# Integral of sech^3 (x)

1. Dec 9, 2011

### sharks

The problem statement, all variables and given/known data
Integral of sech^3 (x)

The attempt at a solution

sech^3 (x) = (sech x)(sech^2 (x)) = (sech x)(1-tanh^2 (x))
And then i expanded by multiplying:

(sech x)(1-tanh^2 (x)) = (sech x) - (sech x)(tanh^2 (x))
The integral of sech x can be found.
But what about the integral of (sech x)(tanh^2 (x))?
I've tried various substitutions, but failed. Let t = tanh x and also let t = sech x.
I also tried this way:
(sech x)(tanh^2 (x)) = (sech x)(1 - sech^2 (x)) = sech x - sech^3 (x)
But then i end up with the original problem again.

2. Dec 9, 2011

### dextercioby

$$\int \frac{d(\sinh x)}{(1+\sinh^2 x)^2} = \int \frac{du}{(1+u^2)^2}$$

Can you solve this ?

3. Dec 9, 2011

### LCKurtz

sech(x)tanh2x = [STRIKE]sech(x)[sech(x)tanh(x)]. Let u=sech(x), what is du[/STRIKE]?

[Edit:] Nevermind. I hurriedly wrote what I wanted it to be instead of what it is

Last edited: Dec 9, 2011
4. Dec 9, 2011

### McAfee

try approaching it from integrating by parts:

∫sec^3(x) dx

u= sec(x) dv=sec^2(x) dx
du=sec(x)tan(x) v=tan(x)

∫sec^3(x) dx = sec(x)*tan(x)-∫sec(x)*tan^2(x) dx
∫sec^3(x) dx = secx*tanx-∫sec^3(x)-sec(x) dx
∫sec^3(x) dx = secx*tanx-∫sec^3(x) dx+∫sec(x) dx

Can you figure out the rest from here?

5. Dec 9, 2011

### sharks

Unfortunately, i must be getting stupider... I can't understand either of you.

@dextercioby In your example, if by letting u = sinh x, then du/dx = cosh x, so... the R.H.S. of your equation should be something different. The numerator sinh x can't be absorbed cleanly into du.

@LCKurtz How can sech(x)tanh^2(x) = sech(x)[sech(x)tanh(x)??
If i start with the R.H.S of this equation, it means sech^2(x)tanh(x).

I think you mean, by using the hyperbolic identity: 1 - tanh^2(x) = sech^2(x)
I would then get tanh^2(x) = 1 - sech^2(x)

This would become upon replacing in sech(x)tanh^2(x):
sech(x)[1 - sech^2(x)]

I don't know where this is going...

@McAfee This problem is not about finding the integral of sec^3(x) but integral of sech^3(x). Thanks though.

6. Dec 9, 2011

### vela

Staff Emeritus
Integration by parts is probably the easiest way to do the problem.

Let u = sinh x. Then du = cosh x dx, so dx = du/cosh x, so your integral becomes
$$\int\text{sech}^3 x\,dx = \int\frac{dx}{\cosh^3 x} = \int\frac{du}{\cosh^4 x}$$Finish changing variables from x to u.

7. Dec 9, 2011

### sharks

OK, i was able to get dextercioby's R.H.S. equation.

Now, i'm unsure about the method of integration. Here are the 2 methods that i'm thinking of. I don't know if both methods are valid in this case though.

Method 1:
Integral of (1 + u^2)^(-2).du which after integration would give: -1/[2u(1 + u^2)] + C

Method 2:
Do partial fractions first and then integrate. This would give:
(Au + B)/(1 + u^2) + (Cu + D)/[(1 + u^2)^(2)]
I hope the format above is correct? I used (Au + B) and (Cu + D) as the numerators since the denominator is quadratic. Otherwise i would have put just A and B instead.

8. Dec 9, 2011

### vela

Staff Emeritus
The first method is wrong. With partial fractions, you'll find A=B=C=0 and D=1. It's already in its simplest form.

I'm not sure what dextercioby had in mind. I usually go the other way. Use the trig substitution to go from the u-form to the trig form, and then use another technique to evaluate the trig integral.

9. Dec 9, 2011

### LCKurtz

It isn't. Just a careless error.

10. Dec 9, 2011

### SammyS

Staff Emeritus
Use the power reduction formula for $\displaystyle\int\text{sech}^m(x)dx\,.$ In this case m=3.

You get $\displaystyle\int\text{sech}^3(x)dx=\frac{1}{2} \tanh(x) \text{sech}(x)+\frac{1}{2} \int \text{sech}(x)\,dx\,.$

To get this result you can do integration by parts with $dv=\text{sech}^2(x)\,dx$ and $u=\text{sech}(x)$ along with some hyperbolic function identities & algebra.

