Question

Integral 1/(x^2 sqrt(4-x^2)) dx

Answer 1

\(x=2\sin\theta\)

Answer 2

@unimatix do you want a regular example where this method (trigonometric substitution) is used?

Answer 3

That would be great honestly.

Answer 4

Ok, I am assuming you are familiar with a regular u-substitution, so
I won't go to deep into explaining it....

\(\large\color{black}{ \displaystyle \int \frac{4}{\sqrt{1+x^2}}dx}\)

\(\large\color{gray}{ \displaystyle x=\tan\theta}\)

The derivative of \(\tan\theta\) is \(\sec^2\theta\), so your \(dx\) component
will be:

\(\large\color{gray}{ \displaystyle dx=\sec^2\theta{~} d\theta}\)

\(\large\color{black}{ \displaystyle \int \frac{4\sec^2\theta}{\sqrt{1+\left(\tan\theta\right)^2}}d\theta}\)

\(\large\color{black}{ \displaystyle \int \frac{4\sec^2\theta}{\sqrt{1+\tan^2\theta}}d\theta}\)

\(\large\color{black}{ \displaystyle \int \frac{4\sec^2\theta}{\sqrt{\sec^2\theta}}d\theta}\)

\(\large\color{black}{ \displaystyle \int \frac{4\sec^2\theta}{\sec\theta}d\theta}\)

\(\large\color{black}{ \displaystyle \int 4\sec\theta{~}d\theta}\)

and then apply the formula for integral of secant

Answer 5

you will then substitute the x back in this example, but what I wanted to show is this mechanism of substituting a trignonometric function instead of x into the integral.

Answer 6

Do you want to come back to our original problem?

Answer 7

\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{1}{x^2\sqrt{4-x^2}}~dx}\)

Try to do this substitution:
\(\large \color{gray}{x=2\sin\theta}\)

TELL ME: what is the derivative of \(2\sin\theta\) ?

Answer 8

(derivative with respect to \(\theta\) )

Answer 9

I can follow what you did after you substituted in. But I can't understand what made you decide to do the trig substitution there.

Answer 10

What made me decide is that I want to make substitution in a way that the part in ssquare root containing x^2+C would become just a single trigonometric function inside the square root.

Answer 11

Derivative should be 2cos right?

Answer 12

yes

Answer 13

2cos\(\theta\)

Answer 14

Yeah that's what I meant, sorry

Answer 15

so I set u = 2sin(theta) and du = 2cos(theta) dx right?

Answer 16

There is a rule for trigonometric substitution
... it will get intuitive after some time, but for now - the rule:
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FORM: (\(a^2\) is just a constant) SUBSTITUTION

\(\Large x^2+a^2{~~~~~~~~~~~~}\Rightarrow {~~~~~~~~~~~~}x=a\tan\theta \)
\(\Large x^2-a^2{~~~~~~~~~~~~}\Rightarrow {~~~~~~~~~~~~}x=a\sec\theta \)
\(\Large a^2-x^2{~~~~~~~~~~~~}\Rightarrow {~~~~~~~~~~~~}x=a\sin\theta \)
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Answer 17

\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{1}{\color{blue}{x}^2\sqrt{4-\color{blue}{x}^2}}~\color{royalblue }{dx}}\)

Try to do this substitution:
\(\large \color{blue}{x=2\sin\theta}\)
\(\large \color{red }{dx=2\cos\theta ~d \theta }\)

\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{1}{\left(\color{blue}{2\sin\theta}\right)^2\sqrt{4-\left(\color{blue}{2\sin\theta}\right)^2}}~\color{red }{2\cos\theta ~d \theta}}\)

Answer 18

\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{2\cos\theta}{4\sin^2\theta\sqrt{4-4\sin^2\theta}} ~d \theta}\)

Answer 19

When you are done reading tell me whether you have followed everything so far, or there are so parts you don't get.

Answer 20

Okay just worked through all the steps you went through

Answer 21

Ok, and now, you can tell this:

\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{2\cos\theta}{4\sin^2\theta\sqrt{4-4\sin^2\theta}} ~d \theta}\)

\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{2\cos\theta}{4\sin^2\theta\sqrt{4\left( 1-\sin^2\theta \right)}} ~d \theta}\)

\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{2\cos\theta}{4\sin^2\theta\cdot (2)\cdot \sqrt{\cos^2\theta }} ~d \theta}\)

Answer 22

\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{2\cos\theta}{4\sin^2\theta\cdot (2)\cdot \sqrt{\cos^2\theta }} ~d \theta}\)

\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{\cos\theta}{4\sin^2\theta \sqrt{\cos^2\theta }} ~d \theta}\)

\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{\cos\theta}{4\sin^2\theta \cos\theta } ~d \theta}\)

\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{1}{4\sin^2\theta } ~d \theta}\)

Answer 23

are you getting it?

