Math Help - Trig substitution

1. Trig substitution

1. Integrate from (square root of 2, divided by 4) to (1/2): (dx) / (x^5*square root of ((16x^2)-(1)).

2. Integrate (x^2)/((36-x^2)^3/2) dx

All of this is by trig substitution and I am not getting anywhere with it

3. Integrate from 3 to 6: sq. rt of (x^2-9) all over x dx.

4. Integrate (8)/(x*sq rt of 5-x^2) dx

5. Integrate (t^5)/(sq rt of t^2+6) dt.

6. Integrate from -pi/2 to zero, (sin(t)) / sq rt of 1+(cos(t))^2 dt.

2. Originally Posted by elocin
1. Integrate from (square root of 2, divided by 4) to (1/2): (dx) / (x^5*square root of ((16x^2)-(1)).

2. Integrate (x^2)/((36-x^2)^3/2) dx

All of this is by trig substitution and I am not getting anywhere with it

3. Integrate from 3 to 6: sq. rt of (x^2-9) all over x dx.

4. Integrate (8)/(x*sq rt of 5-x^2) dx

5. Integrate (t^5)/(sq rt of t^2+6) dt.

6. Integrate from -pi/2 to zero, (sin(t)) / sq rt of 1+(cos(t))^2 dt.

You would probably have more luck in getting help if you showed the work you've done and where you get stuck.

For 6. make the substitution $u = \cos t$: $I = - \int_0^1 \frac{1}{\sqrt{1 + u^2}} \, du$.

Now make the substitution $u = \tan t$ (a better one would involve a hyperbolic function but you said trig substitution so you're stuck with it): $I = - \int_0^{\pi/4} \frac{1}{\cos t} \, dt$.

There are various ways of dealing with this last integral - a trick with algebra that makes it easy and the Weiestrass substitution that makes harder weather of it.

(You should check the details because I'm very careless).

3. Originally Posted by elocin
1. Integrate from (square root of 2, divided by 4) to (1/2): (dx) / (x^5*square root of ((16x^2)-(1)).

2. Integrate (x^2)/((36-x^2)^3/2) dx

All of this is by trig substitution and I am not getting anywhere with it

3. Integrate from 3 to 6: sq. rt of (x^2-9) all over x dx.

4. Integrate (8)/(x*sq rt of 5-x^2) dx

5. Integrate (t^5)/(sq rt of t^2+6) dt.

6. Integrate from -pi/2 to zero, (sin(t)) / sq rt of 1+(cos(t))^2 dt.

These are all trig subs as you noted

if you have something of the form

$a^2-x^2$

then let $x=a\sin(\theta)$

if you have

$a^2+x^2$

let

$x=a\tan(\theta)$

and if you have

$x^2-a^2$

then let

$x=a\sec(\theta)$

4. help! here is what i did

Hi! THank you! For a problem that goes: Integrate (x^3) (sq rt of 36-x^2) dx, I have x=6sin theta. And then I put (6sintheta)^3 times (6costheta) and dx=6costheta d theta. I dont know where to go from that! Thanks!

5. Got it!

Never mind, for the last one, I got it right! yay but here's another: the last one!!

6. Originally Posted by elocin
Hi! THank you! For a problem that goes: Integrate (x^3) (sq rt of 36-x^2) dx, I have x=6sin theta. And then I put (6sintheta)^3 times (6costheta) and dx=6costheta d theta. I dont know where to go from that! Thanks!
$\int\frac{6^4\sin^3(\theta)\cos(\theta)}{6\cos(\th eta)}=6^3\int\sin^3(\theta)d\theta=6^3\int\sin(\th eta)(1-\cos^2(\theta))d\theta$

Now expand and let $u=\cos(\theta)$ in the second integral

7. One more!!

I only have one left!! It is integrate from (radical 2 over 4) to (1/2): dx over (x^5) times sq rt of 16x^2-1. and I thought that x=sectheta but the 16 in front of the x^2 messes me up. SO I was thinking x=(sectheta)/4.

8. Originally Posted by Mathstud28
$\int\frac{6^4\sin^3(\theta)\cos(\theta)}{6\cos(\th eta)}=6^3\int\sin^3(\theta)d\theta{\color{red}=}6^ 3\int\sin(\theta)(1-\cos^2(\theta))d\theta$

Now expand and let $u=\cos(\theta)$ in the second integral
Close...I changed something...

9. Originally Posted by elocin
I only have one left!! It is integrate from (radical 2 over 4) to (1/2): dx over (x^5) times sq rt of 16x^2-1. and I thought that x=sectheta but the 16 in front of the x^2 messes me up. SO I was thinking x=(sectheta)/4.
$16x^2=(4x)^2$

Now we want a one coefficient so it cancels so we let

$4x=\sec(\theta)\Rightarrow{x=\frac{\sec(\theta)}{4 }}$

10. P.s. Thank You!!!

Thank you for the last one, by the way!!

11. Originally Posted by Chris L T521
Close...I changed something...
I think it was apparent I meant =...anyways havent you heard...subtraction is the new equality

12. Ah I see!! THat's what I thought! Nice. Thank yoU!

13. Okay, so for the problem:

For integrating from ((radical 2)/4) to (1/2), the integral of dx over (x^5)*(sq rt of 16x^2-1), I have x=(sectheta)/4. Then I put the dx on the side, and made (1) over (x^5)*(sq rt of 16x^2-1). Then I substituted ((sectheta)/4)^5 for x^5. And then would I substitute (sectheta)/4 for (sq rt of 16x^2-1), I am not sure where to go from there. Also, I would have to change the limits in terms of theta.

15. Originally Posted by elocin
For integrating from ((radical 2)/4) to (1/2), the integral of dx over (x^5)*(sq rt of 16x^2-1), I have x=(sectheta)/4. Then I put the dx on the side, and made (1) over (x^5)*(sq rt of 16x^2-1). Then I substituted ((sectheta)/4)^5 for x^5. And then would I substitute (sectheta)/4 for (sq rt of 16x^2-1), I am not sure where to go from there. Also, I would have to change the limits in terms of theta.
$\int\frac{dx}{x^5\sqrt{16x^2-1}}$

Let $4x=\sec(\theta)\Rightarrow{dx=\frac{\sec(\theta)\t an(\theta)}{4}}$

Giving us

$\int\frac{\frac{1}{4}\sec(\theta)\tan(\theta)}{\fr ac{1}{4^5}\sec^5\tan(\theta)}d\theta=\frac{1}{4^4} \int\cos^4(\theta)d\theta(\theta)$

Now make use of the identity

$\cos^4(\theta)=\left(\frac{1+\cos(2\theta)}{2}\rig ht)^2$

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