Question

Find the integral by substitution (see attached problem)

Answer 1

hmmm

Answer 2

So if we make the u-substitution they suggest, we get this for our du,\[\Large du\quad=\quad\frac{1}{2\sqrt{x+1}}\;dx\]Which changes our integral to,\[\Large \int\limits \frac{1}{\sqrt{3-x}}\left(\frac{1}{2\sqrt{x+1}}\;dx\right)\qquad\to\qquad \int\limits \frac{1}{\sqrt{3-x}}\left(du\right)\]

Answer 3

But then what do we do about that other root? :(
Hmm this is tricky..

Answer 4

\[\Large u=\sqrt{x+1}\]Squaring gives us,\[\Large u^2=x+1\]Subtracting 4 from each side,\[\Large u^2-4=x-3\]Multiplying by -1,\[\Large 4-u^2=3-x\]Square root,\[\Large\sqrt{4-u^2}=\sqrt{x-3}\]

Answer 5

Can we do that maybe? D:

Answer 6

= (3 - x)^(-1/2). Now substitute u = 3 - x and du = -dx to get -u^(-1/2)?

Answer 7

Instead of the u-sub that the suggest we make?
Or another sub?

I don't think that's going to work D:

Answer 8

Understand what I did a sec ago?
That would get us to an integral of:\[\Large \int\limits\frac{1}{\sqrt{4-u^2}}du\]

Answer 9

Which is solvable if you've learned anything about trig subs at this point :)

Answer 10

okay i see what you did now!

Answer 11

which results in arcsin (u/2) and then plug the 'u' back in

Answer 12

yesss, good good.
Remember to plug that 'u' back in before you use your limits of integration. :)

Answer 13

thanks for your help! I appreciate it :)