evaluate the following definite integral

IIT 2010 · #1 $Old$ June 30th, 2009, 09:51 AM

integral(((ln(1+bsinx))/(sinx)),x,-π/2,π/2)

I hope you can understand my way of presenting the question.......

pankaj · #2 $Old$ June 30th, 2009, 10:23 AM

Is this your question
Evaluate:
$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{log_{e}(1+b\sin x)}{\sin x}dx$

running-gag · #3 $Old$ June 30th, 2009, 10:27 AM

Hi

Try to use LATEX code

Do you mean

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\:\frac{\ln(1+b\sin x)}{\sin x}\:dx$

simplependulum · #4 $Old$ July 1st, 2009, 01:01 AM

The answer is $\pi \sin^{-1}{b} ~,~ |b| \leq 1$ Isn't it ?

Moo · #5 $Old$ July 1st, 2009, 02:51 AM

Hello,

$\ln(1+b\sin x)=\sum_{n\geq 1} \frac{b^n \sin^n x}{n}$ (where $|b|<1$ )

So $I=\int_{-\pi/2}^{\pi/2} \sum_{n\geq 1} b^n \cdot \frac{\sin^{n-1} x}{n} ~dx$

Inverting the sum and the integral (using the proper theorems), we have :

$I=\sum_{n\geq 1} \frac{b^n}{n} \int_{-\pi/2}^{\pi/2} \sin^{n-1} x ~dx=\sum_{n\geq 2}\frac{b^{n-1}}{n-1} \cdot J_n$

where $J_n=\int_{-\pi/2}^{\pi/2} \sin^n x ~dx$ (known as Wallis integral in French..couldn't find it in English)

By a common method ( $\sin^n x=\sin^{n-2} x-\cos x \cdot \cos x\sin^{n-2} x$ ) and an integration by parts, we get the recursive relation : $J_n=J_{n-2}+\frac{1}{n-1} \cdot J_n$

$\Rightarrow \boxed{nJ_n=(n-1)J_{n-2}}$

We can deduce this :

$\int_0^{\pi/2} \sin^n x~dx=\begin{cases} \frac{(2p)!}{2^{2p}(p!)^2}\cdot \frac \pi 2 \quad \text{if } n=2p \\ \frac{2^{2p} (p!)^2}{(2p+1)!} \quad \text{if } n=2p+1\end{cases}$

So if n is even, $J_n=\frac{n!}{2^n \left[\left(\tfrac n2\right)!\right]^2} \cdot \pi$

And if n is odd, $J_n=0$

So we have :

$I=\pi\sum_{k\geq 1} \frac{b^{2k-1}(2k)!}{(2k-1)2^{2k}(k!)^2}$

But hey, this is the power series for arcsin !!!

So finally, $\boxed{I=\pi \arcsin(b)}$ (I hope I didn't get wrong in the power series)

Danny · #6 $Old$ July 1st, 2009, 05:14 AM

Here's an alternative way.

Let $I(b) = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\:\frac{\ln(1+b\sin x)}{\sin x}\:dx$ noting that $I(0) = 0$ .

Then $I'(b) = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\:\frac{1}{1+b\sin x}\:dx$ .

Introducing the usual trick $x = 2 \,tan^{-1}z$ , then $I'(b) = 2 \int_{-1}^1 \frac{1}{z^2 + 2bz + 1}\,dz$ which readily integrates. So

$I'(b) = \frac{2}{\sqrt{1-b^2}} \left(\tan^{-1} \frac{b+1}{\sqrt{1-b^2}} - \tan^{-1} \frac{b-1}{\sqrt{1-b^2}} \right) = \frac{2}{\sqrt{1-b^2}} \cdot \frac{\pi}{2} = \frac{\pi}{\sqrt{1-b^2}}$

Integrating wrt $b$ give $I(b) = \pi \sin^{-1} b$ noting the constant of integration is zero due to the fact that $I(0) = 0.$

Same answer but a little different approach.

IIT 2010 · #7 $Old$ July 1st, 2009, 07:38 AM

Quote:

Originally Posted by Danny $View Post$

Here's an alternative way.

Let $I(b) = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\:\frac{\ln(1+b\sin x)}{\sin x}\:dx$ noting that $I(0) = 0$ .

