# Trig substitution

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• June 15th 2008, 12:01 PM
elocin
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.

• June 15th 2008, 04:32 PM
mr fantastic
Quote:

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).
• June 15th 2008, 06:03 PM
Mathstud28
Quote:

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

$x=a\sec(\theta)$
• June 15th 2008, 06:14 PM
elocin
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!
• June 15th 2008, 06:21 PM
elocin
Got it!
Never mind, for the last one, I got it right! yay but here's another: the last one!!
• June 15th 2008, 06:21 PM
Mathstud28
Quote:

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
• June 15th 2008, 06:23 PM
elocin
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.
• June 15th 2008, 06:24 PM
Chris L T521
Quote:

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... :D
• June 15th 2008, 06:25 PM
Mathstud28
Quote:

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

$4x=\sec(\theta)\Rightarrow{x=\frac{\sec(\theta)}{4 }}$
• June 15th 2008, 06:26 PM
elocin
P.s. Thank You!!!
Thank you for the last one, by the way!!
• June 15th 2008, 06:26 PM
Mathstud28
Quote:

Originally Posted by Chris L T521
Close...I changed something... :D

I think it was apparent I meant =...anyways havent you heard...subtraction is the new equality (Cool)
• June 15th 2008, 06:28 PM
elocin
Ah I see!! THat's what I thought! Nice. Thank yoU!
• June 15th 2008, 06:44 PM
elocin
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.
• June 16th 2008, 03:38 PM
elocin
• June 16th 2008, 03:45 PM
Mathstud28
Quote:

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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