Question

What is `intsinh(x)cosh(x)dx`?

Answer 1

`1/4cos2h(x)+C`

Explanation:

`int sinh(x)*cosh(x)*dx`

=`1/2int sin2h(x)*dx`

=`1/4cos2h(x)+C`

Answer 2

`int \ sinhx coshx \ dx = 1/4 \ cosh 2x + C`

Explanation:

We seek:

`I = int \ sinhx coshx \ dx`

Method 1

We can use the identity:

`sinh 2x -= 2sinhx coshx`

Which gives us:

`I = int \ 1/2 sinh 2x \ dx`

`\ \ = 1/2 \ (cosh 2x)/2 + C`

`\ \ = 1/4 \ cosh 2x + C`

Method 2

Using the definitions of `sinhx` and `coshx`

`I = int \ (e^x-e^(-x))/2 (e^x+e^(-x))/2 \ dx`

`\ \ = 1/4 \ int \ (e^x-e^(-x))(e^x+e^(-x)) \ dx`

`\ \ = 1/4 \ int \ e^(2x)-e^(-2x) \ dx`

`\ \ = 1/4 \ {e^(2x)/2+e^(-2x)/2} + C`

`\ \ = 1/4 \ (e^(2x)+e^(-2x))/2 + C`

`\ \ = 1/4 \ cosh2x + C`