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$$\int \tanh(x) - \tanh^3(x)\,dx$$

I get the answer as $\tanh x + c$?

I took out a factor of $\tanh x$, used the identity $1-\tanh^2 x=\text{sech}^2x$, used the substitution of $u=\tanh x$,

and reduced the question to $\int 1 du$.

Is this the way you would normally approach this question?

Cheers.

Gurjinder.B

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  • $\begingroup$ Using the hyperbolic identity and your substitution gives that the integral is $\int \tanh x \operatorname{sech}^2 x \,dx = \int u \,du$, not $\int \,du$. $\endgroup$
    – Travis Willse
    Jan 9, 2016 at 23:07
  • $\begingroup$ Yes it does, sorry, the insert should be $frac1/2tanh^2x+c$ , right? $\endgroup$
    – Gurjinder
    Jan 9, 2016 at 23:11
  • $\begingroup$ That looks good to me. $\endgroup$
    – Travis Willse
    Jan 9, 2016 at 23:36
  • $\begingroup$ Check your substitutions: You should get $-\frac{1}{2} \text{sech}^2(x)$. $\endgroup$
    – David G. Stork
    Jan 9, 2016 at 23:44

1 Answer 1

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The derivative of $f(x)=\tanh x$ is $$ f'(x)=\frac{\cosh^2x-\sinh^2x}{\cosh^2x}=1-\tanh^2x $$ so the integral is $$ \int f(x)f'(x)\,dx=\frac{1}{2}(f(x))^2+c=\frac{1}{2}\tanh^2x+c $$

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