Then $I'(b) = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\:\frac{1}{1+b\sin x}\:dx$ .

Introducing the usual trick $x = 2 \,tan^{-1}z$ , then $I'(b) = 2 \int_{-1}^1 \frac{1}{z^2 + 2bz + 1}\,dz$ which readily integrates. So

$I'(b) = \frac{2}{\sqrt{1-b^2}} \left(\tan^{-1} \frac{b+1}{\sqrt{1-b^2}} - \tan^{-1} \frac{b-1}{\sqrt{1-b^2}} \right) = \frac{2}{\sqrt{1-b^2}} \cdot \frac{\pi}{2} = \frac{\pi}{\sqrt{1-b^2}}$

Integrating wrt $b$ give $I(b) = \pi \sin^{-1} b$ noting the constant of integration is zero due to the fact that $I(0) = 0.$

Same answer but a little different approach.

hi sir,
plz explain what I'(b) actually is and how do we differentiate a portion within the integration symbol with constant limits of integration........
I'm a precollege student preparing for an entrance exam.We were not introduced to these type of problems,still I have got such a problem in a test.I hope sincerely you people will definitely help me out in these kind of problems
thank you very much.

IIT 2010 · #8 $Old$ July 1st, 2009, 07:42 AM

Quote:

Originally Posted by Moo $View Post$

Hello,

$\ln(1+b\sin x)=\sum_{n\geq 1} \frac{b^n \sin^n x}{n}$ (where $|b|<1$ )

So $I=\int_{-\pi/2}^{\pi/2} \sum_{n\geq 1} b^n \cdot \frac{\sin^{n-1} x}{n} ~dx$

Inverting the sum and the integral (using the proper theorems), we have :

$I=\sum_{n\geq 1} \frac{b^n}{n} \int_{-\pi/2}^{\pi/2} \sin^{n-1} x ~dx=\sum_{n\geq 2}\frac{b^{n-1}}{n-1} \cdot J_n$

where $J_n=\int_{-\pi/2}^{\pi/2} \sin^n x ~dx$ (known as Wallis integral in French..couldn't find it in English)

By a common method ( $\sin^n x=\sin^{n-2} x-\cos x \cdot \cos x\sin^{n-2} x$ ) and an integration by parts, we get the recursive relation : $J_n=J_{n-2}+\frac{1}{n-1} \cdot J_n$

$\Rightarrow \boxed{nJ_n=(n-1)J_{n-2}}$

We can deduce this :

$\int_0^{\pi/2} \sin^n x~dx=\begin{cases} \frac{(2p)!}{2^{2p}(p!)^2}\cdot \frac \pi 2 \quad \text{if } n=2p \\ \frac{2^{2p} (p!)^2}{(2p+1)!} \quad \text{if } n=2p+1\end{cases}$

So if n is even, $J_n=\frac{n!}{2^n \left[\left(\tfrac n2\right)!\right]^2} \cdot \pi$

And if n is odd, $J_n=0$

So we have :

$I=\pi\sum_{k\geq 1} \frac{b^{2k-1}(2k)!}{(2k-1)2^{2k}(k!)^2}$

But hey, this is the power series for arcsin !!!

So finally, $\boxed{I=\pi \arcsin(b)}$ (I hope I didn't get wrong in the power series)

sir can you plz explain me the first step??
thanks in advance...........

Moo · #9 $Old$ July 4th, 2009, 06:38 AM

Quote:

Originally Posted by IIT 2010 $View Post$

hi sir,
plz explain what I'(b) actually is and how do we differentiate a portion within the integration symbol with constant limits of integration........
I'm a precollege student preparing for an entrance exam.We were not introduced to these type of problems,still I have got such a problem in a test.I hope sincerely you people will definitely help me out in these kind of problems
thank you very much.

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

Quote:

Originally Posted by IIT 2010 $View Post$

sir can you plz explain me the first step??
thanks in advance...........

Well, it's a basic power series (google for it)

We know (geometric series) that if $|y|<1$ , then $\frac{1}{1+y}=\sum_{k\geq 0} y^k$

Integrate from 0 to x :
$\ln(1+x)=\int_0^x \frac{1}{1+y} ~dy=\sum_{k\geq 0} \frac{x^{k+1}}{k+1}=\sum_{k\geq 1}\frac{x^k}{k}$