Question

@lgbasallote Calculus - Parts - Integration

Answer 1

\[\int\limits_{0}^{1/2} (arcsinx)/(\sqrt{1-x^2}dx\]

Answer 2

I know I need to u-sub but I'm really bad at it

Answer 3

are you sure this is integration by parts? im thinking trig sub but not that sure

Answer 4

I'm sure

Answer 5

well then you're familiar with LIATE right?
L - logarithms
I - inverse trig
A - algebraic functions
T - trig functions
E - exponential functions

this dictates the order of priority for u

Answer 6

I've never heard that but that seems very helpful!

Answer 7

oh i see...if you need more info on it...check this out (it's a tutorial i wrote about LIATE) http://openstudy.com/study#/updates/4f9df983e4b000ae9ed2688c

anyway in this problem you have an inverse function (which is arcsin x) and an algebraic expression (that denominator) therefore according to LIATE you will use arcsin x as u

Answer 8

Yes, my teacher never taught me this stuff
Is there anyway you could possibly do the problem out for me as an example please? I have the test tomorrow... @lgbasallote

Answer 9

hmm...i have thought about it..and this is a simple u-sub...NOT integration by parts

Answer 10

derivative of arcsin x = \(\frac{1}{\sqrt{1-x^2}}\)

Answer 11

@lgbasallote I'm still confused, could you please write out the whole problem, you don't need to explain the process, but if I saw the steps I would be able to understand it better

Answer 12

let me explain by taking the derivative of arcsin x

let y = arcsin x

take the sin of both sides

sin y = x

now i'll make a figure you get what that means right??

now i take the derivative of both sides

\[\cos y (\frac{dy}{dx}) = 1\] now i isolate dy/dx by dividing both sides by cos y \[\frac{dy}{dx} = \frac{1}{\cos y}\] now refer to the figure \(\cos y = \sqrt{1-x^2}\) therefore \[\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}\]

Answer 13

Okay I understand that part @lgbasallote

Answer 14

okay so if i let u = arcsin x then du will be that...you get that right?

Answer 15

Yes, I understand

Answer 16

then you can rephrase the question into \[\int udu\] and that is simple integration