Integration using Leibnitz theorem

[Titan97]

Homework Statement

Prove that $$\int_{-\pi/2}^{\pi/2}\frac{log(1+b\sin x)}{\sin x}dx=\pi \arcsin(b)$$
Where $|b|\le1$

Homework Equations

$$\frac{d}{dx}\big(\int_a^bf(t)dt\big)=\int_a^b\frac{\partial f(t)}{\partial x}dx$$

The Attempt at a Solution

Let $$f(b)=\int_{-\pi/2}^{\pi/2}\frac{log(1+b\sin x)}{\sin x}dx$$
Using Leibnitz theorem, $$f'(b)=\int_{-\pi/2}^{\pi/2}\frac{1}{1+b\sin x}dx$$
Now, $\sin(x)=\frac{2\tan(x/2)}{1+\tan^2(x/2)}$
After simplifying and substituting $\tan(x/2)=u$, I got
$$f'(b)=\frac{2}{\sqrt{1-b^2}}\big[\arctan{\frac{\sqrt{1+b}}{\sqrt{1-b}}}+\arctan{\frac{\sqrt{1-b}}{\sqrt{1+b}}}\big]$$
Now, for $b=0$, its clear that $f(b)=0$.
So $$f(b)=\frac{2}{\sqrt{1-b^2}}\big[\arctan{\frac{\sqrt{1+b}}{\sqrt{1-b}}}+\arctan{\frac{\sqrt{1-b}}{\sqrt{1+b}}}\big]$$
After putting $b=\cos\phi$ since $|b|\le 1$, I got $f(b)=\frac{\pi}{\sin({\arccos(b)})}$ which is wrong.
Also, the new $f(b)$ does not satisfy $f(0)=0$

Is my method correct?

 

[andrewkirk]

How did you get this:

$$
f'(b)=\int_{-\pi/2}^{\pi/2}\frac{1}{1+b\sin x}dx
\ \ ?$$

That's not what I get when I take $\frac{\partial}{\partial b}$ of the integrand in $f(b)$ .

 

[Titan97]

Sorry. Its $log(1+b\sin x)$. I have edited the question.

 

[Umrao]

You made a mistake.
$$f'(b)=\frac{2}{\sqrt{1-b^2}}\big[\arctan{\frac{\sqrt{1+b}}{\sqrt{1-b}}}+\arctan{\frac{\sqrt{1-b}}{\sqrt{1+b}}}\big] $$
After this is your mistake.
$$ \rightarrow \ f(b)=\frac{2}{\sqrt{1-b^2}}\big[\arctan{\frac{\sqrt{1+b}}{\sqrt{1-b}}}+\arctan{\frac{\sqrt{1-b}}{\sqrt{1+b}}}\big]$$

It is $$f'(b)$$ and NOT $$f(b)$$

Okay I think I started nitpicking, I am sorry. I think you messed up the last Integration part.

$$ \int f'(b) db = \int ((\frac{2}{\sin \phi})\arctan(\tan (\frac{\pi}{2}- \phi) + \arctan (\tan \phi)) \times (-sin \phi) d(\phi)$$

Integrate. Put f(0) = 0.

 

[Titan97]

Thank you for pointing out the error. So,
$$f'(cos\phi)=\frac{2}{\sin\phi}\cdot(\arctan\cot\phi-\arctan\tan\phi)=\frac{\pi}{\sin\phi}$$
Integrating $d(f(\cos\phi))=g(\phi)d\phi$,
$$f(\cos\phi)=\pi\log\tan\frac{\phi}{2}+C$$
$$f(b)=\pi\log\sqrt{\frac{1-b}{1+b}}$$
Now I get a different answer.

 

[Umrao]

Thank you for pointing out the error. So,
$$f'(cos\phi)=\frac{2}{\sin\phi}\cdot(\arctan\cot\phi-\arctan\tan\phi)=\frac{\pi}{\sin\phi}$$
Integrating $d(f(\cos\phi))=g(\phi)d\phi$,
$$f(\cos\phi)=\pi\log\tan\frac{\phi}{2}+C$$
$$f(b)=\pi\log\sqrt{\frac{1-b}{1+b}}$$
Now I get a different answer.