Answer 24

Yeah I get that.

Answer 25

Can you finish the problem up?

Answer 26

What function has a derivative of csc²(\(\theta\)) ?

Answer 27

Is it cot(theta)?

Answer 28

or I think -cot(theta)

Answer 29

YES, -cot(\(\theta\))

Answer 30

-1/4 * cot(theta) ?

Answer 31

- 1/(4 tan(theta)) ?

Answer 32

\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{1}{4\sin^2\theta } ~d \theta}\)

\(\large\color{black}{\displaystyle\frac{1}{4}\int\limits_{~}^{~}\csc^2\theta ~d \theta}\)

\(\large\color{black}{\displaystyle-\frac{1}{4} \cot(\theta)+{\rm C} }\)

and then you had that
\(\large x=2\sin\theta \)

so put back for x.


\(\large\color{black}{\displaystyle-\frac{1}{4} \frac{\cos\theta}{\sin\theta}+{\rm C} }\)

\(\large\color{black}{\displaystyle-\frac{1}{2} \frac{\cos\theta}{2\sin\theta}+{\rm C} }\)

\(\large\color{black}{\displaystyle-\frac{1}{2} \frac{\sqrt{\cos^2\theta}}{2\sin\theta}+{\rm C} }\)

\(\large\color{black}{\displaystyle-\frac{1}{2} \frac{\sqrt{1-\sin^2\theta}}{2\sin\theta}+{\rm C} }\)

\(\large\color{black}{\displaystyle-\frac{1}{4} \frac{\sqrt{4-4\sin^2\theta}}{2\sin\theta}+{\rm C} }\)

\(\large\color{black}{\displaystyle-\frac{1}{4} \frac{\sqrt{4-(2\sin\theta)^2}}{2\sin\theta}+{\rm C} }\)

\(\large\color{black}{\displaystyle\frac{-\sqrt{4-x^2}}{4x}+{\rm C} }\)

Answer 33

it was kind of long to put in for x....

And the check is:
http://www.wolframalpha.com/input/?i=derivative+of+%28-%E2%88%9A%284-x%C2%B2%29%29%2F%284x%29

Answer 34

Sorry it's taking me a little while to go through it.

Answer 35

I just rearranged the answer in terms of 2sin(\(\theta\)),
and substituted for x.

Answer 36

I'm confused as to how you got the third to last step

Answer 37

\(\large\color{black}{\displaystyle-\frac{1}{2} \frac{\sqrt{1-\sin^2\theta}}{2\sin\theta}+{\rm C}=-\frac{1}{4} \frac{\sqrt{4-4\sin^2\theta}}{2\sin\theta}+{\rm C} }\)

Answer 38

Are you referring to that?

Answer 39

Yeah

Answer 40

I think I've figured it out. That's tricky though.

Answer 41

The last line is the final step right?

Answer 42

\(\large\color{black}{\displaystyle-\frac{1}{2} \frac{\sqrt{1-\sin^2\theta}}{2\sin\theta}+{\rm C}=}\)


\(\large\color{black}{\displaystyle-\frac{1}{2} \frac{\sqrt{\left(1-\sin^2\theta \right)}}{2\sin\theta}+{\rm C} }\)


\(\large\color{blue}{\displaystyle-\frac{1}{2} \frac{\sqrt{\left(1/4 \right){~}(4){~}\left(1-\sin^2\theta \right)}}{2\sin\theta}+{\rm C} }\)


\(\large\color{black}{\displaystyle-\frac{1}{2} \frac{\sqrt{(1/4)}\cdot \sqrt{{~}4{~}\left(1-\sin^2\theta \right)}}{2\sin\theta}+{\rm C} }\)


\(\large\color{black}{\displaystyle-\frac{1}{2} \frac{(1/2)\cdot \sqrt{{~}4-4\sin^2\theta}}{2\sin\theta}+{\rm C} }\)


\(\large\color{black}{\displaystyle-\frac{1}{4} \frac{\sqrt{{~}4-\left(2\sin\theta\right)^2}}{2\sin\theta}+{\rm C} }\)

Answer 43

By the way, I made a mistake:

\(\large\color{black}{\displaystyle-\frac{1}{4} \frac{\sqrt{4-(2\sin\theta)^2}}{2\sin\theta}+{\rm C} }\)

\(\large\color{red}{\displaystyle\frac{-\sqrt{4-x^2}}{8x}+{\rm C} }\)

Answer 44

the red is correct.

Answer 45

Oh I should've seen that.