# Thread: evaluate the following definite integral

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

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

Evaluate:
$\displaystyle \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{log_{e}(1+b\si n x)}{\sin x}dx$

3. Hi

Try to use LATEX code

Do you mean

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

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

5. 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)

6. 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.

7. Originally Posted by Danny
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.

8. Originally Posted by Moo
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)
sir can you plz explain me the first step??

9. Originally Posted by IIT 2010
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

Originally Posted by IIT 2010
sir can you plz explain me the first step??
We know (geometric series) that if $\displaystyle |y|<1$, then $\displaystyle \frac{1}{1+y}=\sum_{k\geq 0} y^k$
$\displaystyle \ln(1+x)=\int_0^x \frac{1}{1+y} ~dy=\sum_{k\geq 0} \frac{x^{k+1}}{k+1}=\sum_{k\geq 1}\frac{x^k}{k}$