integrate $\int e^{-\cosh4x}\sinh4x~dx$

Question

$$\int e^{-\cosh4x}\sinh4x~dx$$

I tried to do it by parts but first i am unsure what $e^{-\cosh4x}$ would integrate to? Also is the correct method by parts? Any suggestions to get me started?

Answer 1

Use a u substitution.

$u = cosh(4x)$

$du = 4 sinh(4x)dx$

So $\int e^{-cosh(4x)}sinh(4x)dx = \frac14 \int e^{-u}du = -\frac 14e^{-u}$

Which is, substituting back in

$-\frac 14e^{-cosh(4x)}$

Answer 2

$$\dfrac{d(\cosh4x)}{dx}=\dfrac{d\left(\dfrac{e^{4x}+e^{-4x}}2\right)}{dx}$$

$$=4\cdot\dfrac{(e^{4x}-e^{-4x})}2=4\sinh4x$$

So, set $\displaystyle\cosh4x=u$