Integration With Trigonometric Functions

[raveyp]

I'm stuck on this integration question? Am I missing a rule in trigonometry?

 

[raveyp]

Is it because if you integrate sin you get cosine - and the cosine of pi/3 or pi/6 is equal to 1?

 

[pka]

Is it because if you integrate sin you get cosine - and the cosine of pi/3 or pi/6 is equal to 1?

First make these changes $\displaystyle \sin \left( {\frac{{\pi }}{3}} \right) = \frac{{\sqrt 3 }}{2}\,\&\;\sin \left( {\frac{\pi }{6}} \right) = \frac{1}{2}$
Then we have $\displaystyle \int( {\frac{{\sqrt 3 }}{2}\,x + \frac{1}{2}{x^2})dx}$

SEE HERE

 

[raveyp]

Ahh yeah of course! Unit circle stuff!! Been ages since I've done that.. Thanks pka!

 

[Dr.Peterson]

I'm stuck on this integration question? Am I missing a rule in trigonometry?

There are two errors you made.

One is an error of omission: you didn't do what pka showed, namely to actually find the values of the sine.

The other is an error of commission: you somehow changed sin(pi/3) to sin(pi/6), possibly by thinking you could "distribute" the 1/2 inside the sine. That's just wrong! There is nothing you can do outside a sine that will change the argument inside the sine. If you had evaluated after doing this, you would have the wrong values.

 

[raveyp]

There are two errors you made.

One is an error of omission: you didn't do what pka showed, namely to actually find the values of the sine.

The other is an error of commission: you somehow changed sin(pi/3) to sin(pi/6), possibly by thinking you could "distribute" the 1/2 inside the sine. That's just wrong! There is nothing you can do outside a sine that will change the argument inside the sine. If you had evaluated after doing this, you would have the wrong values.

Ah okay, thanks Dr. Peterson. Helpful as always.