Question

Evaluate the indefinite integral.

Answer 1

\[\int\limits u \sqrt{1-u ^{2}}du\]

Answer 2

let x=1-u^2

Answer 3

also can substitute u=sin a
\[\int\limits u \sqrt{1-u^2}du =\int\limits \sin a \sqrt{1-\sin^2a} \cos a da =\int\limits \sin a \cos ^2 a\ da\]

Answer 4

why did you substitute u for sin a?

Answer 5

that is a calculus 2 technique called trigonometric substitution, but looking at your problem I'm guessing you are in calc I and should not yet worry about those tricks

Answer 6

\[\sqrt{1-u^2}\\1-u^2 \ge0 \\so \\u^2 \le 1 \\-1 \le u \le 1\\and\\ -1 \le \sin a \le 1\]

Answer 7

@TuringTest how would i solve this problem?

Answer 8

x=1-u^2
dx=?

Answer 9

do you remember how to take differentials?

Answer 10

dx = -2u^3 du

Answer 11

exponents tend to go down during differentiation, not up

Answer 12

oh is -2u du

Answer 13

right, so du=?

Answer 14

1/2 u

Answer 15

well look, you want to substitute for the part outside the radical, so I should have perhaps asked what
udu= ???
remember: we are changing everything to be in terms of x right now, so don't drop dx, we need it

Answer 16

du=-1/2u dx

Answer 17

yes, but again, look at what we want to get rid of: udu
if you make the current substitution you will have u's and x's in the same problem!

Answer 18

I dont understand what you mean?

Answer 19

this is what we have
I told you to substitute x=1-u^2, so we have to put everything in terms of x now, including having a dx at the end.
if I substitute x=1-u^2, i have this: