An integration problem using trigonometric substitution

[mech-eng]

Homework Statement

I cannot deal with this integration. $\int\sqrt{1+y^2}$

Relevant Equations

$\int\sqrt{1+y^2}$
$tan\theta=3y$

This is the integral I try to take. $\int\sqrt{1+9y^2}$ and $9y^2=tan^2\theta$ so the integral becomes $\int\sqrt{1+tan^2\theta}=\sqrt {sec^2\theta}$. Now I willl calculate dy.

$tan\theta=3y$ and $y=\frac {tan\theta}3$ and $dy=\frac{1+tan^2\theta}3$

Here is where I can only reach. I can't go any further.

Thanks.

 

[vela]

This is the integral I try to take. $\int\sqrt{1+9y^2}$ and $9y^2=tan^2\theta$ so the integral becomes $\int\sqrt{1+tan^2\theta}=\sqrt {sec^2\theta}$. Now I willl calculate dy.

$tan\theta=3y$ and $y=\frac {tan\theta}3$ and $dy=\frac{1+tan^2\theta}3$

You need a factor of $d\theta$ on the last part.

Here is where I can only reach. I can't go any further.

Why not? What's stopping you?

 

[tnich]

It seems to me that your integral should include a $dy$. You can easily state $dy$ in terms of $sec\theta$. You are almost there already.
Have you thought about the limits of integration?

 

[Orodruin]

Maybe it is only me, but I would use a hyperbolic substitution rather than a trigonometric one ...

 

[ehild]

I would try integration by parts
$\int(\sqrt{1+y^2}) (1)dy$

 

[mech-eng]

$tan\theta=3y$; so $\frac{1+tan^2d\theta}3=dy$ which is also equal to $\frac {sec^2\theta}3d\theta=dy$

Now I obtained $d\theta$ and I will write the integral transformation.

$\int{ \sqrt{1+tan^2\theta} \frac {1+tan^2\theta}3d\theta }$

$\int\sqrt{sec^2\theta}\frac{sec^2\theta}3 d\theta$

But here I am confused by this integral. I write everything. I cannot do next step.

 

[Ray Vickson]

$tan\theta=3y$; so $\frac{1+tan^2d\theta}3=dy$ which is also equal to $\frac {sec^2\theta}3d\theta=dy$

Now I obtained $d\theta$ and I will write the integral transformation.

$\int{ \sqrt{1+tan^2\theta} \frac {1+tan^2\theta}3d\theta }$

$\int\sqrt{sec^2\theta}\frac{sec^2\theta}3 d\theta$

But here I am confused by this integral. I write everything. I cannot do next step.

Well, if a trigonometric substitution gets you nowhere, why not take the advice of Ordruin in #4 and use a hyperbolic substitution instead?

 

[ehild]

$\int\sqrt{sec^2\theta}\frac{sec^2\theta}3 d\theta$

But here I am confused by this integral. I write everything. I cannot do next step.

Is not it
$\int\frac{|sec^3(\theta)|}{3} d\theta$

 

[mech-eng]

Is not it
$\int\frac{|sec^3(\theta)|}{3} d\theta$

Yes, that is exactly right. But I can't see the next step. And integrating an absolute value is a little confusing to me.

 

[ehild]

Yes, that is exactly right. But I can't see the next step. And integrating an absolute value is a little confusing to me.

Just ignore the absolute value sign for the time being.
There is a method to integrate rational functions of sin(x), cos(x), tan(x), by expressing them in terms of tan(x/2), and then using the substitution u=tan(x/2), so x=2 arctan(u),
dx = 2/(1+u²)du.
You will get a rational function to integrate.
$\cos(x)=\frac{1-u^2}{1+u^2}$
$\sin(x)=\frac{2u}{1+u^2}$
(check!)
Now you have the integral
$\int {\frac{1}{\cos^3\theta}d\theta}$
which becomes $\int{(\frac{1+u^2}{1-u^2})^3\frac{2}{1+u^2}du}$.
You are able to integrate it by decomposing into partial fractions.
This method is a bit complicated, so I suggest to use @Orodruin's hyperbolic substitution, or integrating by parts.

 

[tnich]

You can also integrate $\sec^3\theta$ by parts. This breaks it down into functions that you probably already know how to integrate.

 

[Orodruin]

@mech-eng You have reacted to the latest posts, does that mean that you have solved your issues? If so it is informative to tell us about it and if not we need to know where you are still stuck. I still suggest that the easiest route is a hyperbolic substitution.

 

[mech-eng]

I will turn back. I still look at some concepts and complete my integration knowledge. Sorry for such delays. For exampl I don't know how to take integral of $\sec^3\theta$ by parts.

 

[Mark44]

I will turn back. I still look at some concepts and complete my integration knowledge. Sorry for such delays. For exampl I don't know how to take integral of $\sec^3\theta$ by parts.

Your textbook might have this integral, either in a table of integrals or as an example of using integration by parts.

 

[Ray Vickson]

I will turn back. I still look at some concepts and complete my integration knowledge. Sorry for such delays. For exampl I don't know how to take integral of $\sec^3\theta$ by parts.

