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Mathematics 18 Online
OpenStudy (anonymous):

Step by step integration of 1/(5-4x-x^2)^(3/2)

ganeshie8 (ganeshie8):

looks the thing inside is a perfect square

ganeshie8 (ganeshie8):

start by writing it as square and see if u can cancel out radical

OpenStudy (anonymous):

It doesn't work, you need to do trig substitution but I can't figure out the details of how that works yet.

ganeshie8 (ganeshie8):

ahh ok its not a perfect square

ganeshie8 (ganeshie8):

but still completing the square is the first thing to do here

ganeshie8 (ganeshie8):

after completing the square, u wil see wat u need to substitute

ganeshie8 (ganeshie8):

\(\large \int \frac{1}{(5-4x-x^2)^{\frac{3}{2}}} dx\)

OpenStudy (anonymous):

I have integral (1/(9-(x+2)^2)^(3/2))

OpenStudy (anonymous):

And then you use sin for trig substitution but that's where I got stuck

ganeshie8 (ganeshie8):

\(\large \int \frac{1}{(5-4x-x^2)^{\frac{3}{2}}} dx\) \(\large \int \frac{1}{(9 - (x+2)^2)^{\frac{3}{2}}} dx\)

ganeshie8 (ganeshie8):

substitute \(x+2 = 3\sin \theta\)

OpenStudy (anonymous):

Could you explain why you use that?

ganeshie8 (ganeshie8):

good question :) plugging that gives me 9sin^2 term,

ganeshie8 (ganeshie8):

9 - 9sin^2

ganeshie8 (ganeshie8):

i can pull out 9, and simplify the remaining this to cos^2

ganeshie8 (ganeshie8):

and cancel out the radical !

ganeshie8 (ganeshie8):

plugin first, then u wil understand

OpenStudy (anonymous):

Oh okay, thanks a lot! :)

ganeshie8 (ganeshie8):

u wil get use to these after doing 2-3 problems

ganeshie8 (ganeshie8):

if u dont want to guess like this, there is another way

ganeshie8 (ganeshie8):

let me show u that as well

ganeshie8 (ganeshie8):

\(\large \int \frac{1}{(5-4x-x^2)^{\frac{3}{2}}} dx\) \(\large \int \frac{1}{(9 - (x+2)^2)^{\frac{3}{2}}} dx\)

ganeshie8 (ganeshie8):

factor out 9 from the bottom

ganeshie8 (ganeshie8):

\(\large \int \frac{1}{(5-4x-x^2)^{\frac{3}{2}}} dx\) \(\large \int \frac{1}{(9 - (x+2)^2)^{\frac{3}{2}}} dx\) \(\large \int \frac{1}{[9 (1 - \frac{(x+2)^2}{9}]^{\frac{3}{2}}} dx\)

ganeshie8 (ganeshie8):

\(\large \int \frac{1}{(5-4x-x^2)^{\frac{3}{2}}} dx\) \(\large \int \frac{1}{(9 - (x+2)^2)^{\frac{3}{2}}} dx\) \(\large \int \frac{1}{[9 (1 - \frac{(x+2)^2}{9}]^{\frac{3}{2}}} dx\) \(\large \int \frac{1}{[9 (1 - (\frac{x+2}{3})^2]^{\frac{3}{2}}} dx\)

ganeshie8 (ganeshie8):

Now you're forced to substitute the thing inside square to sin or cos right ?

ganeshie8 (ganeshie8):

\(\frac{x+2}{3} = \sin\theta\)

ganeshie8 (ganeshie8):

which gives the same substitution :)

OpenStudy (anonymous):

Thank you so much :)

ganeshie8 (ganeshie8):

do watever way makes more sene to u

ganeshie8 (ganeshie8):

np.. u wlc :)

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