$\displaystyle

\int \limits_0^{\frac{\\pi}{2}} \frac{cosx}{1+sin^2(x)}dx

$

Printable View

- Jan 26th 2009, 03:36 PMsaiyanmx89Evaluate the integral.
$\displaystyle

\int \limits_0^{\frac{\\pi}{2}} \frac{cosx}{1+sin^2(x)}dx

$ - Jan 26th 2009, 03:39 PMJester
- Jan 26th 2009, 03:47 PMsaiyanmx89
why does u=sinx? how did you get sinx?

- Jan 26th 2009, 03:48 PMmollymcf2009
$\displaystyle sin^2(x)$ is the same thing as $\displaystyle (sinx)^2$

So, (sinx) is what you are using for your u-substitution - Jan 26th 2009, 03:50 PMsaiyanmx89
thank you!!!

- Jan 26th 2009, 03:53 PMsaiyanmx89
So, would the answer be:

arctan(sinx) +c ??? - Jan 26th 2009, 04:33 PMmollymcf2009
No!! This is a definite integral! It will not have a " +C" only indefinite integrals use the +C. You should get a real number answer after you integrate and plug in your limits.

Also, look at danny's post. When he did the u-substitution, he changed his limits based on "u" so you don't have to plug "sinx" back into the problem. You can just evaluate it by plugging your new limits in for "u"

Do you understand the u-substitution part and why the limits changed?

Ask if you need clarification on how to do that! - Jan 26th 2009, 05:16 PMsaiyanmx89
So, the answer would be found from:

arctan(1)-arctan(0)

??

I'm still new to changing the limits. I understand how to change them but, I usually don't know when they need to be changed... - Jan 26th 2009, 06:21 PMmr fantastic
- Jan 26th 2009, 07:01 PMmollymcf2009
No problem! u-substitution can be tricky to learn!

You have probably been taught that when you use u-substitution, that you plug the "u" value back in to your problem before you integrate it. It is okay to do that and you can get the right answer, but you can save yourself a lot of time and paper, by changing your limits of integration based on "u" and just integrating "u" instead of whatever "u" was.

Here are the steps one by one:

$\displaystyle \int\limits_{0}^{\frac{\pi}{2}} \frac{cos x}{1 + sin^2 x} dx$

$\displaystyle \int\limits_{0}^{\frac{\pi}{2}} \frac{cos x}{1 + (sin x)^2} dx$

Let u = sin x

du = cos x dx * This will get rid of the cos x on the top and also your dx

$\displaystyle \int\limits_{0}^{\frac{\pi}{2}} \frac{1}{1 + (u)^2} du$

**NOW** This is where you would usually replace the "u" with "sin x"

But, instead of doing that, you can create new limits based on "u." You do this by plugging the original limits into your u= sin x equation and solving for "u"

So,

u = sin x * Your original limits were 0 and $\displaystyle \frac{\pi}{2}$

Plug those in to "x" and solve for "u" Those values will be your new limits of integration

$\displaystyle u = sin(0) = 0 $ *this is your new lower limit

$\displaystyle u = sin (\frac{\pi}{2}) = 1$ * this is your new upper limit

So now you rewrite your integral like this:

$\displaystyle \int\limits_{0}^{1} \frac{1}{1 + (u)^2} du$

Then you integrate as usual but in terms of du!!

$\displaystyle \int\limits_{0}^{1} \frac{1}{1 + u^2} du$

$\displaystyle = [\arctan(u)] ^1_0 = \arctan (1) - \arctan (0)$

Hope that helps you! Sorry it took me so long! I am SLOW at typing LaTex!