Don't put $$\arctan (tan (\frac{\pi}{2} - \phi))$$ as $$(\arctan\cot\phi)$$ I wrote $$\arctan (tan (\frac{\pi}{2} - \phi))$$ deliberately, cause I have some devious plans for it. Apply some manipulation
(like taking out ph...)

 

[Titan97]

When I substituted $b=cos\phi$, I got $\arctan\cot\phi$ and not $\arctan\tan(\frac{\pi}{2}-\phi)$
Then $\arctan x= \frac{\pi}{2}-\cot^{-1}x$.

 

[Umrao]

When I substituted $b=cos\phi$, I got $\arctan\cot\phi$ and not $\arctan\tan(\frac{\pi}{2}-\phi)$
Then $\arctan x= \frac{\pi}{2}-\cot^{-1}x$.

I see the problem there.

1. Make a right angled triangle, take an angle as phi.
2. Since sum of angles of triangle is pi hence the other angle has to be pi/2 - phi.
3. Find tan of phi is perpendicular upon base [Let's say = AB/BC, ABC is triangle and B is right angle and C is phi].
4. cot of (pi/2 -phi) is base upon perpendicular [From above we get cot (pi/2-phi) = AB/BC]
5. From point 3. and 4. we have tan (phi) = cot (pi/2 -phi).
6. I leave to you to prove cot (phi) = tan (pi/2 -phi).

So even if you are ready to accept point 6 without proving it, you get, the manipulation I did. Can you get the answer from where I left? It is almost done, I am surprised that you did it so far and something so trivial is stopping from getting the answer. I think you should revise some basic trigonometry.

 

[Titan97]

Proof:
$tan\phi=x$
$cot(\frac{\pi}{2}-\phi)=tan\phi=x$
$\frac{\pi}{2}-\phi=\cot^{-1}x$
$\phi=\frac{\pi}{2}-\cot^{-1}x$
$\tan^{-1}x=\frac{\pi}{2}-\cot^{-1}x$

 

[Umrao]

Proof:
$tan\phi=x$
$cot(\frac{\pi}{2}-\phi)=tan\phi=x$
$\frac{\pi}{2}-\phi=\cot^{-1}x$
$\phi=\frac{\pi}{2}-\cot^{-1}x$
$\tan^{-1}x=\frac{\pi}{2}-\cot^{-1}x$

Why did you put that proof here? Are you even reading what I am writing? Look I don't understand what are you trying to say. If your confusion is cleared then good, if it is still there, then I am going to ask you to elaborate it, clearly. And by the way why did you put that proof here?

 

[Umrao]

I hope you do know $$\tan^{-1} (cot \phi) = \tan^{-1}(tan (\frac{\pi}{2}-\phi))= \frac{\pi}{2}-\phi$$

 

[Titan97]

This is my confusion:

∫f′(b)db=∫((2sinϕ)arctan(tan(π2−ϕ)+arctan(tanϕ))×(−sinϕ)d(ϕ)

$$f'(\cos\phi)=\frac{\pi}{\sin\phi}$$
Integrating with respect to $d\phi$ I get wrong answer. But in the quoted message, why did you take $f'(b)$ when the function is $f'(\cos\phi)$?
Can't you integrate directly?

 

[Umrao]

Thank you for pointing out the error. So,
$$f'(cos\phi)=\frac{2}{\sin\phi}\cdot(\arctan\cot\phi-\arctan\tan\phi)=\frac{\pi}{\sin\phi}$$
Integrating $d(f(\cos\phi))=g(\phi)d\phi$,
$$f(\cos\phi)=\pi\log\tan\frac{\phi}{2}+C$$
$$f(b)=\pi\log\sqrt{\frac{1-b}{1+b}}$$
Now I get a different answer.

