1. ## Integration by Substitution

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.

2. 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...

3. 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
$=-arctan(cos(x))+C$

4. 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).

5. 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.

6. 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!