#\int\sinh^2(x)\cosh(x)dx#?

Question

Mostly not sure what to use for u-substitution. :\

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Answer 1 · 2017-09-17T16:48:21

see below

If the question was meant to be #I=intsinh^2xcoshxdx#

by inspection the answer would be

#1/3sinh^3x+C#

by substitution

# u=sinhx=>du=coshxdx#

#I=intu^2cancel(coshu)xx(du)/cancel(coshu)#

#=intu^2du=1/3u^3+C#

#=1/3sinh^3x+C#