What is a more efficient approach for finding the integral of tanhx?

[azatkgz]

What's wrong with my solution?

\int \sqrt {\tanh x}dx

for u = \tanh x\rightarrow du = \frac {dx}{\cosh^2x}


\int \sqrt {\tanh x}dx = \int\sqrt {u}\cosh^2x dx = \int\frac {\sqrt {u}dx}{1 - u} = \int \frac {du}{2(1 + \sqrt {u})} - \int \frac {du}{2(1 - \sqrt {u})}

for v = (1 + \sqrt {u})\rightarrow dv = \frac {du}{2\sqrt {u}}and for z = (1 - \sqrt {u})\rightarrow dz = - \frac {du}{2\sqrt {u}}


\int\frac {dv(v - 1)}{v} + \int\frac {dz(1 - z)}{z} = v - \ln v + \ln z - z

\int \sqrt {\tanh x}dx = \ln (\frac {1 - \sqrt {\tanh x}}{1 + \sqrt {\tanh x}}) + 2\sqrt {\tanh x}

 

[Tolya]

I=\int \sqrt{tanhx}dx

u=tanhx

dx=cosh^2xdu

I=\int \sqrt{u}cosh^2xdu=\int \frac{\sqrt{u}du}{1-u^2}

Notice, that cosh^2x=\frac{1}{1-tanh^2x}

Then, let t=\sqrt{u} \frac{dt}{du}=\frac{1}{2\sqrt{u}}=\frac{1}{2t}.

I=2\int \frac{t^2dt}{1-t^4}=\int \frac{dt}{1-t^2} + \int \frac{dt}{1+t^2}

So, your problem is: when you substitute dx you write the same dx in integral. But it is wrong! You must write du :)

Also look at cosh^2x=\frac{1}{1-tanh^2x}

 

[rocomath]

it would be faster to make a u^2 substitution