Question

the integral of (x/(sq root(16-x^2)) dx

Answer 1

16-x^2 = t

Answer 2

Use the above substitution, or you can even use 16-x^2=t^2

Answer 3

2
∫ √(16 - x²) dx =
0

let's find the antiderivative:

∫ √(16 - x²) dx =

let x = 4 sin u → sin u = (x/4) → u = arcsin(x/4)

dx = 4 cos u du

thus, substituting, you get:

∫ √(16 - x²) dx = ∫ √[16 - (4 sin u)²] 4 cos u du =

∫ √(16 - 16 sin²u) 4 cos u du =

factor out 16:

∫ √[16(1 - sin²u)] 4 cos u du =

∫ 4 √(1 - sin²u) 4 cos u du =

replace (1 - sin²u) with cos²u:

∫ 4 √(cos²u) 4 cos u du =

∫ 4 cos u 4 cos u du =

∫ 16 cos²u du =

recall the double-angle identity: cos²θ = [1 + cos(2θ)]/2

thus:

∫ 16 {[1 + cos(2u)]/2} du =

∫ 8 [1 + cos(2u)] du =

split it and pull out the constants:

8 ∫ du + 8 ∫ cos(2u) du =

8u + 8 (1/2)sin(2u) + C =

according to the double-angle identities, rewrite sin(2u) as 2sin u cos u:

8u + 8 (1/2)2sin u cos u + C =

8u + 8sin u cos u + C

summing up. you have:

x = 4 sin u → sin u = (x/4) → u = arcsin(x/4)

hence
cos u = √(1 - sin²u) = √[1 - (x/4)²] = √[1 - (x²/16)] = √[(16 - x²)/16)] = [√(16 - x²)]/4

thus, substituting back, you get:

8u + 8sin u cos u + C = 8arcsin(x/4) + 8(x/4) {[√(16 - x²)]/4} + C =

8arcsin(x/4) + (8/16) x√(16 - x²) + C =

8arcsin(x/4) + (1/2) x√(16 - x²) + C

then evaluate the definite integral, yielding:

2
∫ √(16 - x²) dx = [8arcsin(2/4) + (1/2)2√(16 - 2²)] - [8arcsin(0/4) + (1/2) 0√(16 - 0²)] =
0

8arcsin(1/2) + √(16 - 4) - (0 + 0) =

8(π/6) + √12 =

(4/3)π + 2√3 (≈ 7.6528)


:-) hope it helps...
Bye!

Answer 4

Where did you get sin and cos from?

Answer 5

How do you do the Ctrl+C and Ctrl+V rohan?

Answer 6

\[\int\limits\frac{x}{\sqrt{16-x^{2}}} dx\]
let u = 16-x^2 => du/-2x
\[\int\limits\frac{x}{\sqrt{u}} *\frac{du}{-2x} =>-\frac{1}{2} \int\limits\frac{1}{\sqrt{u}} du\]
the integrate it..

Answer 7

See Meme agrees with me, and Mimi_x3 is always right.

Answer 8

lol, just wanted to elaborate..

Answer 9

\[-\sqrt{16-x^2}\]

Answer 10

you can check @Ishaan94 in yahoo answer some one asked this question 3 years ago

Answer 11

2 me

Answer 12

-1/2ln |16-x^2| +c ?