What I fail to understand is why, oh why, you refuse to look at the simplest solution, but insist, stubbornly, to stick to just about the hardest way so solve the problem! If your aim is to develop skills at integration, the absolutely best way is to learn to be flexible: if one way is too hard or is taking too long, try something else.

Several of us have suggested you look at hyberbolic substitution instead of a trigonometric one. I can even offer a hint: the identity $\cosh^2 u - \sinh^2 u = 1$ allows us to write $1+y^2 = \cosh^2 u$ when we take $y = \sinh u.$

 

[mech-eng]

There is a worked example in Calculus with Analytical Geometry by Howard Anton, 3rd edition. It introduces "reduction formulas" as this form.

$\int sec^nx~~dx=\frac {sec^{n-2}tanx}{n-1} + \frac{n-2}{n-1} \int sec^{n-2}xdx$

And in a previous section, the author gives it as an exercise to derive the reduction formula.

https://books.google.com.tr/books?i...ng powers of secant and tangent anton&f=false
And I can't also figure out why my latex code does not work.

Using the formula: $\frac13 \int sec^3 \theta=\frac13 \frac{sec\theta tan\theta}{n-1}+ \frac{n-2}{n-1} \int{sec \theta dx}$

 

[tnich]

There is a worked example in Calculus with Analytical Geometry by Howard Anton, 3rd edition. It introduces "reduction formulas" as this form.

$\int sec^nx~~dx=\frac {sec^{n-2}tanx}{n-1} + \frac{n-2}{n-1} \int sec^{n-2}xdx$

And in a previous section, the author gives it as an exercise to derive the reduction formula.

https://books.google.com.tr/books?i...ng powers of secant and tangent anton&f=false
And I can't also figure out why my latex code does not work.

Using the formula: $\frac13 \int sec^3 \theta=\frac13 \frac{sec\theta tan\theta}{n-1}+ \frac{n-2}{n-1} \int{sec \theta dx}$

That is getting you closer to an answer, though as others point out, it may be easier to use hyperbolic functions. If you continue this route, you are missing a factor of $\frac 1 3$ on the last term.

 

[mech-eng]

I will arrive at a result but now the problem is that after $tan\theta=3y$ substitution, how can I express $sec\theta$ in terms of y?

Thank you.

 

[Orodruin]

I will arrive at a result but now the problem is that after $tan\theta=3y$ substitution, how can I express $sec\theta$ in terms of y?

Thank you.

Trigonometric identities. How is sec related to tan?

 

[mech-eng]

Trigonometric identities. How is sec related to tan?

Since sec=$\frac 1{cos}~~and~~tan=\frac {sin}{cos}$ then $tan={sinsec}$ but I don't know how to use this for $y=3tan \theta$

 

[Orodruin]

No, you need a trigonometric identity involving just tan and sec. Hint: You have already used it in this thread.

 

[mech-eng]

No, you need a trigonometric identity involving just tan and sec. Hint: You have already used it in this thread.

This is might be the identiy required: $\sec^2\theta=1+tan^2\theta$ but I can't use it to manipulate $sec\theta tan\theta$. I need more guidance.

 

[Orodruin]

but I can't use it to manipulate $sec\theta tan\theta$

Why not? It is perfectly possible. You know what $\sec\theta$ is in terms of $\tan\theta$ and you know what $\tan\theta$ is in terms of $y$, so where is the problem?

 

[vela]

This is might be the identiy required: $\sec^2\theta=1+tan^2\theta$ but I can't use it to manipulate $sec\theta tan\theta$. I need more guidance.

Sometimes if you can't see how to proceed, it's worth it just to try something and see where it leads. It seems like you didn't even give it a shot here.

Anyway, here's an easy way to figure out what $\sec \theta$ is in terms of $y$. The substitution was $\tan \theta = 3y = \frac{y}{1/3}$, which corresponds to the right triangle below.

It's straightforward to write $\sec\theta$ as a ratio of two sides of the triangle.

 

[HappyS5]

I am a bit confused. Why are people substituting $3y = tan(\theta)$

I would substitute $y=tan(\theta)$, and the deivative is $\frac{dy}{dx} = tan^2(\theta) + 1 = sec^2(\theta)$. The final setup $dy = sec^2(\theta) d \theta$.

$\int \sqrt(1 + y^2) dy = \int \sqrt(1 + tan^2(\theta)) \cdot sec^2(\theta) d\theta = \int sec^3(\theta) d\theta$

Now, I would use a reduction formula or integration by parts to solve the problem futher. The reduction formula can be derived by the integration by parts method.

 

[Orodruin]

I am a bit confused. Why are people substituting $3y = tan(\theta)$

I would substitute $y=tan(\theta)$, and the deivative is $\frac{dy}{dx} = tan^2(\theta) + 1 = sec^2(\theta)$. The final setup $dy = sec^2(\theta) d \theta$.

$\int \sqrt(1 + y^2) dy = \int \sqrt(1 + tan^2(\theta)) \cdot sec^2(\theta) d\theta = \int sec^3(\theta) d\theta$

Now, I would use a reduction formula or integration by parts to solve the problem futher. The reduction formula can be derived by the integration by parts method.

