# FP2: Quick Integration QuestionWatch

Announcements
#1
Q:

I got but the answer states that it is . Where did the come from?
0
11 years ago
#2
(Original post by Wish)
Q:

I got but the answer states that it is . Where did the come from?
Dividing by the derivative of the inner function. (2x)
0
11 years ago
#3
you have to say that 2x=u

then dx/du = 1/2

so you intergrate: 0.5[1/(1-u^2)]du

and you end up with the right answer
0
11 years ago
#4
to integrate it you use root(a^2 - x^2). In this case x^2 is 4x^2 and hence, x is 2x, so you have to divide the integral by its derivate which is 2.
0
#5
Is that always the case? For all the ones provided in the formula book. If it has a constant cooefficient then we have to have a fraction on the square root at the front.
0
11 years ago
#6
This is how i do it, maybe you could find it useful...
you take the 4 out of the root i.e. 1 / sqrt(4 (1/4)+ x^2) which becomes 1/(2 sqrt( (1/4)+(x^2) )... then take the 1/2 out since its a constant and then integrate 1 / sqrt ( (1/4) + (x^2) ) and multiply the (1/2) at the end...
doing it this way always works for me
0
X

new posts
Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### University open days

Thu, 24 Oct '19
• Cardiff University
Sat, 26 Oct '19
• Brunel University London
Sat, 26 Oct '19

### Poll

Join the discussion

Yes (94)
24.42%
No (291)
75.58%