- #1
chuachinghong
- 8
- 0
Well as I am practising for my coming test, I encountered this question:
Integrate
f(x) = absolute [(sin X)^3 * (cos X)^15]dx
within the interval of [0,2pi]
I tried simplifying this integral into...
f(x) = absolute[ ((cos X)^15)*(sin X) -((cos X)^17)*(sin X))] dx
within the interval of [0,2pi]
Also, I break up the integral into 4 parts of pi/2 each i.e [0,pi/2], [pi/2,pi], [pi,3pi/2] and [3pi/2, 2pi]
My question is, when I am calculating the interval of the Second Quadrant, which is from [pi/2, pi]. What should my f(x) look like after taking out the absolute sign?
Hope you can help me with this. Thank you
Integrate
f(x) = absolute [(sin X)^3 * (cos X)^15]dx
within the interval of [0,2pi]
I tried simplifying this integral into...
f(x) = absolute[ ((cos X)^15)*(sin X) -((cos X)^17)*(sin X))] dx
within the interval of [0,2pi]
Also, I break up the integral into 4 parts of pi/2 each i.e [0,pi/2], [pi/2,pi], [pi,3pi/2] and [3pi/2, 2pi]
My question is, when I am calculating the interval of the Second Quadrant, which is from [pi/2, pi]. What should my f(x) look like after taking out the absolute sign?
Hope you can help me with this. Thank you
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