Solving the Integral of sech^3 (x)

[DryRun]

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.

 

[dextercioby]

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

Can you solve this ?

 

[LCKurtz]

(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

 

[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?

 

[DryRun]

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]

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.

 

[DryRun]

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]

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]

@LCKurtz How can sech(x)tanh^2(x) = sech(x)[sech(x)tanh(x)??

It isn't. Just a careless error.

 

[SammyS]

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.

 

[DryRun]

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:
http://s2.ipicture.ru/uploads/20111210/gBo5g6t3.jpg

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

 

[vela]

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

 

[DryRun]

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?

 

[Karamata]

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

 

[DryRun]

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.

 

[Karamata]

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

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

 

[DryRun]

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.

 

[Karamata]

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

 

[DryRun]

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

 

[Karamata]

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

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

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

This isn't true.

 

[DryRun]

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]

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?

 

[DryRun]

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]

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

 

[DryRun]

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.

 

[vela]

Put this

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

together with this

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

To get
\begin{align*}
\int \text{sech}^3 x\,dx &= \int \frac{d(\sinh x)}{(1+\sinh^2 x)^2} = \int \frac{du}{(1+u^2)^2} \\
&= \frac{\arctan u}{2} + \frac{\sin(\arctan u)\cos(\arctan u)}{2}\end{align*}Forget the rest of the stuff.

Draw a right triangle with the leg opposite θ having length u and the adjacent leg having length 1. Then you'd have tan θ = u/1 = u. What's the length of the hypotenuse of this triangle in terms of u? Once you have that, write down algebraic expressions for sin(arctan u) = sin θ and cos(arctan u) = cos θ in terms of u.

Finally, undo the substitution u = sinh x to obtain the final answer. Don't forget to tack on the arbitrary constant of integration at the end.

 

[vela]

In the other approach, you had
$$\int\frac{1}{1+u^2}\,du = \frac{u}{1+u^2}+2\int\frac{1}{1+u^2}\,du - 2\int\frac{1}{(1+u^2)^2}\,du$$
If you let $I = \int\frac{1}{(1+u^2)^2}\,du$, you can say
$$\int\frac{1}{1+u^2}\,du = \frac{u}{1+u^2}+2\int\frac{1}{1+u^2}\,du - 2I$$Now you solve this for I and evaluate the remaining integrals, you'll find I equals the same expression in terms of u that you'll get after simplifying sin(arctan u) and cos(arctan u).

 

[DryRun]

Draw a right triangle with the leg opposite θ having length u and the adjacent leg having length 1. Then you'd have tan θ = u/1 = u. What's the length of the hypotenuse of this triangle in terms of u? Once you have that, write down algebraic expressions for sin(arctan u) = sin θ and cos(arctan u) = cos θ in terms of u.

Finally, undo the substitution u = sinh x to obtain the final answer. Don't forget to tack on the arbitrary constant of integration at the end.

The hypotenuse of the triangle is: $$\sqrt{1 + u^2}$$

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

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

$$\frac{\arctan u}{2} + \frac{u}{2(1 + u^2)} = \frac{\arctan (sinh x)}{2} + \frac{\sinh x}{2(1 + \sinh^2 x)} = \frac{\arctan (\sinh x)}{2} + \frac{(sech x)(\tanh x)}{2} + C$$

OK, this is very close to the final answer, but this part does not match the answer in my notes:
$$\frac{\arctan (\sinh x)}{2}$$
It should be:
$$\arctan (\tanh \frac{1}{2}x)$$

 

[dextercioby]

They should be equal up to a numerical constant.

 

[DryRun]

Is there a way to convert one into the other? I ask because i should give the answer exactly.

 

[dextercioby]

If 2 functions have the same derivative, then they must be equal up to a constant...