Fp2 integration question

Original post by Music99

Any hints?

Do you get twice what the answer should be by any chance?

Reply 2

justinawe

19

Original post by Music99

Unparseable latex formula:

[color="#444444"][font="verdana"]\displaystyle\int^\frac{\sqrt3}{4}_0 \frac{2xsin^{-1}2x}{\sqrt(1-4x^2)}\ dx [/font].[br][br]The square root in the denominator is over everything. So In the previous part of the question I worked out that the derivative of [latex] y= x-\sqrt(1-x^2)sin^{-1}x [/latex] is [latex] \frac{xsin^{-1}x}{\sqrt(1-x^2)} [/latex].[br][br]so I called the [/color]

y= x-\sqrt(1-x^2)sin^{-1}x

f(x), so then what I want to integrate is essentially f(2x), so then I just replaced all the x values in y=... by 2x, but then when I substitute in the limits I get the wrong answer. Any hints?

Try a substitution that gets your integral in the form of the derivative you found.

Reply 3

OP

Original post by ghostwalker

Do you get twice what the answer should be by any chance?

I end up with

\frac{\sqrt3}{2}-\frac{\sqrt10}{4}(\frac{\pi}{3})

. the root 10 bit seems really out of place, but I double checked it and it gives the correct value for what I have as the integral.

Reply 4

Original post by Music99

I end up with

\frac{\sqrt3}{2}-\frac{\sqrt10}{4}(\frac{\pi}{3})

. the root 10 bit seems really out of place, but I double checked it and it gives the correct value for what I have as the integral.

I got

\dfrac{\sqrt{3}}{4}-\dfrac{\pi}{12}

What's the correct answer?

Reply 5

BabyMaths

Original post by ghostwalker

I got

\dfrac{\sqrt{3}}{4}-\dfrac{\pi}{12}

What's the correct answer?

That is.

Reply 6

OP

Original post by ghostwalker

I got

\dfrac{\sqrt{3}}{4}-\dfrac{\pi}{12}

What's the correct answer?

That is. Maybe I just subbed it in wrong or something, I'm confused now :P.

Reply 7

Original post by BabyMaths

That is.

Thanks.

Original post by Music99

That is. Maybe I just subbed it in wrong or something, I'm confused now :P.

Post working?

Reply 8

OP

Original post by ghostwalker

Thanks.

Post working?

im on my phone so can't latex, but what I did was in the equation with y=... Everywhere there is an x I replaced it with a 2x then subbed in the limits.

Reply 9

Original post by Music99

im on my phone so can't latex, but what I did was in the equation with y=... Everywhere there is an x I replaced it with a 2x then subbed in the limits.

Safest way would be to use a subsitution, u=2x, so du/dx =2, and convert the limits. Then you'll have the exact form for the integral.

Reply 10

OP

Original post by ghostwalker

Safest way would be to use a subsitution, u=2x, so du/dx =2, and convert the limits. Then you'll have the exact form for the integral.

Ah okay, also on questions like this one how do you know what substitution to use? I mean I know it comes from experience but is there a general rule of thumb.

Reply 11