Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 544 by 123456 last updated on 26/Jan/15

∫_0 ^(2π) sin x sinh x dx=?

$$\underset{\mathrm{0}} {\overset{\mathrm{2}\pi} {\int}}\mathrm{sin}\:{x}\:\mathrm{sinh}\:{x}\:{dx}=? \\ $$

Answered by prakash jain last updated on 26/Jan/15

∫sin x sinh xdx  =sin x cosh x −∫cos x cosh x dx  =sin x cosh x − [cos x sinh x +∫sin x sinh x dx]  =(1/2)(sin x cosh x−cos x sinh x)  putting the limits  (1/2)(−sinh 2π)  =− ((sinh 2π)/2)= − sinh π cosh π

$$\int\mathrm{sin}\:{x}\:\mathrm{sinh}\:{xdx} \\ $$ $$=\mathrm{sin}\:{x}\:\mathrm{cosh}\:{x}\:−\int\mathrm{cos}\:{x}\:\mathrm{cosh}\:{x}\:{dx} \\ $$ $$=\mathrm{sin}\:{x}\:\mathrm{cosh}\:{x}\:−\:\left[\mathrm{cos}\:{x}\:\mathrm{sinh}\:{x}\:+\int\mathrm{sin}\:{x}\:\mathrm{sinh}\:{x}\:{dx}\right] \\ $$ $$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{sin}\:{x}\:\mathrm{cosh}\:{x}−\mathrm{cos}\:{x}\:\mathrm{sinh}\:{x}\right) \\ $$ $${putting}\:{the}\:{limits} \\ $$ $$\frac{\mathrm{1}}{\mathrm{2}}\left(−\mathrm{sinh}\:\mathrm{2}\pi\right) \\ $$ $$=−\:\frac{\mathrm{sinh}\:\mathrm{2}\pi}{\mathrm{2}}=\:−\:\mathrm{sinh}\:\pi\:\mathrm{cosh}\:\pi \\ $$

Terms of Service

Privacy Policy