# Integral of sech^3 (x)

Gold Member
Homework Statement
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.

## Answers and Replies

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

Can you solve this ?

LCKurtz
Homework Helper
Gold Member
(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.
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

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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?

Gold Member
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.

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

@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.
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.

Gold Member
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.

vela
Staff Emeritus
Homework Helper
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.

LCKurtz
Homework Helper
Gold Member
@LCKurtz How can sech(x)tanh^2(x) = sech(x)[sech(x)tanh(x)??
It isn't. Just a careless error.

SammyS
Staff Emeritus
Homework Helper
Gold Member
Homework Statement
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.
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.

Gold Member
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.

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

Gold Member
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?
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.
Could you please explain how to go about it using your method?

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$$\int \frac{du}{(1+u^2)^2}$$
But now how to proceed next?
Solve $$\int \frac{du}{(1+u^2)}$$ where $$t=\frac{1}{(1+u^2)}$$ and $${dv}={du}$$ (Integration by parts)

Gold Member
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.

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

Gold Member
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.

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To continue from above:

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

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

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Gold Member
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:
$$\frac{\arctan u}{2} + \frac{sin (2\arctan u)}{4}$$

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

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

$$\int \frac{du}{(1+u^2)^2} = \arctan^2 u$$
This isn't true.

Gold Member
I'm not familiar with this formula:
$$\arctan {u}=\frac{u}{u^2+1}+2\left( \arctan {u} + C -\int \frac{du}{(1+u^2)^2}\right)$$
Is it a standard formula that i have to learn by heart to solve these kinds of problems. My notes have no such formula. What if i use a substitution to solve it, like i did in my previous reply above?

vela
Staff Emeritus
Homework Helper
Sharks, I think you need to start afresh to understand what you're doing here.

In post 14, Karamata suggested you integrate 1/(1+u2) by parts, which you did, and you found
$$\int\frac{1}{1+u^2}\,du = \frac{u}{1+u^2}+2\int\frac{u^2}{(1+u^2)^2}\,du$$Then Karamata told you to slightly rewrite the integrand on the second integral to get
\begin{align*}
\int\frac{1}{1+u^2}\,du &= \frac{u}{1+u^2}+2\int\frac{u^2+1-1}{(1+u^2)^2}\,du \\
&= \frac{u}{1+u^2}+2\left[\int\frac{u^2+1}{(1+u^2)^2}\,du - \int\frac{1}{(1+u^2)^2}\,du\right] \\
&= \frac{u}{1+u^2}+2\int\frac{1}{1+u^2}\,du - 2\int\frac{1}{(1+u^2)^2}\,du
\end{align*}Note that the last integral on the righthand side of the equation is the one you're trying to solve for. Can you pick it up from here?

Gold Member
OK, i believe i've solved the right-most integral above, in post #19 by using this substitution:
$$u = \tan \theta$$
Or maybe it's wrong?

vela
Staff Emeritus
Homework Helper
Try using the identity $\sin 2\theta = 2\sin \theta \cos \theta$ to simplify
$$\frac{\sin(2\arctan u)}{4}$$

Gold Member
Then, it expands to become:
$$\frac{\sin(2\arctan u)}{4} = \frac{\sin(\arctan u)\cos(\arctan u)}{2}$$

So, now
$$\int \frac{du}{(1+u^2)^2} = \frac{\arctan u}{2} + \frac{\sin(\arctan u)\cos(\arctan u)}{2}\; .......equation 1$$

Then,
$$\int\frac{u^2}{(1+u^2)^2} = \arctan u - \frac{\arctan u}{2} - \frac{\sin(\arctan u)\cos(\arctan u)}{2}$$

Continuing the substitution and moving up the integration levels:
$$\int \frac{du}{(1+u^2)} = \frac{u}{(1+u^2)} + 2\int\frac{u^2}{(1+u^2)^2}$$

$$\int \frac{du}{(1+u^2)} = \frac{u}{(1+u^2)} + 2\arctan u - \arctan u - \sin(\arctan u)\cos(\arctan u)$$

$$\int \frac{du}{(1+u^2)} = \frac{u}{(1+u^2)} + \arctan u - \sin(\arctan u)\cos(\arctan u) \; .......equation 2$$

OK, now, i've reached post #14 where i need to find:
$$\int \frac{du}{(1+u^2)^2}$$

I'm confused how to proceed, since equation 1 above, is the exact thing that i'm trying to find, or should i derive it, using equation 2?

Also, in post #20, Karamata used BRB and TUT. What are these? I'm sorry if this is a stupid question, but i have no clue.

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