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OpenStudy (anonymous):
Find ∫sin (πx/2)
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OpenStudy (amistre64):
what does sin integrate to?
OpenStudy (anonymous):
-cos
OpenStudy (amistre64):
then we know this at least has to come from -cos(pi x/2) taket he derivative of that to see what we need to include.
OpenStudy (anonymous):
Theres my problem we get sin(pix/2) times (pi/2)
OpenStudy (anonymous):
So I need something that cancles out the (pi/2)
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OpenStudy (amistre64):
then we need to find a pi/2 to attach to our integral so that we have exactly that :)
OpenStudy (amistre64):
does 2/pi cancel out pi/2?
OpenStudy (anonymous):
Yeah
OpenStudy (amistre64):
then lets us that :)
OpenStudy (amistre64):
\[\frac{d}{dx}[-\frac{2}{pi}cos(\frac{pi}{2}x)]\to\ \frac{2pi}{2pi}sin(\frac{pi}{2}x)\]
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OpenStudy (anonymous):
Thank you, I have to admit that was fun
OpenStudy (amistre64):
lol :)
OpenStudy (amistre64):
when the problems arent so basic, they use something called a u-sub technique to clean things up
OpenStudy (amistre64):
\[\int sin(pix/2)dx\]
\[pix/2 = u\]
\[\frac{d}{dx}[pix/2 = u]\to\ \frac{dx}{dx}pi/2 = \frac{du}{dx}\]
\[dx=\frac{2}{pi}du\]
substitute the parts now
\[\int sin(pix/2)dx\]\[\int sin(u)\frac{2}{pi}du\to\ \frac{2}{pi} \int sin(u)du\]
OpenStudy (anonymous):
o ok I didnt think to use the u sub when I was looking at it
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OpenStudy (amistre64):
i did, but felt it rather pointless :) when they are this basic, you need to develop a foresight
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