11. Dec 9, 2011

### sharks

OK, this is getting immensely complicated! I'm still at undergraduate level and haven't yet done anything pertaining to the power reduction formulae.

So, here is the answer from my notes:

Hopefully, this will help all of us to trace a better and simpler route to solve this problem.

12. Dec 9, 2011

### vela

Staff Emeritus
It only seems immensely complicated because you keep ignoring the suggestion to use integration by parts.

13. Dec 11, 2011

### sharks

I took a break, and i apologize for being a little bit frustrated with this problem.

From this point: $$\int \frac{d(\sinh x)}{(1+\sinh^2 x)^2} = \int \frac{du}{(1+u^2)^2}$$
I expressed it in partial fractions (can confirm the results that A=B=C=0 and D=1), and got $$\int \frac{du}{(1+u^2)^2}$$
But now how to proceed next?
Could you please explain how to go about it using your method?

Last edited: Dec 11, 2011
14. Dec 11, 2011

### Karamata

Solve $$\int \frac{du}{(1+u^2)}$$ where $$t=\frac{1}{(1+u^2)}$$ and $${dv}={du}$$ (Integration by parts)

15. Dec 12, 2011

### sharks

Hi Karamata

OK, to solve the problem you typed above, here is what i have so far:

$$\frac{u}{(1+u^2)} + 2\int\frac{u^2}{(1+u^2)^2}$$

Now, to evaluate $$\int\frac{u^2}{(1+u^2)^2}$$

If i break the integral using partial fractions, i believe i will get the same answer, as it's already in its simplest form.

So, i integrate by parts again.

Last edited: Dec 12, 2011
16. Dec 12, 2011

### Karamata

$$\frac{u^2}{(1+u^2)^2}=\frac{u^2+1-1}{(1+u^2)^2}=\frac{u^2+1}{(1+u^2)^2}-\frac{1}{(1+u^2)^2}$$.....

17. Dec 12, 2011

### sharks

To continue from above:

$$\int\frac{1 + u^2}{(1 + u^2)^2} = \int\frac{1}{1 + u^2} = \arctan u$$

But $$\int\frac{1}{(1 + u^2)^2}$$ is a bit tricky. I cannot simplify it further using partial fractions.

I suppose i'll have to use substitution.

Let $$u = tan \theta$$ Not sure though.

Last edited: Dec 12, 2011
18. Dec 12, 2011

### Karamata

$$\int\frac{1}{1 + u^2}= \arctan u$$

I don't get it.

$$\int {\frac{du}{1+u^2}}=\frac{u}{u^2+1}+2\left( \arctan {u} + C -\int \frac{du}{(1+u^2)^2}\right)$$
Or $$\arctan {u}=\frac{u}{u^2+1}+2\left( \arctan {u} + C -\int \frac{du}{(1+u^2)^2}\right)$$
And now, $$\int \frac{du}{(1+u^2)^2}=$$

Last edited: Dec 13, 2011
19. Dec 12, 2011

### sharks

I had made a mistake when evaluating.
$$\int\frac{1}{1 + u^2}= \arctan u$$
I edited my post as you were replying. Thank you for pointing it out.

In evaluating the integral below, surely, it can't be as simple as this (but i'll take a chance as i can't see what substitution to make):
$$\int \frac{du}{(1+u^2)^2} = \arctan^2 u$$
It's probably wrong though.

If i use the substitution:
Let $$u = \tan \theta$$

Then the integral evolves into:
$$\int \frac{du}{sec^4 \theta}$$
But
$$\frac{du}{d\theta} = sec^2 \theta$$
So, the integral becomes
$$\int \frac{d\theta}{sec^2 \theta} = \int cos^2 \theta.d\theta$$

Next, i use double angle formula to convert the above integral into:
$$\int \frac{1}{2}.d\theta + \int \frac{cos 2\theta}{2}.d\theta$$
Then, after integration, i get:
$$\frac{\theta}{2} + \frac{sin 2\theta}{4}$$
From
$$u = \tan \theta$$
I get
$$\theta = \arctan u$$ which i substitute into the integration results, and i get:
$$\frac{\arctan u}{2} + \frac{sin (2\arctan u)}{4}$$

Last edited: Dec 12, 2011
20. Dec 12, 2011

### Karamata

$$\arctan {u}=\frac{u}{u^2+1}+2\left( \arctan {u} + C -\int \frac{du}{(1+u^2)^2}\right)$$
And now, $$\int \frac{du}{(1+u^2)^2}=$$

$$\hbox{TUT}=\hbox{BRB}+2\left(\hbox{TUT}-\int \frac{du}{(1+u^2)^2}\right)$$
$$\int \frac{du}{(1+u^2)^2}=\frac{1}{2}(\hbox{BRB}+\hbox{TUT})$$

This isn't true.