Integral

**joatmon** · April 12th 2011, 09:09 PM

Evaluate:

$\displaystyle \int^{\frac{\pi}{2}}_0 [\frac{d}{dx} (\sin(\frac{x}{2}) \cos(\frac{x}{3})]dx$

Here is what I did:

$\displaystyle \int^{\frac{\pi}{2}}_0 \cos(\frac{x}{2}) \cos(\frac{x}{3})dx + \int^{\frac{\pi}{2}}_0 \sin(\frac{x}{2})(-\sin(\frac{x}{3}))dx$ (using the product rule to evaluate d/dx and separating the resultant integrals)

$\displaystyle \int^{\frac{\pi}{2}}_0 \cos(\frac{x}{2}) \cos(\frac{x}{3})dx - \int^{\frac{\pi}{2}}_0 \sin(\frac{x}{2})\sin(\frac{x}{3})dx$ (just rewriting to pull out the minus sign)

Assuming that this is right so far, I don't know how to evaluate these two integrals. I'm thinking that maybe I can use FTC to simplify this process somehow?

Thanks.

**pickslides** · April 12th 2011, 09:16 PM

hmmmm,

consider $\displaystyle \frac{d}{dx}\left(\sin\left(\frac{x}{2}\right)\rig ht) =\frac{1}{2}\cos\left(\frac{x}{2}\right)$

**joatmon** · April 12th 2011, 09:45 PM

You're right, but I'm still stuck. I just forgot to apply the chain rule on the derivatives, but all that does is adds a constant to each integral (I think):

$\displaystyle \int^{\frac{\pi}{2}}_0 \cos(\frac{x}{2}) \cos(\frac{x}{3})(\frac{1}{2})dx + \int^{\frac{\pi}{2}}_0 \sin(\frac{x}{2})(-\sin(\frac{x}{3}))(\frac{1}{3})dx$

$\displaystyle \int^{\frac{\pi}{2}}_0 \cos(\frac{x}{2}) \cos(\frac{x}{3})(\frac{1}{2})dx - \int^{\frac{\pi}{2}}_0 \sin(\frac{x}{2})\sin(\frac{x}{3})(\frac{1}{3})dx$

I don't see a U substitution here. Am I approaching this correctly?

**Prove It** · April 12th 2011, 09:51 PM

Isn't the integral of the derivative of a function just that function itself?

**pickslides** · April 12th 2011, 10:06 PM

Originally Posted by joatmon

I don't see a U substitution here. Am I approaching this correctly?

You can continue this way, no u-sub needed, just use,

$\displaystyle \sin a \sin b = \frac{1}{2}(\cos (a+b)-\cos (a-b))$

$\displaystyle \cos a \cos b = \frac{1}{2}(\cos (a+b)+\cos (a-b))$

but it should match up with what Hayden has said in post #4

**joatmon** · April 12th 2011, 10:17 PM

That's what I was wondering, but I am used to seeing it the other way around (with the derivative on the outside), so I didn't think that was legal. Thanks!

**joatmon** · April 12th 2011, 10:28 PM

Sorry, but I am still confused. If the integral and the derivative "cancel each other out", then do I just evaluate the function at the upper limit like this:

$\displaystyle \sin(\frac{\pi}{4}) \cos(\frac{\pi}{6})$

**joatmon** · April 12th 2011, 10:35 PM

I got it. That's the answer. To finish this off:

$\displaystyle \sin(\frac{\pi}{4}) \cos(\frac{\pi}{6})$

$\displaystyle \frac{\sqrt{2}}{2}\frac{\sqrt{3}}{2} = \frac{\sqrt{6}}{4}$

Thanks, everyone.

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