Can someone tell me the antiderivative to the square root of (25-(x^2))?

Question

If you could, please find the antiderivative using the "power rule" of antiderivatives. Please show all your work step by step and clearly, if you do I promise to select you as best answer

Answer 1

∫ √(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

I hope it helps...

Bye!