Wolf72
Explorer
kingpaul said:But math is deeply rooted in philosophy and faith, it'd be good from a historical perspective.
I think you're just trying to get me to have a Calculas siezure

kingpaul said:But math is deeply rooted in philosophy and faith, it'd be good from a historical perspective.
Actually, no. Math is based entirely on faith. What is the most basic mathematical question? What is 1 + 1? 2. Why is this? We take this on faith that it is true. Taking the class filled with the proofs of calculus just goes to show how much we take for granted in the mathematical arena.Wolf72 said:I think you're just trying to get me to have a Calculas siezure![]()
Sir Whiskers said:Okay, enough metaphysics for one day...![]()
Dinkeldog said:Just remember: regardless of what you get on the AP test, take Calculus in college. This goes especially if you're looking at a technical degree. You're not suffering through a year's worth of classes you could avoid (thereby speeding up graduation by that much--logical fallacy, as graduation is never sped up, only delayed), you're padding your GPA with a year's worth of A's.![]()
orbitalfreak said:(1)
∫ (sin<sup>3</sup>x)(cos<sup>2</sup>x)dx
= ∫ (sinx)(sin<sup>2</sup>x)(cos<sup>2</sup>x)dx
= ∫ (sinx)(1-cos<sup>2</sup>x)(cos<sup>2</sup>x)dx
= ∫ (sinx)(cos<sup>2</sup>x - cos<sup>4</sup>x)dx
Let u = cosx
then du = (-sinx)dx
So,
∫ (sinx)(cos<sup>2</sup>x - cos<sup>4</sup>x)dx
= -∫ u<sup>2</sup> - u<sup>4</sup> dx
= - ((1/3)u<sup>3</sup> - (1/5)u<sup>5</sup>) +C
clean it up, supstituting back for u, the final answer is
(1/5)cos<sup>5</sup>x - (1/3)cos<sup>3</sup>x + C
//
(2)
∫<sub>0</sub><sup>1</sup> 1/(√(4-x<sup>2</sup>)) dx
This is a standard integral of the form
∫<sub>0</sub><sup>1</sup> 1/(√(a<sup>2</sup>-x<sup>2</sup>)) dx
= arcsin (x/a) + C
in this case, a<sup>2</sup> = 4, so a = 2. Substitute this, and you get
arcsin(x/2)¦<sub>0</sub><sup>1</sup>
= arcsin(1/2) - arcsin(0/2)
= arcsin(1/2)
= π/6 <-- That's a (pi over six), if it doesn't show up.
//