Really struggling with a problem here:
I need to find $\int \tanh^5 2x \textrm d x$ - absolutely no idea how to do it.
I tried splitting into $\tanh^2 2x,\tanh^2 2x, \tanh 2x$, and tried using double angle formulas. A complete solution would be really helpful as it also puts me on track for some other problems.
On
First write the integral as $\int \tanh 2x \, dx = \frac{1}{2} \int \tanh y \, dy$. Write $\tanh y = \frac{\sinh y}{\cosh y}$ so looking at $\int \frac{\sinh^5 y}{\cosh^5 y} \, dy$, next put $\sinh^5 y= \sinh^4 y \times \sinh y$, then two further steps: $\sinh y \, dy$ can be related to $d \cosh y$ and then think of replacing $\sinh^4 y$ in terms of powers of $\cosh y$.
On
Hint:
Substitute $2x=t$ than use the fact that: $$ \int \tanh^n(t)dt=\int \tanh^{n-2}(t)dt- \frac{\tanh^{n-1}(t)}{n-1} $$
( This is essentially the same as @Archis answer).
On
Do a u-substitution.
$$\frac{1}{2}\int \tanh^5 \space u \space du \space = -\frac{\tanh^4 u}{8} - \frac{\tanh^2 u}{4} + \frac{\log(\cosh(u))}{2} + c$$
From the identity $$\int \tanh^p \space x \space dx = -\frac{\tanh^{p-1}x}{p-1} + \int \tanh^{p-2}x \space dx$$
An alternative method:
$$\frac{1}{2}\int \tanh^{5}u \space du = -\frac{1}{2}\bigg(\sum_{k=1}^{n}\frac{\tanh^{6-2k} \space u}{5-2k} + u\bigg) $$
These come from Table of Integrals, Series, and Products, Seventh Edition, Chapter 2, section 2.424, examples 1 & 2.
I am not much in hyperbolic functions but use reduction formula ( I am writing this on basis of reduction of $\tan^n(x)$ so $\int\tanh^5(x)dx=\int\tanh^2(x)\tanh^3(x)dx=\int\tanh^3(x)\ sech^2(x)dx-\int\tanh^3(x)dx$ now in first integral put $tanh=u$ so $\ sech^2(x)dx=du$ repeat same reduction for $\tanh^3(x)$ hope you get it now