How to solve using Leibniz Rule

[Md. Abde Mannaf]

Homework Statement

Homework Equations

The Attempt at a Solution

here is i am still stuck.

 

[STEMucator]

I believe there are quite a few typos in the problem statement if you are intending to use Leibniz rule. I for one have never seen an $\text{In}$ function before.

The Leibniz rule in one dimension would be:

$$\frac{d}{dx} \int_{a(x)}^{b(x)} f(t) dt = f(b(x)) b'(x) - f(a(x)) a'(x)$$

EDIT: It would be more appropriate to call this the fundamental theorem.

 

[Md. Abde Mannaf]

at first i am use substitution. some one said

Don't use substitution, use Leibniz rule with α as the second variable.

what am i do now?

 

[STEMucator]

I can't see any really obvious substitutions here, but you should check out:

http://en.wikipedia.org/wiki/Leibniz_integral_rule#Formal_statement

See if that changes your answer.

 

[Md. Abde Mannaf]

i know the Leibniz's rule. i solve many mathematical term . bt i cann't solve this math with Leibniz rule. i am still stuck
here. here limit is constant so 2nd and 3rd term will be zero.

 

[Md. Abde Mannaf]

any one solve this mathematical term and prove it??

 

[Ray Vickson]

i know the Leibniz's rule. i solve many mathematical term . bt i cann't solve this math with Leibniz rule. i am still stuck
here. here limit is constant so 2nd and 3rd term will be zero.

Your application of Leibnitz' rule is correct. You could try to change variables to $u = \cos(x)$ in your integral.

 

[certainly]

if you are intending to use Leibniz rule. I for one have never seen an In\text{In} function before.

See this, example 3.
Now let's do that daunting looking integral. Like I said before, this is a good problem.
First write $\int_0^{\pi} \frac{1}{\alpha}-\frac{1}{(\alpha)(1+\alpha cos(x))} dx$ for the original integral.
However $\int \frac{1}{1+\alpha cos(x)} dx$ itself is not nice, in-fact it's pretty nasty. It will be an interesting exercise to do, but even after that there is a good bit to do for the original problem. So I would recommend you use Wolfram for that for now, and proceed.

 

[Md. Abde Mannaf]

i could not solve this math above analysis . i am trying to my best. but i am fail every time. and again try...

please see this and give me more idea to solve

 

[ehild]

Homework Statement

Homework Equations

The Attempt at a Solution

here is i am still stuck.

You can even cheat :) using wolframalpha.com.
The substitution tan(x/2)=u works well if you have a rational expression of trigonometric functions. Try.

 

[ehild]

I believe there are quite a few typos in the problem statement if you are intending to use Leibniz rule. I for one have never seen an $\text{In}$ function before.

The Leibniz rule in one dimension would be:

$$\frac{d}{dx} \int_{a(x)}^{b(x)} f(t) dt = f(b(x)) b'(x) - f(a(x)) a'(x)$$

EDIT: It would be more appropriate to call this the fundamental theorem.

The derivation of the OP is correct. Read Leibnitz Rule http://en.wikipedia.org/wiki/Leibniz_integral_rule

 

[certainly]

In regards to post #9:- I'm not entirely sure what you're trying to do here. Did you have a look at the links I provided in post #8. I think your trying to come up with a value for $\int_0^{\pi} \frac{1}{1+\alpha cos(x)} dx$ for $|\alpha|<1$using $\int_0^{\pi} \frac{1}{\alpha-cos(x)} dx=\frac{\pi}{\sqrt{\alpha^2-1}}$ for $|\alpha|>1$, however that + sign in the original integral makes a good difference.

You can even cheat :) using wolframalpha.com.

Does no one read my posts :-)

 

[Ray Vickson]

i could not solve this math above analysis . i am trying to my best. but i am fail every time. and again try...

please see this and give me more idea to solve

You can write your integral for $f'(\alpha)$ as $\int_0^{\pi/2} + \int_{\pi/2}^{\pi}$, then change variables to $x \leftarrow \pi -x$ in the second integral, to get
$$f'(\alpha) = \int_0^{\pi/2} \left[ \frac{\cos(x)}{1 + \alpha \cos(x)} - \frac{\cos(x)}{1 - \alpha \cos(x)} \right] \, dx \\ = -\int_0^{\pi/2} \frac{2 \alpha \cos^2(x)}{1 - \alpha^2 \cos^2(x)} \, dx$$
Changing variables to $\tan(x) = y$ produces an integral in $y$ that can be split up into partial fractions, giving two well-known and easily do-able integrations.

 

[ehild]

Now let's do that daunting looking integral. Like I said before, this is a good problem.
First write $\int_0^{\pi} \frac{1}{\alpha}-\frac{1}{(\alpha)(1+\alpha cos(x))} dx$ for the original integral.
However $\int \frac{1}{1+\alpha cos(x)} dx$ itself is not nice, in-fact it's pretty nasty.

$\int \frac{1}{1+\alpha cos(x)} dx$ is not that nasty.
Use the identity $\cos(x)=\frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}$. Substitute u=tan(x/2), x=2arctan(u), $dx=\frac{2}{1+u^2}du$. The integration limits become 0-->infinite.

 

[SammyS]

There is an interesting Wikipedia entry for tangent half-angle formulas

 

[certainly]

$\int \frac{1}{1+\alpha cos(x)} dx$ is not that nasty.
Use the identity $\cos(x)=\frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}$. Substitute u=tan(x/2), x=2arctan(u), $dx=\frac{2}{1+u^2}du$. The integration limits become 0-->infinite.

Yeah, your right. I didn't actually do the integral, I only saw the answer and thought, this might take some time.