$\Rightarrow{I}=\frac{\mathrm{1}}{\mathrm{14}}\left[\left(\frac{{e}^{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} +{e}^{−{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} }{\mathrm{2}}\right)^{\mathrm{14}} −\mathrm{1}\right]− \\ $$$$\frac{\mathrm{1}}{\mathrm{12}}\left[\left(\frac{{e}^{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} +{e}^{−\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} }{\mathrm{2}}\right)^{\mathrm{12}} −\mathrm{1}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{14}}\left[\left(\frac{\mathrm{1}+\sqrt{\mathrm{2}}+\sqrt{\mathrm{2}}−\mathrm{1}}{\mathrm{2}}\:\right)^{\mathrm{14}} −\mathrm{1}\right] \\ $$$$−\frac{\mathrm{1}}{\mathrm{12}}\left[\left(\frac{\mathrm{1}+\sqrt{\mathrm{2}}+\sqrt{\mathrm{2}}−\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{12}} −\mathrm{1}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{14}}\left(\mathrm{127}\right)−\frac{\mathrm{1}}{\mathrm{12}}\left(\mathrm{63}\right) \\ $$$$=\frac{\mathrm{107}}{\mathrm{28}}\:\:\checkmark\: \\ $$$$ \\ $$$$ \\ $$