Derivative substitution/reduction formula

[Sebastian L]

Hi, would someone be able to help me with this question?
Thanks very much in advance

 

[BigBeachBanana]

[math]\frac{\tanh^{n-1}(x)}{n-1}= \int \tanh^{n-2}(x) \text{sech}^{n-2}(x)\,dx=\dots[/math]Substitute and continue...

 

[AvgStudent]

[math]\frac{\tanh^{n-1}(x)}{n-1}= \int \tanh^{n-2}(x) \text{sech}^{n-2}(x)\,dx=\dots[/math]Substitute and continue...

How did you come up with that? I'm curious.

 

[Deleted member 4993]

How did you come up with that? I'm curious.

Equivalent relation in real domain:

 

[BigBeachBanana]

How did you come up with that? I'm curious.

Start by taking the derivative.
[math]\frac{d}{dx}\frac{\tanh^{n-1}x}{n-1}[/math]Then integrate both sides.

 

[AvgStudent]

Start by taking the derivative.
[math]\frac{d}{dx}\frac{\tanh^{n-1}x}{n-1}[/math]Then integrate both sides.

Oh! Nice trick. I was able to solve the OP's question.

 

[Steven G]

The standard way is to use integration by parts.

 

[BigBeachBanana]

[math]\frac{\tanh^{n-1}(x)}{n-1}= \int \tanh^{n-2}(x) \text{sech}^{\red{n-2}}(x)\,dx=\dots[/math]

There's an error in post#2.
Correction:
[math]\frac{\tanh^{n-1}(x)}{n-1}= \int \tanh^{n-2}(x) \text{sech}^{2}(x)\,dx=\dots[/math]Substitute and continue...