Integral, hmm where did i mess up?

[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

http://www.mathlinks.ro/Forum/latexrender/pictures/a/f/f/affbb15186b375255a388df46ab0b2701a8798dc.gif

 

[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.

 

[D H]

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

 

[rocomath]

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.

if i have

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

isn't that the power rule? so

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

 

[hotcommodity]

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

 

[rocomath]

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

why not? and what should it be?

 

[hotcommodity]

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

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

 

[hotcommodity]

why not? and what should it be?

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

 

[rocomath]

i simplified

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

 

[hotcommodity]

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

 

[rocomath]

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

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

 

[D H]

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

Ok.

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

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

my last part is wrong or perhaps it's equivalent, but not obvious to me

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

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

 

[rocomath]

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

IN WORK.

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

 

[D H]

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.

 

[rocomath]

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

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.

 

[rocomath]

http://www.mathlinks.ro/Forum/latexrender/pictures/4/1/0/410c7ab5ddcc2586a7250b56ba46df50369f6fbf.gif

 

[lazypast]

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

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

can someone link me to a site describing well the transformation of integral limits, I've been shown a lot but never understood it well, thanks.

anyway i got up to $$\int \frac {d \theta}{sec^4 \theta}$$
i don't know what limits to put in there

 

[rocomath]

all you do is evaluate your current limits with your substitution

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

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.

 

[D H]

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),

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

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

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

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

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

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

 

[D H]

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