integral(((ln(1+bsinx))/(sinx)),x,-π/2,π/2)
I hope you can understand my way of presenting the question.......
Hello,
$\displaystyle \ln(1+b\sin x)=\sum_{n\geq 1} \frac{b^n \sin^n x}{n}$ (where $\displaystyle |b|<1$)
So $\displaystyle I=\int_{-\pi/2}^{\pi/2} \sum_{n\geq 1} b^n \cdot \frac{\sin^{n-1} x}{n} ~dx$
Inverting the sum and the integral (using the proper theorems), we have :
$\displaystyle I=\sum_{n\geq 1} \frac{b^n}{n} \int_{-\pi/2}^{\pi/2} \sin^{n-1} x ~dx=\sum_{n\geq 2}\frac{b^{n-1}}{n-1} \cdot J_n$
where $\displaystyle J_n=\int_{-\pi/2}^{\pi/2} \sin^n x ~dx$ (known as Wallis integral in French..couldn't find it in English)
By a common method ($\displaystyle \sin^n x=\sin^{n-2} x-\cos x \cdot \cos x\sin^{n-2} x$) and an integration by parts, we get the recursive relation : $\displaystyle J_n=J_{n-2}+\frac{1}{n-1} \cdot J_n$
$\displaystyle \Rightarrow \boxed{nJ_n=(n-1)J_{n-2}}$
We can deduce this :
$\displaystyle \int_0^{\pi/2} \sin^n x~dx=\begin{cases} \frac{(2p)!}{2^{2p}(p!)^2}\cdot \frac \pi 2 \quad \text{if } n=2p \\ \frac{2^{2p} (p!)^2}{(2p+1)!} \quad \text{if } n=2p+1\end{cases}$
So if n is even, $\displaystyle J_n=\frac{n!}{2^n \left[\left(\tfrac n2\right)!\right]^2} \cdot \pi$
And if n is odd, $\displaystyle J_n=0$
So we have :
$\displaystyle I=\pi\sum_{k\geq 1} \frac{b^{2k-1}(2k)!}{(2k-1)2^{2k}(k!)^2}$
But hey, this is the power series for arcsin !!!
So finally, $\displaystyle \boxed{I=\pi \arcsin(b)}$ (I hope I didn't get wrong in the power series)
Here's an alternative way.
Let $\displaystyle
I(b) = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\:\frac{\ln(1+b\sin x)}{\sin x}\:dx$ noting that $\displaystyle I(0) = 0$.
Then $\displaystyle I'(b) = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\:\frac{1}{1+b\sin x}\:dx$.
Introducing the usual trick $\displaystyle x = 2 \,tan^{-1}z$, then $\displaystyle I'(b) = 2 \int_{-1}^1 \frac{1}{z^2 + 2bz + 1}\,dz$ which readily integrates. So
$\displaystyle I'(b) = \frac{2}{\sqrt{1-b^2}} \left(\tan^{-1} \frac{b+1}{\sqrt{1-b^2}} - \tan^{-1} \frac{b-1}{\sqrt{1-b^2}} \right) = \frac{2}{\sqrt{1-b^2}} \cdot \frac{\pi}{2} = \frac{\pi}{\sqrt{1-b^2}}$
Integrating wrt $\displaystyle b$ give $\displaystyle I(b) = \pi \sin^{-1} b$ noting the constant of integration is zero due to the fact that $\displaystyle I(0) = 0.$
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.
http://en.wikipedia.org/wiki/Leibniz_integral_rule
Well, it's a basic power series (google for it)
We know (geometric series) that if $\displaystyle |y|<1$, then $\displaystyle \frac{1}{1+y}=\sum_{k\geq 0} y^k$
Integrate from 0 to x :
$\displaystyle \ln(1+x)=\int_0^x \frac{1}{1+y} ~dy=\sum_{k\geq 0} \frac{x^{k+1}}{k+1}=\sum_{k\geq 1}\frac{x^k}{k}$