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## dfresenius 2 years ago Trig substitutions with integrals. Really basic question.

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1. dfresenius

$\int\limits_{2\sqrt2}^{4}\frac{ 1 }{ (t^3)\sqrt(t^2-1)dt }$

2. dfresenius

so, I know how to do the entire problem, but, when it comes to plugging in the cos(4) and cos(2root2) i get lost.

3. dfresenius

|dw:1360033290591:dw|

4. zepdrix

You plugged everything in correctly, I think maybe you just made a tiny boo boo at the end. See how you're squared secant that you're left with is in the denominator?

5. dfresenius

oh ya cos^2x

6. zepdrix

Oh the limits? yah let's not forget that the original limits were in terms of t. We can't plug those into our result. We have 2 options. Plug the limits into our original substitution that we made, to find new limits of integration. Or we can back-substitute and plug the t values in.

7. dfresenius

lets find t now, thats where Im lost

8. dfresenius

the other problem i had was real easy, root 2 and 2

9. dfresenius

now im not sure how to use these weird ones

10. zepdrix

$\large t=\sec \theta$The lower limit is $$2\sqrt2$$. giving us,$\large \frac{2}{\sqrt2}=\sec \theta$ Which is a special angle, maybe it just doesn't look like it :)$\large \frac{2}{\sqrt2}=\frac{1}{\cos \theta} \theta \qquad \rightarrow \qquad \cos \theta = \frac{\sqrt2}{2}$

11. dfresenius

|dw:1360034188203:dw|

12. dfresenius

ahhhh! algebra bites me in the retrice

13. zepdrix

Solution looks good so far :)

14. dfresenius

|dw:1360034363382:dw|

15. zepdrix

Umm if I'm remembering my angles correctly, I think that's at $$\dfrac{\pi}{4}$$. Sine and Cosine both measure sqrt2/2 when theta is pi/4.

16. zepdrix

Doesn't sound familiar? :D

17. dfresenius

ya

18. dfresenius

hmm now im lost on the 4

19. zepdrix

$\large 4=\sec \theta \qquad \rightarrow \qquad \cos \theta=1/4$Hmmmmm..

20. zepdrix

$\large \cos \theta=\left(\frac{1}{2}\right)^2$Hmm no I don't think that's going to get us in the right direction...

21. zepdrix

$\large \cos \theta=1/4 \qquad \rightarrow \qquad \theta=\arccos\left(\frac{1}{4}\right)$I think we would have to leave it like this.

22. dfresenius

|dw:1360034799626:dw|

23. zepdrix

Hmm I'm not quite sure what you did there :o

24. dfresenius

not sure if that is legal, i multiplied both sides by sec^2/sec^2

25. zepdrix

But it looks like you canceled how a secant and a secant squared somehow :o

26. dfresenius

ya thats dumb my bad

27. dfresenius

|dw:1360035089559:dw|

28. zepdrix

no :c i dont think that works either. I'm pretty sure we have to leave it as the arccosine.

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