Question

How is the integral of `int x / ( sinh(x) + cosh(x) ) dx`?

Answer 1

`-xe^(-x)-e^(-x)+C`

Explanation:

`int x/(sinhx+coshx)*dx`

=`int x/(e^x)*dx`

=`int xe^(-x)*dx`

=`x*(-e^(-x))-int (-e^(-x))*dx`

=`-xe^(-x)+int e^(-x)*dx`

=`-xe^(-x)-e^(-x)+C`

Answer 2

Given: `int x / ( sinh(x) + cosh(x) ) dx`

Multiply the integrand by 1 in the form of `(sinh(x)-cosh(x))/(sinh(x)-cosh(x))`:

`int x / ( sinh(x) + cosh(x) ) dx = int x / ( sinh(x) + cosh(x) )(sinh(x)-cosh(x))/(sinh(x)-cosh(x)) dx`

The denominator is multiplied using the difference of two squares pattern and the numerator is multiplied using the distributive property:

`int x / ( sinh(x) + cosh(x) ) dx = int (xsinh(x)-xcosh(x)) / ( sinh^2(x) - cosh^2(x)) dx`

The identity `cosh^2(x)- sinh^2(x)=1` tells us that the denominator is -1:

`int x / ( sinh(x) + cosh(x) ) dx = int (xsinh(x)-xcosh(x)) / -1 dx`

Eliminate the denominator by changing signs in the numerator:

`int x / ( sinh(x) + cosh(x) ) dx = int xcosh(x)-xsinh(x) dx`

Separate into two integrals:

`int x / ( sinh(x) + cosh(x) ) dx = int xcosh(x) dx- int xsinh(x) dx`

Both integrals are trivial integrations by parts:

`int x / ( sinh(x) + cosh(x) ) dx = (xsinh(x) - cosh(x)) - (xcosh(x)-sinh(x))+C`

Regroup:

`int x / ( sinh(x) + cosh(x) ) dx = xsinh(x)- xcosh(x)+ sinh(x)- cosh(x) +C`