Evaluate the integral.

**saiyanmx89** · January 26th 2009, 03:36 PM

$<br /> \int \limits_0^{\frac{\\pi}{2}} \frac{cosx}{1+sin^2(x)}dx<br />$

**Danny** · January 26th 2009, 03:39 PM

Originally Posted by saiyanmx89

$<br /> \int \limits_0^{\frac{\\pi}{2}} \frac{cosx}{1+sin^2(x)}dx<br />$

Let $u = \sin x, \;\;\; \text{so} \;\;\; du = \cos x \, dx$

so (note the change of limits)

$\int_0^1 \frac{1}{1 + u^2} \; du$

You can do it from here.

**saiyanmx89** · January 26th 2009, 03:47 PM

why does u=sinx? how did you get sinx?

**mollymcf2009** · January 26th 2009, 03:48 PM

$sin^2(x)$ is the same thing as $(sinx)^2$

So, (sinx) is what you are using for your u-substitution

**saiyanmx89** · January 26th 2009, 03:50 PM

thank you!!!

**saiyanmx89** · January 26th 2009, 03:53 PM

So, would the answer be:

arctan(sinx) +c ???

**mollymcf2009** · January 26th 2009, 04:33 PM

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!

**saiyanmx89** · January 26th 2009, 05:16 PM

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...

**mr fantastic** · January 26th 2009, 06:21 PM

Originally Posted by saiyanmx89

So, the answer would be found from:

arctan(1)-arctan(0)

??
[snip]

Yes.

**mollymcf2009** · January 26th 2009, 07:01 PM

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:

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

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

Let u = sin x

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

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

**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 $\frac{\pi}{2}$
Plug those in to "x" and solve for "u" Those values will be your new limits of integration

$u = sin(0) = 0$ *this is your new lower limit
$u = sin (\frac{\pi}{2}) = 1$ * this is your new upper limit

So now you rewrite your integral like this:

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

Then you integrate as usual but in terms of du!!
$\int\limits_{0}^{1} \frac{1}{1 + u^2} du$

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

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

Thread: Evaluate the integral.

LinkBack

Thread Tools

Display

Evaluate the integral.

Similar Math Help Forum Discussions

Use Cauchy's integral theorem for derivatives to evaluate the contour integral.

Graph the region of integration and evaluate the double integral (integral from 0 to

Using a given property of an integral to evaluate and integral

Evaluate the integral using the FTC I

How do you evaluate this integral?

Search Tags