## unimatix one year ago Integral 1/(x^2 sqrt(4-x^2)) dx

1. SolomonZelman

$$x=2\sin\theta$$

2. SolomonZelman

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

3. unimatix

That would be great honestly.

4. SolomonZelman

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

5. SolomonZelman

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.

6. SolomonZelman

Do you want to come back to our original problem?

7. SolomonZelman

$$\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{1}{x^2\sqrt{4-x^2}}~dx}$$ Try to do this substitution: $$\large \color{gray}{x=2\sin\theta}$$ TELL ME: what is the derivative of $$2\sin\theta$$ ?

8. SolomonZelman

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

9. unimatix

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.

10. SolomonZelman

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.

11. unimatix

Derivative should be 2cos right?

12. SolomonZelman

yes

13. SolomonZelman

2cos$$\theta$$

14. unimatix

Yeah that's what I meant, sorry

15. unimatix

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

16. SolomonZelman

There is a rule for trigonometric substitution ... it will get intuitive after some time, but for now - the rule: ☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼ FORM: ($$a^2$$ is just a constant) SUBSTITUTION $$\Large x^2+a^2{~~~~~~~~~~~~}\Rightarrow {~~~~~~~~~~~~}x=a\tan\theta$$ $$\Large x^2-a^2{~~~~~~~~~~~~}\Rightarrow {~~~~~~~~~~~~}x=a\sec\theta$$ $$\Large a^2-x^2{~~~~~~~~~~~~}\Rightarrow {~~~~~~~~~~~~}x=a\sin\theta$$ ☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼

17. SolomonZelman

$$\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{1}{\color{blue}{x}^2\sqrt{4-\color{blue}{x}^2}}~\color{royalblue }{dx}}$$ Try to do this substitution: $$\large \color{blue}{x=2\sin\theta}$$ $$\large \color{red }{dx=2\cos\theta ~d \theta }$$ $$\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{1}{\left(\color{blue}{2\sin\theta}\right)^2\sqrt{4-\left(\color{blue}{2\sin\theta}\right)^2}}~\color{red }{2\cos\theta ~d \theta}}$$

18. SolomonZelman

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

19. SolomonZelman

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

20. unimatix

Okay just worked through all the steps you went through

21. SolomonZelman

Ok, and now, you can tell this: $$\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{2\cos\theta}{4\sin^2\theta\sqrt{4-4\sin^2\theta}} ~d \theta}$$ $$\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{2\cos\theta}{4\sin^2\theta\sqrt{4\left( 1-\sin^2\theta \right)}} ~d \theta}$$ $$\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{2\cos\theta}{4\sin^2\theta\cdot (2)\cdot \sqrt{\cos^2\theta }} ~d \theta}$$

22. SolomonZelman

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

23. SolomonZelman

are you getting it?

24. unimatix

Yeah I get that.

25. SolomonZelman

Can you finish the problem up?

26. SolomonZelman

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

27. unimatix

Is it cot(theta)?

28. unimatix

or I think -cot(theta)

29. SolomonZelman

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

30. unimatix

-1/4 * cot(theta) ?

31. unimatix

- 1/(4 tan(theta)) ?

32. SolomonZelman

$$\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{1}{4\sin^2\theta } ~d \theta}$$ $$\large\color{black}{\displaystyle\frac{1}{4}\int\limits_{~}^{~}\csc^2\theta ~d \theta}$$ $$\large\color{black}{\displaystyle-\frac{1}{4} \cot(\theta)+{\rm C} }$$ and then you had that $$\large x=2\sin\theta$$ so put back for x. $$\large\color{black}{\displaystyle-\frac{1}{4} \frac{\cos\theta}{\sin\theta}+{\rm C} }$$ $$\large\color{black}{\displaystyle-\frac{1}{2} \frac{\cos\theta}{2\sin\theta}+{\rm C} }$$ $$\large\color{black}{\displaystyle-\frac{1}{2} \frac{\sqrt{\cos^2\theta}}{2\sin\theta}+{\rm C} }$$ $$\large\color{black}{\displaystyle-\frac{1}{2} \frac{\sqrt{1-\sin^2\theta}}{2\sin\theta}+{\rm C} }$$ $$\large\color{black}{\displaystyle-\frac{1}{4} \frac{\sqrt{4-4\sin^2\theta}}{2\sin\theta}+{\rm C} }$$ $$\large\color{black}{\displaystyle-\frac{1}{4} \frac{\sqrt{4-(2\sin\theta)^2}}{2\sin\theta}+{\rm C} }$$ $$\large\color{black}{\displaystyle\frac{-\sqrt{4-x^2}}{4x}+{\rm C} }$$

33. SolomonZelman

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

34. unimatix

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

35. SolomonZelman

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

36. unimatix

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

37. SolomonZelman

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

38. SolomonZelman

Are you referring to that?

39. unimatix

Yeah

40. unimatix

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

41. unimatix

The last line is the final step right?

42. SolomonZelman

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

43. SolomonZelman

By the way, I made a mistake: $$\large\color{black}{\displaystyle-\frac{1}{4} \frac{\sqrt{4-(2\sin\theta)^2}}{2\sin\theta}+{\rm C} }$$ $$\large\color{red}{\displaystyle\frac{-\sqrt{4-x^2}}{8x}+{\rm C} }$$

44. SolomonZelman

the red is correct.

45. unimatix

Oh I should've seen that.

46. unimatix

|dw:1441678847661:dw|

47. unimatix

ahh the dx should be 2 cos(theta)

48. SolomonZelman

well, that is how I dx should be: 2cos(theta) d(theta)

49. SolomonZelman

and that is exactly how I started to substitute $$2\sin\theta$$ for $$x$$.

50. unimatix

Oh sorry. I don't know what I was thinking . We just finished it.

51. unimatix

Thank you so much. I'm going to go try and work through it a second time on my own. THANK YOU. THANK YOU. THANK YOU.

52. SolomonZelman

Sure, remember the formas though ☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼ FORM: ($$a^2$$ is just a constant) SUBSTITUTION $$\Large x^2+a^2{~~~~~~~~~~~~}\Rightarrow {~~~~~~~~~~~~}x=a\tan\theta$$ $$\Large x^2-a^2{~~~~~~~~~~~~}\Rightarrow {~~~~~~~~~~~~}x=a\sec\theta$$ $$\Large a^2-x^2{~~~~~~~~~~~~}\Rightarrow {~~~~~~~~~~~~}x=a\sin\theta$$ ☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼ Like 4+x², is 2²+x² ---> form 1, x=atan($$\theta$$).

53. unimatix

Will do!

54. SolomonZelman

Ok, yw!