Step by step integration of 1/(5-4x-x^2)^(3/2)
looks the thing inside is a perfect square
start by writing it as square and see if u can cancel out radical
It doesn't work, you need to do trig substitution but I can't figure out the details of how that works yet.
ahh ok its not a perfect square
but still completing the square is the first thing to do here
after completing the square, u wil see wat u need to substitute
\(\large \int \frac{1}{(5-4x-x^2)^{\frac{3}{2}}} dx\)
I have integral (1/(9-(x+2)^2)^(3/2))
And then you use sin for trig substitution but that's where I got stuck
\(\large \int \frac{1}{(5-4x-x^2)^{\frac{3}{2}}} dx\) \(\large \int \frac{1}{(9 - (x+2)^2)^{\frac{3}{2}}} dx\)
substitute \(x+2 = 3\sin \theta\)
Could you explain why you use that?
good question :) plugging that gives me 9sin^2 term,
9 - 9sin^2
i can pull out 9, and simplify the remaining this to cos^2
and cancel out the radical !
plugin first, then u wil understand
Oh okay, thanks a lot! :)
u wil get use to these after doing 2-3 problems
if u dont want to guess like this, there is another way
let me show u that as well
\(\large \int \frac{1}{(5-4x-x^2)^{\frac{3}{2}}} dx\) \(\large \int \frac{1}{(9 - (x+2)^2)^{\frac{3}{2}}} dx\)
factor out 9 from the bottom
\(\large \int \frac{1}{(5-4x-x^2)^{\frac{3}{2}}} dx\) \(\large \int \frac{1}{(9 - (x+2)^2)^{\frac{3}{2}}} dx\) \(\large \int \frac{1}{[9 (1 - \frac{(x+2)^2}{9}]^{\frac{3}{2}}} dx\)
\(\large \int \frac{1}{(5-4x-x^2)^{\frac{3}{2}}} dx\) \(\large \int \frac{1}{(9 - (x+2)^2)^{\frac{3}{2}}} dx\) \(\large \int \frac{1}{[9 (1 - \frac{(x+2)^2}{9}]^{\frac{3}{2}}} dx\) \(\large \int \frac{1}{[9 (1 - (\frac{x+2}{3})^2]^{\frac{3}{2}}} dx\)
Now you're forced to substitute the thing inside square to sin or cos right ?
\(\frac{x+2}{3} = \sin\theta\)
which gives the same substitution :)
Thank you so much :)
do watever way makes more sene to u
np.. u wlc :)
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