TO ALL OF YOU WHO UNDERESTIMATE ME

[chillintoucan28]

Here it is, for all of you who claimed that I could barely pass an algebra class, I solved the antiderivative to the square root of 25-x^2

THIS IS THE RIGHT WAY TO SOLVE IT:

? ?(25 - x²) dx =

let: x = 5 sin u ? dx = 5 cos u du
sin u = (x/5) ? u = arcsin(x/5)

then substitute, yielding:

? ?(25 - x²) dx = ? ?[25 - (5 sin u)²] 5 cos u du =

? [?(25 - 25 sin²u)] 5 cos u du =

factor out 25:

? {?[25(1 - sin²u)]} 5 cos u du =

? [5?(1 - sin²u)] 5 cos u du =

take out the constants:

25 ? [?(cos²u)] cos u du =

25 ? cos u cos u du =

25 ? cos²u du =

according to half-angle identities, replace cos²u with (1/2)[1 + cos(2u)]:

25 ? (1/2)[1 + cos(2u)] du =

(25/2) ? [1 + cos(2u)] du =

(25/2) ? du + (25/2) ? cos(2u) du =

(25/2)u + (25/2) (1/2)sin(2u) + C =

(25/2)u + (25/4) sin(2u) + C
that, due to double-angle identities, is the same as:

(25/2)u + (25/4) (2 sin u cos u) + C =

(25/2)u + (25/2) sin u cos u + C

now recall:
x = 5 sin u ?
sin u = (x/5) ?
u = arcsin(x/5)
thus cos u = ?(1 - sin²u) = ?[1 - (x/5)²] = ?[1 - (x²/25)] = ?[(25 - x²) /25] =
(1/5)?(25 - x²)

thus, substituting back, you get:

(25/2)u + (25/2) sin u cos u + C = (25/2) arcsin(x/5) + (25/2)(x/5)(1/5)?(25 - x²) + C
and therefore:

? ?(25 - x²) dx = (25/2)arcsin(x/5) + (1/2)x?(25 - x²) + C


Ha, I guess this is my last post, due to the moderators request, but instead of you guys showing me the wrong way to solve a problem I have shown you the right way.

 

[Deleted member 4993]

Nice copy from:

http://answers.yahoo.com/question/index ... 920AAaJyFJ

 

[stapel]

TO ALL OF YOU WHO UNDERESTIMATE ME: Here it is....I solved the antiderivative to the square root of 25-x^2....I have shown you the right way [to solve this exercise]

Nice copy from [Yahoo Answers]

If this isn't a great demonstration of why it's generally a bad practice to provide complete worked solutions, especially for "students" who have refused to show any effort at all, I can't imagine what would be! :shock:

Eliz.