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Integral of sinx from 0 to pi/2?
My calculator gives me a value of 1. But how is that possible?
The integral of sinx is -cosx, and cos(pi/2) is 0.
Does the integral of sinx have a special property that I am missing?
3 Answers
- Anonymous8 years agoFavorite Answer
When intergrating sinx from 0 to pi/2, you have to plug in BOTH 0 and pi/2 to get the answer.
int(sinx, 0, pi/2) = [-cosx], 0, pi/2 => (-cos(0))-(cos(pi/2)) = (1)-(0)= 1
- 8 years ago
The radius of the unit circle is 1. The circumference is 2 pi.
sin(x) is in the 1st quadrant, which gives you a positive value of 1.
If you integrate cos(pi/2), you essentially just integrate the x-axis itself, which gives you an area of 0.
- ?Lv 45 years ago
?cosx dx / (sinx + a million) enable sinx + a million = t => cosx dx = dt => crucial = ?dt / t = ln l t l + c = ln (sinx + a million) + c [be conscious: Modulus sign isn't used as sinx + a million ? 0] Plugging limits 0 to ?/2, = ln [sin(?/2) + a million] - ln1 = ln(2).
