Integration by Substitution

**thejabronisayz** · February 10th 2008, 07:26 PM

Hello, I'm having trouble doing this one integration problem and it needs to be done using substitution and only substitution (no integration by parts). I can pretty much figure it out by looking at it that it needs to be arctan(cos x) + C, but I need to show my work and I can't figure out how to get that answer using substitution. I would think that it would make sense to make u = 1 + (cosx)^2, but when I do that it doesn't seem to work. Can anyone help? Thanks!

sin[x] / (1 + (cos[x])^2)

To be clearer, it's sine x over the quantity: one plus cosine squared x. Cosine is being squared, not the x inside of it, and not the entire quantity.

**Jhevon** · February 10th 2008, 07:43 PM

Originally Posted by thejabronisayz

Hello, I'm having trouble doing this one integration problem and it needs to be done using substitution and only substitution (no integration by parts). I can pretty much figure it out by looking at it that it needs to be arctan(cos x) + C, but I need to show my work and I can't figure out how to get that answer using substitution. I would think that it would make sense to make u = 1 + (cosx)^2, but when I do that it doesn't seem to work. Can anyone help? Thanks!

sin[x] / (1 + (cos[x])^2)

To be clearer, it's sine x over the quantity: one plus cosine squared x. Cosine is being squared, not the x inside of it, and not the entire quantity.

the substitution u = cos(x) will suffice...

**angel.white** · February 10th 2008, 07:47 PM

Originally Posted by thejabronisayz

Hello, I'm having trouble doing this one integration problem and it needs to be done using substitution and only substitution (no integration by parts). I can pretty much figure it out by looking at it that it needs to be arctan(cos x) + C, but I need to show my work and I can't figure out how to get that answer using substitution. I would think that it would make sense to make u = 1 + (cosx)^2, but when I do that it doesn't seem to work. Can anyone help? Thanks!

sin[x] / (1 + (cos[x])^2)

To be clearer, it's sine x over the quantity: one plus cosine squared x. Cosine is being squared, not the x inside of it, and not the entire quantity.

$\int \frac{1}{1+cos^2(x)}sin(x)dx$

> substitute:
> $tan(u) = cos(x)$
> $u = arctan(cos(x))$
> $-sec^2(u)du = sin(x)dx$

$=\int \frac 1{1+tan^2(x)}*(-sec^2(u))du$

$=-\int \frac {sec^2(u)}{1+tan^2(x)}du$

$=-\int \frac {sec^2(u)}{sec^2(u)}du$

$=-\int du$

Integrate
$=-u + C$

Unsubstitute
$=-arctan(cos(x))+C$

**thejabronisayz** · February 11th 2008, 11:35 AM

I appreciate the responses, however, I don't think they can help me in my situation. Making u = cos x does not work (I've tried), and even though the second reply does work, I cannot even fathom thinking of making tan(u) = cosx and then making u = arctan(cosx). Are there any other ways, or is the answer given in the second reply the only way? (I'm not trying to be critical of the replies at all; I greatly appreciate them. I am just trying to find the way that I could use on an exam).

**galactus** · February 11th 2008, 11:59 AM

It works very well.

$\frac{d}{dx}[cos(x)]=-sin(x)$

So, if you let $u=cos(x), \;\ -du=sin(x)$

Just makes the subs:

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

This is a well-known integral. This is one we mostly look up or remember. You are not expected to derive it each time it occurs.
What is the derivative of $tan^{-1}(u)$ ?.

I bet it's $\frac{1}{1+u^{2}}$

Now, resub and get $-tan^{-1}(cos(x))$ as was shown before.

That's as far as you need to take it.

**thejabronisayz** · February 11th 2008, 07:40 PM

You're right. I'm sorry about that, I accidentally was integrating [cos(x)]^2 and not just the plain cos(x). It works. Thank you!

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