Question

Can some one help me with a fairly lengthy integration question? Compute the intergal of 1/(5-4x-x^2)^(5/2)...I have completed the square and been able to factor out the 9^(5/2) to leave the denominator as ((x-2)^2-1)^(5/2). My solution set tells me to substitute (x-2) as a sin function and I'm totally lost from there

Answer 1

\[x-2=\sec u~~\Rightarrow~~dx=\sec u\tan u~du\]
So the integral becomes
\[\int\frac{\sec u\tan u~du}{(\sec^2u-1)^{5/2}}\]

Answer 2

(5-4x-x^2)^(5/2) is not ((x-2)^2-1)^(5/2)

Answer 3

^^ I was operating under the assumption it was - shame on me for not checking OP's work :P

Answer 4

lol its ok, you would have figured out :P

Answer 5

\[\begin{align*}5-4x-x^2&=5-(x^2+4x)\\
&=5-(x^2+4x+4-4)\\
&=5-((x+2)^2-4)\\
&=5-(x+2)^2+4\\
&=9-(x+2)^2
\end{align*}\]
You can't just factor out the 9 like that...

Answer 6

are you sure your set tells x-2 as sin ?
or x+2 as sin ?

Answer 7

\[\int\frac{dx}{\left(9-(x+2)^2\right)^{5/2}}\]
The substitution would then be
\[x+2=3\sin u~~\Rightarrow~~dx=3\cos u~du\]
and so
\[\int\frac{3\cos u~du}{\left(9-(3\sin u)^2\right)^{5/2}}\]
NOW you can factor out the 9:
\[\int\frac{3\cos u~du}{\left(9-9\sin^2 u\right)^{5/2}}\\
\int\frac{3\cos u~du}{9^{5/2}\left(1-\sin^2 u\right)^{5/2}}\\
\]

Answer 8

I can't see the sign change when I completing the square...

Answer 9

show your work, we'll help you find the error

Answer 10

I just saw it. I was convinced that was wrong on the solution set. At that point it tells me to set x+2=3sin theta. Where is the 3 coming from?

Answer 11

for
\(\Large a^2-x^2\)
we plug in x = a sin theta

Answer 12

I can see it now. Thank you. I've been staring at this thing for a couple days and could not find the mistake!

Answer 13

glad we could help ^_^
welcome :)

Answer 14

I've got another thats bugging me if you are still here....

Answer 15

\[\int\limits_{1}^{4}\sqrt{t}\ln t dt\]

Answer 16

i am here :)

Answer 17

integration by part

Answer 18

Let me show you where I'm stuck\[= \ln t \times2/3t ^{(3/2})-\int\limits_{1}^{4}2/3t ^{(3/2)}*(1/t)*t ^{(1/2)}dt\]

Answer 19

My solution set shows that it should be \[\int\limits_{1}^{4} (2/3) t ^{(3/2)}*t ^{-1}dt\]

Answer 20

how did u get that extra t^(1/2)
?

Answer 21

\(= \ln t \times2/3t ^{(3/2)}-\int\limits_{1}^{4}(2/3)t ^{(3/2)}\times (1/t)dt\)

and 1/t is \(t^{-1}\)

Answer 22

I am wondering the same thing...

Answer 23

The formula for integration by parts says \[\int\limits_{a}^{b}uv \prime=uv -\int\limits_{a}^{b}u primev\]

Answer 24

Ive got u=lnt and u prime =1/t v prime sqrt v and v=2/3(t)^3/2

Answer 25

correct

Answer 26

Oh dear... I just my mistake....its the end of the quarter and I'm fried!!!! Lol. Just typing it and looking at it with you helped me. THANKS!!!

Answer 27

I just randomly added that t^(1/2). WOW!!!!

Answer 28

lol yeah, ans welcome :)