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# Definite integral of trig function

AP.CALC:
FUN‑6 (EU)
,
FUN‑6.B (LO)
,
FUN‑6.C (LO)
Sal finds the definite integral of 9sin(x) between 11π/2 and 6π.

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• at i don't understand why he can just subtract?
• That's a good question Vivian! See, he is not actually subtracting; he is putting it in terms that are easier to understand. Cosine is a periodic function with a period of 2π. This means that every 2π, the function repeats. This makes f(0) = f(2π) = f(4π) and etc. It is a ton easier to know the cosine of a more familiar value, namely those that are 0<x<2π. Sal simply rewrote 11π/2 as 3π/2 + 4π. Just like f(0) = f(2π), we can also say that f(3π/2) = f(7π/2) = f(11/2) and etc. I hope that helped!
• Is it acceptable to state that the antiderivative of sin(x) is -cos(x) instead of multiplying the sin(x) and 9 by -1? I believe it results in the same answer in this case.
• I can see no reason not to, as long as you keep careful track of the signs.
(I do it that way as well.)
• So you can integrate in radians too?
• Actually it is easier to differentiate and integrate using radians instead of degrees. The formulas for derivatives and integrals of trig functions would become more complicated if degrees instead of radians are used (example: the antiderivative of cos(x) is sin(x) + C if radians are used, but is (180/pi)sin(x) + C if degrees are used). This is one of the main reasons why radian measurement is taught in trigonometry.
The simplicity of using radians instead of degrees should come as no surprise. Whereas the choice of 360 degrees in a circle is arbitrary, the fact that there are 2*pi radians (radiuses) in a circle is a consequence of the fact that the circumference is 2*pi*r. So radians, unlike degrees, is a natural unit of angle measurement.

Have a blessed, wonderful day!
• This feels wrong to me.

I understand that cos(6pi) = cos(0), and cos(11pi/2) == cos(3pi/2), but those are just the point values of the the functions. When we integrate, we're taking tiny slices and adding them together. Don't have to do this over the entire range 6pi to 11pi/2?

It seems like this shortcut is disregarding the fact that this is a sum.
• Because cos(x) is a constant oscillating function, the areas will keep on being the same because the area under a curve is - and above is + so because in intervals of pi/2, it is equal to each other if the intervals stay the same even if the bounds change
• When do we know to use either radian or degree mode on our calculator when calculating integrals?
(1 vote)
• When calculating integrals, always use radians.
• In this example Sal evaluates from 11/2pi to 6pi. In earlier videos on integration with non trig functions we saw that evaluating an integral from, for example 10 to 2 (where 10 is at the buttom and 2 at the top) required us to swap the signs for the final result.

Does this not apply to integrals of periodic functions?
(1 vote)
• I think you only need to do that if you're using Riemann sums.