The integrand is $\sqrt{1+9y^2}$, not $\sqrt{1+y^2}$.

 

[mech-eng]

Now it seems that I can solve $\sec^3\theta$ hence I can solve the question. This is my result I have arrived at. Would you please check it?

$sec\theta=\frac {\sqrt {\frac13^2~+y^2}} {\frac13}$

$\frac13\int sec^3\theta=\frac13 3\sqrt{{\frac13}^2+y^2}~~3y+\frac13 \frac12 ln |sec\theta+tan\theta|$

Substituting the values obtained and cancelling out,

$\frac13\int sec^3\theta=3\sqrt{\frac13^2+y^2}~~y+\frac16ln|\frac {\sqrt{\frac13^2+y^2}}{\frac13}|$

In the end, is this the solution?

P.S: When I asked this question, I thought this was a very straightforward and short question with a few easy steps.

 

[HappyS5]

The integrand is $\sqrt{1+9y^2}$, not $\sqrt{1+y^2}$.

Thanks.

In that case, I would solve it the following way:

$\int \sqrt{(1+(3y)^2)} dy \ and \ 3y = tan(\theta) \ and \ dy = \frac{sec^2(\theta) d\theta}{3}$

This gives $\int \sqrt{(1 + tan^2(\theta))} \cdot \frac{sec^2(\theta)}{3} d\theta$ or $\frac {1}{3} \cdot \int sec^3(\theta) d\theta$

I would then solve the above by the reduction formula or integration by parts. As mentioned before, the reduction formula can be derived by integration by parts.

 

[vela]

$\frac13\int sec^3\theta=3\sqrt{\frac13^2+y^2}~~y+\frac16ln|\frac {\sqrt{\frac13^2+y^2}}{\frac13}|$

In the end, is this the solution?

Differentiate it and see if you recover what you started with.

 

[Orodruin]

I still recomment the hyperbolic substitution ...

 

[HappyS5]

Now it seems that I can solve $\sec^3\theta$ hence I can solve the question. This is my result I have arrived at. Would you please check it?

$sec\theta=\frac {\sqrt {\frac13^2~+y^2}} {\frac13}$

$\frac13\int sec^3\theta=\frac13 3\sqrt{{\frac13}^2+y^2}~~3y+\frac13 \frac12 ln |sec\theta+tan\theta|$

Substituting the values obtained and cancelling out,

$\frac13\int sec^3\theta=3\sqrt{\frac13^2+y^2}~~y+\frac16ln|\frac {\sqrt{\frac13^2+y^2}}{\frac13}|$

In the end, is this the solution?

P.S: When I asked this question, I thought this was a very straightforward and short question with a few easy steps.

Remember the integration constant C.

Here is what I get:$\frac{1}{3} \int sec^3(\theta) d\theta = \frac{1}{6} (9y \cdot \sqrt{(\frac{1}{3})^2 + y^2}$ $+ \ln \left|3 \big(\sqrt{(\frac{1}{3})^2 + y^2} + y) \right|)$+ C

 

[Ray Vickson]

Now it seems that I can solve $\sec^3\theta$ hence I can solve the question. This is my result I have arrived at. Would you please check it?

$sec\theta=\frac {\sqrt {\frac13^2~+y^2}} {\frac13}$

$\frac13\int sec^3\theta=\frac13 3\sqrt{{\frac13}^2+y^2}~~3y+\frac13 \frac12 ln |sec\theta+tan\theta|$

Substituting the values obtained and cancelling out,

$\frac13\int sec^3\theta=3\sqrt{\frac13^2+y^2}~~y+\frac16ln|\frac {\sqrt{\frac13^2+y^2}}{\frac13}|$

In the end, is this the solution?

P.S: When I asked this question, I thought this was a very straightforward and short question with a few easy steps.

You can check it for yourself: just differentiate the "answer" and see if it delivers the original integrand. This is something that you should get in the habit of doing, every time.

 

[ehild]

The original problem

$\int\sqrt{1+9y^2}dy$ can be solved without any substitution , just integrating by parts: $\int{\sqrt{1+9y^2}(1)dy}=\sqrt{1+9y^2}y-\int{\frac {9y^2}{\sqrt{1+9y^2}}dy}$ Add and subtract 1 to the numerator of the fraction in the integral, it decomposes into two terms, one is the original integral, the other integral is a basic one.

 

[mech-eng]

The original problem

$\int\sqrt{1+9y^2}dy$ can be solved without any substitution , just integrating by parts: $\int{\sqrt{1+9y^2}(1)dy}=\sqrt{1+9y^2}y-\int{\frac {9y^2}{\sqrt{1+9y^2}}dy}$ Add and subtract 1 to the numerator of the fraction in the integral, it decomposes into two terms, one is the original integral, the other integral is a basic one.

I don't understand what you do here mostly because there is no fraction in the original problem. It is just a square root of square of an expression of y.

 

[ehild]

I don't understand what you do here mostly because there is no fraction in the original problem. It is just a square of an expression of y.

It is the integral in the second line.