Kein Zurück: Indefinite Integrals

Here is a problem on the first homework for my Calculus II class. I thought it was pretty nice so here you go!

Use $\cos2\theta%20=1-2\sin%20^{2}\theta%20=2\cos%20^{2}-1\;\;$ to help evaluate the integrals:

$(a)\;\int \sin^{2}(\frac{x}{2})dx\;\;\;\;\;\;and\;\;\;\;\;\;(b)\;\int\cos^{2}(\frac{x}{2})dx$

To be able to evaluate this integral using the stipulations above, we have to rewrite it in such a way that looks more familiar. One way this can be done is by rewriting the integrand using $\theta$ as a substitution for x/2.

$\int \sin^{2}\theta dx$
Because we have exchanged x/2 for theta we also have to change dx to $d\theta$ .

$\theta=\frac{x}{2}\;\;\;\;\;\;\;d\theta=\frac{1}{2}dx\;\;\rightarrow\;\;dx=2d\theta$

Substitute then we will have:

$\int 2\sin^{2}\theta d\theta$
Now doesn't this look familiar?

If it does look familiar, you should notice that it is a part of the identity:
$\cos2\theta=1-2sin^{2}\theta$

Do some rearranging and we get this:
$1-\cos2\theta=2sin^{2}\theta$

Now we can rewrite the integral again and get:
$\int1-\cos2\theta d\theta$

Okay don't get too excited! We're not done yet! Remember that
$\theta=\frac{x}{2}\;\;\;\;and\;\;\;\;d\theta=\frac{1}{2}dx$

Plug these back into the equation, do the cancellations and we get:
$\frac{1}{2}\int 1-\cos x\; dx$

Integrate to get the final answer:

$\int \sin^{2}(\frac{x}{2})\;dx=\frac{1}{2}(x-\sin x)+C$

That was fun right?! I thought so! :]

I'm not going to go through the second integral because it's basically the same thing. Try it for yourself and you should get the answer:

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

Kein Zurück

Saturday, June 27, 2009

Indefinite Integrals

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