Question Number 87861 by Rio Michael last updated on 06/Apr/20
$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\int}}\:\mathrm{tanh}\:\mathrm{2}{x}\:{dx} \\ $$
Commented byMJS last updated on 06/Apr/20
![tanh 2x =((e^(4x) −1)/(e^(4x) +1)) ∫tanh 2x dx= [t=e^(4x) → dx=(dt/(4e^(4x) ))] =(1/4)∫((t−1)/(t(t+1)))dt=(1/2)∫(dt/(t+1))−(1/4)∫(dt/t)= =(1/2)ln (t+1) −(1/4)ln t =−x+(1/2)ln (e^(4x) +1) +C](Q87868.png)
$$\mathrm{tanh}\:\mathrm{2}{x}\:=\frac{\mathrm{e}^{\mathrm{4}{x}} −\mathrm{1}}{\mathrm{e}^{\mathrm{4}{x}} +\mathrm{1}} \\ $$ $$\int\mathrm{tanh}\:\mathrm{2}{x}\:{dx}= \\ $$ $$\:\:\:\:\:\left[{t}=\mathrm{e}^{\mathrm{4}{x}} \:\rightarrow\:{dx}=\frac{{dt}}{\mathrm{4e}^{\mathrm{4}{x}} }\right] \\ $$ $$=\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{t}−\mathrm{1}}{{t}\left({t}+\mathrm{1}\right)}{dt}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{{t}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{dt}}{{t}}= \\ $$ $$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left({t}+\mathrm{1}\right)\:−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:{t}\:=−{x}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left(\mathrm{e}^{\mathrm{4}{x}} +\mathrm{1}\right)\:+{C} \\ $$
Commented byRio Michael last updated on 06/Apr/20
$$\:\mathrm{sir}\:\mathrm{can}\:\mathrm{we}\:\mathrm{use}\:''\:\boldsymbol{\mathrm{integration}}\:\boldsymbol{\mathrm{by}}\:\boldsymbol{\mathrm{parts}}''? \\ $$
Commented byMJS last updated on 07/Apr/20
$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think}\:\mathrm{that}\:\mathrm{would}\:\mathrm{be}\:\mathrm{easier} \\ $$
Commented byRio Michael last updated on 07/Apr/20
$$\mathrm{i}\:\mathrm{will}\:\mathrm{try}\:\mathrm{it}\:\mathrm{out}\:\mathrm{sir} \\ $$