Question

Integrate by trig sub.

\[\int\limits_{}^{}dx/x^{2}\sqrt{4-x^2}\] let x=2sintheta

Answer 1

\[\int\limits_{}^{}dx/x^{2}\sqrt{4-x^2}\]

Answer 2

Do you know how to do a \(u\) sub? Trig sub isn't really much different.

Answer 3

Find dx

Answer 4

Yes, I know how to do it all. I do all the sub, and after doing it all, i can get it down to .25\[.25\int\limits_{}^{}1/\sin^2\theta\]

Answer 5

now i have to go back into terms of x, and thats where im having trouble

Answer 6

You have to do the integral first.

Answer 7

What is \(\frac{d(\tan(\theta))}{d\theta}\)? What is \(\frac{d(\cot(\theta))}{d\theta}\)?

Answer 8

1/sin^2theta = csc^2theta. integral of csc^2theta = -1cot(theta)

Answer 9

Okay, so before undoing the trig sub, what is the anti-derivative in terms of theta?

Answer 10

\[\int\limits_{}^{}dtheta=\theta\]

Answer 11

No.

Answer 12

Isn't it \(-0.25\cot(\theta) +C\)

Answer 13

yes, thats right

Answer 14

\[\int\limits \frac{dx}{x^2\sqrt{4-x^2}}= \frac{1}{4}\frac{(x-2)(x+2)}{x \sqrt{4-x^2}}+c\]

Answer 15

need to do it through trig sub mathsmind :<

Answer 16

that's what i got, solving on piece of paper

Answer 17

Now you need to remember that \(\cot = \cos/\sin\) and that \(\cos = \sqrt{1-\sin^2}\)

Answer 18

oh ok that would make life easier like a piece of cherry on top of the cake

Answer 19

answer in the back of the book is \[-\sqrt{4-x^2}/4x+c\]

Answer 20

yes that is right and my answer is right too...

Answer 21

So basically: \[
2\cot(\theta) = \frac{\sqrt{4-(2\sin(\theta))^2}}{2\sin(\theta)}
\]

Answer 22

because you are asked to use trig sub

Answer 23

wio i dont have 2cot though, i have -.25cot so far and im trying to turn that into what i just posted

Answer 24

\(-0.25 = 2 \times -0.125\)

Answer 25

\( -0.25\cot = -0.125 \cdot 2\cot\)

Answer 26

Could somebody just solve the problem for me, with the final answer being in terms of x and i can see where im going wrong? i essentially have majority of the problem right, but im messing up somewhere in the end

Answer 27

wait am solving this on a piece of paper be patient

Answer 28

i will be.

Answer 29

oh wow. i figured it out haha. i realized where i was making my mistake :<

Answer 30

ok do you agree that the final answer would be -cos(theta)/sin(theta)

Answer 31

all you have to do is sub things back in

Answer 32

so in our assumptions x^2=4sin^(theta)

Answer 33

also we said 4-4sin^(theta)=4cos^(theta)

Answer 34

i was forgetting the basic identity that 1/tan=cotan. using that identity, i was able to figure out where i was making the mistake, thanks mathsmind and wio

Answer 35

you're welcome

Answer 36

all you have to do is move backwords sub x's back in you'll get the exact answer in the book, i used partial fraction to solve the above problem which will lead to the same answer after simplification

Answer 37

\[
\begin{split}
\int\frac{dx}{x^{2}\sqrt{4-x^2}} &= \int\frac{2\cos(\theta)d\theta}{4\sin^{2}(\theta)\sqrt{4-4\sin^{2}(\theta)}} \\
&= \int \frac{d\theta}{4\sin^2(\theta) } \\
&=\int \frac{1}{4}\csc^2(\theta)d\theta \\
&= -\frac{\cot(\theta)}{4}+C\\
&=-\frac{1}{8}\frac{2\cos(\theta)}{2\sin(\theta)} +C \\
&= -\frac{1}{8}\frac{\sqrt{4-4\sin^2(\theta)}}{2\sin(\theta)} \\
&= -\frac{\sqrt{4-x^2}}{8x}
\end{split}
\]

Answer 38

\(+C\)