# Integral, hmm where did i mess up?

1. Nov 10, 2007

### rocomath

[SOLVED] Integral, hmm ... where did i mess up?

ok, i typed this integral into mathematica, my last part is wrong or perhaps it's equivalent, but not obvious to me

thanks for the help!

in mathematic ... (x^2 -1)/(x^2 + 1)^3

Last edited: Nov 10, 2007
2. Nov 10, 2007

### hotcommodity

I believe where you have $$\int^{1}_{0} \frac{tan^2 \theta}{sec^4 \theta} - \frac{1}{sec^4 \theta}$$, you should have $$sec^5 \theta$$, instead of to the 4th power.

3. Nov 10, 2007

### D H

Staff Emeritus
You forgot to transform the integration limits when you switch from integrating wrt x to integrating wrt theta.

4. Nov 10, 2007

### rocomath

if i have

$$(\sec^{2}{\theta})^{3}$$

isn't that the power rule? so

$$\sec^{6}{\theta}$$

5. Nov 10, 2007

### hotcommodity

Oops, yes you're correct. But it still wouldn't be to the 4th power.

6. Nov 10, 2007

### rocomath

why not? and what should it be?

7. Nov 10, 2007

### hotcommodity

But it looks like he changed it back to "x," so it wouldn't matter right?

8. Nov 10, 2007

### hotcommodity

Well if $$x = tan \theta$$, and $$tan^2 \theta + 1 = sec^2 \theta$$, then you have $$(sec^2 \theta)^3$$ in the denominator.

9. Nov 10, 2007

### rocomath

i simplified

$$\int[\frac{tan^{2}{\theta}\sec^{2}{\theta}}{\sec^{6}{\theta}}-\frac{\sec^{2}{\theta}}{\sec^{6}{\theta}}]d{\theta}$$

10. Nov 10, 2007

### hotcommodity

OK ok, I see, let me go back and re-work it.

11. Nov 10, 2007

### rocomath

thanks a lot :-] i just don't know wth is wrong with that last part, lol.

12. Nov 10, 2007

### D H

Staff Emeritus
Roco, you did the trig part right. The problem is the integration limits.

Ok.

He made the same mistake twice, yet another example of two wrongs making a right. I answered this question,

Upon switching the integration variable from $x$ to $\theta$, roco, you should have switch the integration limits as well:

$$\int_0^1 \frac{x^2-1}{(x^2+1)^3}dx$$

$$\int_{\tan^{-1} 0}^{\tan^{-1} 1} 1+2\cos2\theta + \cos4\theta d\theta$$

Evaluating the limits, $\tan^{-1} 0=0$ but $\tan^{-1} 1=\pi/4$, not 1.

BTW, this integrates directly to

$$-\;\frac 1 4 \left.\left(\theta + \sin2\theta+\frac 1 4\sin4\theta\right)\right|_0^{\pi/4}$$

Last edited: Nov 10, 2007
13. Nov 10, 2007

### rocomath

why would they change? i re-substituted or whatever you call it. hmm

14. Nov 10, 2007

### D H

Staff Emeritus
I updated my IN WORK post. Do you see now why you have to pay attention to the integration limits? When you do so, you get the correct results.

15. Nov 10, 2007

### rocomath

but that's not what i'm trying to do. when using trig subst. i don't HAVE TO switch limits do i?

i fix my limits for other problems, but i'm just trying to work this one all the way.

16. Nov 10, 2007

### rocomath

17. Nov 11, 2007

### lazypast

can someone link me to a site describing well the transformation of integral limits, ive been shown alot but never understood it well, thanks.

anyway i got up to $$\int \frac {d \theta}{sec^4 \theta}$$
i dont know what limits to put in there

18. Nov 11, 2007

### rocomath

$$\int_0^1 \frac{x^2-1}{(x^2+1)^3}dx$$

my limits are from 0 to 1

my chosen substitution is tangent, so i set my current limits equal to my substitution

tangent = 0 and tangent = 1

where does tangent equal 0 and 1? solve for that, and those are your new limits, respectively

thus

$$\int_{0}^{\frac{\pi}{4}}(1+2\cos{2\theta}+\cos{4\theta})d\theta$$

this is done so that there is no need to re-substitute.

19. Nov 11, 2007

### D H

Staff Emeritus
Suppose you can rewrite some function $f(x)$ as $h(g(x))g'(x)$. This is what makes u-substitution viable for solving a integration problem. Applying the u-substitution $u=g(x)$ to the indefinite integral of f(x),

$$\int f(x) dx = \int h(g(x))g'(x)dx \begin{matrix} \\[-8pt] \longrightarrow \\[-10pt] ^{u=g(x)} \end{matrix} \int h(u) du$$

In applying the u-substitution to a definite integral, you must apply the substitution to the integration limits as well as to the integrand:

$$\int_{x=a}^{x=b} f(x) dx = \int_{x=a}^{x=b} h(g(x))g'(x)dx \begin{matrix} \\[-8pt] \longrightarrow \\[-10pt] ^{u=g(x)} \end{matrix} \int_{u=g(a)}^{u=g(b)} h(u) du$$

In this case, the u-substitution is $x=\tan\theta$ or $u=\tan^{-1}x$. Applying the u-substitution to this problem,

$$\int_{x=0}^{x=1} \frac{x^2-1}{(x^2+1)^3}dx\, \begin{matrix} \\[-8pt] \longrightarrow \\[-8pt] ^{\theta=\tan^{-1}(x)} \end{matrix}\, \int_{\theta=\tan^{-1} (0)}^{\theta=\tan^{-1} (1)} 1+2\cos2\theta + \cos4\theta d\theta$$

Finally $\tan^{-1}(0) = 0$ and $\tan^{-1}(1) = \pi/4$.

20. Nov 11, 2007

### D H

Staff Emeritus
Spam! Somebody please delete the previous post (#20). As that will make this post #20, you can delete it, too.