What is #intxsinh(x)dx#?

Question

851 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-08T11:27:13

#xcoshx-sinhx+c#

#I=intu(du)/(dx)dx=uvintv(du)/(dx)dx#

#I=intxsinhxdx#

#u=x=>(du)/(dx)=1#

#(dv)/(dx)=sinhx=>v=coshx#

#I=xcoshx-intcoshxdx#

#I=xcoshx-sinhx+c#