Finally, deciphered. Yet another mistake in calculation and hopefully this will be my last thread to this post.
Mistake -1. $f'(cos\phi)=\frac{2}{\sin\phi}\cdot(\arctan\cot\phi-\arctan\tan\phi)=\frac{\pi}{\sin\phi}$. Since R.H.S. is correct so you made a typo on computer here (started nitpicking again, sorry ) It is $f'(cos\phi)=\frac{2}{\sin\phi}\cdot(\arctan\cot\phi+\arctan\tan\phi)=\frac{\pi}{\sin\phi}$ Notice the plus sign after tan inverse of cot of phi.

Mistake -2(Actual one)

This is my confusion:

$$f'(\cos\phi)=\frac{\pi}{\sin\phi}d\phi$$
Integrating with respect to $d\phi$ I get wrong answer. But in the quoted message, why did you take $f'(b)$ when the function is $f'(\cos\phi)$?
Can't you integrate directly?

I was typing while you posted this message but thanks to it I can clearly tell you where you made the mistake, otherwise it was getting very hard to do so with

$d(f(\cos\phi))=g(\phi)d\phi$

You can do it with cos as function too
Note: In the quote you have written incorrectly $f'(\cos\phi)=\frac{\pi}{\sin\phi}d\phi$ no $d\phi$
$$f'(\cos\phi)=\frac{\pi}{\sin\phi}$$
$$\rightarrow f'(\cos\phi)d(cos\phi)=\frac{\pi}{\sin\phi}d(cos\phi)$$
$$\rightarrow \int f'(\cos\phi)d(cos\phi)= \int \frac{\pi}{\sin\phi}d(cos\phi)$$
$$\rightarrow \int f'(\cos\phi)d(cos\phi)= \int \frac{\pi}{\sin\phi} \times (-\sin\phi)d(\phi)$$
$$\rightarrow f(cos\phi) = -\pi (\phi) +c$$
$$\rightarrow f(b)=-\pi (\cos^{-1}b) +c$$
$$\rightarrow f(b)=-\pi (\frac {\pi}{2} - \sin ^{-1}b) +c$$
$$\rightarrow f(b) =-\frac {\pi^2}{2}+ (\pi\sin^{-1}b) +c$$ and
$f(0)=0$But there is more to it. We both made a common mistake which does not interfere with answer.
We both got $$\arctan cot\phi + \arctan tan\phi$$ It should be $$\arctan cot\frac{\phi}{2} + \arctan tan\frac{\phi}{2}$$.
Though summation remains same but a we will have to accept that we got lucky and in some other place in some other question we will face the consequence if this mistake is not eradicated. We both got to work on it.

I hope I answered your question, so please this question Why were you confused that I wrote $$\arctan cot\phi + \arctan tan\phi = \arctan tan (\frac{\pi}{2} - \phi)+ \arctan tan\phi$$.

And how did you quote me your last reply, you have quoted me entirely wrong. It was 2/sin($\phi$) not 2sin($\phi$) same with ($\pi$)/2

 

[Titan97]

I wrote $\phi/2$ in notebook. I was typing in early morning just after I woke up. It was basically a communication problem.
The answer given in the textbook however is $\pi\sin^{-1}b$.
But I know how to solve it now.

 

[Umrao]

I wrote $\phi/2$ in notebook. I was typing in early morning just after I woke up. It was basically a communication problem.
The answer given in the textbook however is $\pi\sin^{-1}b$.
But I know how to solve it now.

Then why did you bother me with all the latex. I wasted 1.5 hours

 

[Titan97]

You could have avoided LaTeX if you wanted to. I never forced you. Besides, the answer is still wrong.

 

[Umrao]

You could have avoided LaTeX if you wanted to. I never forced you. Besides, the answer is still wrong.

Why don't you tell me where

 

[Titan97]

$$f(b)=\pi\sin^{-1}b$$
That's the answer given in the textbook.

 

[Umrao]

$$f(b)=\pi\sin^{-1}b$$
That's the answer given in the textbook.

Put the damn value of c to get that. f(0) = 0

 

[Umrao]

Put the damn value of c to get that. f(0) = 0

c = (pi^2)/2