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

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

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?

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

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

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

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

So, set $\displaystyle\cosh4x=u$

• wouldn't \dfrac look better/be more readable? – Tim Seguine Jan 16 '14 at 17:03
• @TimSeguine, adjusted:) But would you mind sharing the difference between $\\dfrac,\\frac$? – lab bhattacharjee Jan 16 '14 at 17:05
• \dfrac keeps a constant font size, \frac makes successive fractions in the numerator or denominator smaller and smaller. – Tim Seguine Jan 16 '14 at 17:07
• dfrac: $\dfrac{x}{y}$ vs. frac: $\frac{x}{y}$; more here: f.kth.se/~ante/latex.php – BaronVT Jan 16 '14 at 17:07
• @TimSeguine, thanks for sharing the knowledge.Shall use it whenever applicable – lab bhattacharjee Jan 16 '14 at 17:08