1. ## Integral

Hello All

I could do with some help on this integral. I didn't have any problems with part a) and b), I'm stuck on part c).

Bobak

Edit: the integral is $\int \frac{\sin^2((k+1)x)}{\sin^2x}$

2. Originally Posted by bobak
Hello All

I could do with some help on this integral. I didn't have any problems with part a) and b), I'm stuck on part c).

Bobak
Hello,

I am sorry, I hate this notation. Is this

$\sin^2(k+1)\cdot{x}$

Or

$\sin^2((k+1)x)$
?

3. Originally Posted by Mathstud28
Hello,

I am sorry, I hate this notation. Is this

$\sin^2(k+1)\cdot{x}$

Or

$\sin^2((k+1)x)$
?
the second one

4. Hello

Originally Posted by bobak
Hello All

I could do with some help on this integral. I didn't have any problems with part a) and b), I'm stuck on part c).

Bobak
yay ! I think I found it ! ^^

$\frac{sin[(2k+1)x]}{sin(x)}=\frac{sin(2kx)cos(x)+cos(2kx)sin(x)}{sin (x)}$

$=\frac{2cos(kx)sin(kx)cos(x)+sin(x)[cos^2(kx)-sin^2(kx)]}{sin(x)}$

Multiply above and below by sin(x) :

$=...$ (skipped the step... I guess you can do it )

Rearranging :

$=\frac{2{\color{red}sin(x)cos(kx)} \cdot {\color{blue}sin(kx)cos(x)}+[{\color{red}sin(x)cos(kx)}]^2-[sin(x)sin(kx)]^2}{sin^2(x)}$

Complete the square in red :

$=\frac{\left(\overbrace{{\color{red}sin(x)cos(kx)} +{\color{blue}sin(kx)cos(x)}}^{sin[(k+1)x]}\right)^2-[{\color{blue}sin(kx)cos(x)}]^2-[sin(x)sin(kx)]^2}{sin^2(x)}$

$\frac{sin[(2k+1)x]}{sin(x)}=\frac{sin^2[(k+1)x]}{sin^2(x)}-\frac{[sin(kx)cos(x)]^2+[sin(x)sin(kx)]^2}{sin^2(x)}$

----------------------------

$\frac{[sin(kx)cos(x)]^2+[sin(x)sin(kx)]^2}{sin^2(x)}$

$=\frac{sin^2(kx)cos^2(x)+sin^2(x)sin^2(kx)}{sin^2( x)}=\frac{sin^2(kx)[cos^2(x)+sin^2(x)]}{sin^2(x)}=\boxed{\frac{sin^2(kx)}{sin^2(x)}}$

----------------------------

Integrating :

-------> $\frac{\pi}{2}=\overbrace{\int_0^{\pi /2} \frac{sin[(2k+1)x]}{sin(x)} \ dx}^{\frac{\pi}{2}}=\int_0^{\pi /2} \frac{sin^2[(k+1)x]}{sin^2(x)} \ dx - \int_0^{\pi /2} \frac{sin^2(kx)}{sin^2(x)} \ dx \quad \square$

5. Thanks for that Moo. Much apperiaceted.

The last part is pretty straightforward

Bobak

6. Originally Posted by bobak
Thanks for that Moo. Much apperiaceted.

The last part is pretty straightforward

Bobak
I edited many times my post, so I'm not sure what you're talking about
Refresh ^^

7. Originally Posted by Moo
I edited many times my post, so I'm not sure what you're talking about
Refresh ^^
Yeah I noticed you editing it, I got the general idea thought, I made the error of trying to work form the result backwards, Which didn't work out very well for me.

Bobak

8. Originally Posted by bobak
Hello All

I could do with some help on this integral. I didn't have any problems with part a) and b), I'm stuck on part c).

Bobak

Edit: the integral is $\int \frac{\sin^2((k+1)x)}{\sin^2x}$
In case you want to see an optional solution to the integral in part b) (not much to do with that exercise)

$
I_n = \int_{ - \pi }^\pi {\frac{{\sin \left[ {\left( {2n + 1} \right) \cdot x} \right]}}
{{\sin \left( x \right)}}dx} = 2\pi ,\forall n \in \mathbb{N}
$

By Euler's Formula: $
\sin \left( {n \cdot x} \right) = \tfrac{{e^{i \cdot n \cdot x} - e^{ - i \cdot n \cdot x} }}
{{2 \cdot i}};\sin \left( {n \cdot x} \right) = \tfrac{{e^{i \cdot x} - e^{ - i \cdot x} }}
{{2 \cdot i}}
$
(we're not using contours)

Thus $
I_n = \int_{ -\pi }^{\pi } {\frac{{e^{i \cdot
{\left( {2n + 1} \right)}
\cdot x} - e^{ - i \cdot
{\left( {2n + 1} \right)}
\cdot x} }}
{{e^{i \cdot x} - e^{ - i \cdot x} }}dx}
$

Now recall the formula: $
x^n - y^n = \left( {x - y} \right) \cdot \sum\limits_{k = 0}^{n - 1} {x^k \cdot y^{n - 1 - k} }
$
we have: $
e^{i \cdot \left( {2n + 1} \right) \cdot x} - e^{ - i \cdot \left( {2n + 1} \right) \cdot x} = \left( {e^{i \cdot x} - e^{ - i \cdot x} } \right) \cdot \sum\limits_{k = 0}^{2n} {e^{2 \cdot i \cdot x \cdot \left( {k - n} \right)} }

$

Thus: $
I_n = \int_{ - \pi }^\pi {\left( {\sum\limits_{k = 0}^{2n} {e^{2 \cdot i \cdot x \cdot \left( {k - n} \right)} } } \right)dx} = \sum\limits_{k = 0}^{2n} {\int_{ - \pi }^\pi {e^{2 \cdot i \cdot x \cdot \left( {k - n} \right)} dx} }

$

Note that: $
\left\{ \begin{gathered}
{\text{If }}k \ne n \Rightarrow \int_{ - \pi }^\pi {e^{2 \cdot i \cdot x \cdot \left( {k - n} \right)} dx} = 0 \hfill \\
{\text{If }}k = n \Rightarrow \int_{ - \pi }^\pi {e^{2 \cdot i \cdot x \cdot \left( {k - n} \right)} dx} = 2\pi \hfill \\
\end{gathered} \right.
$
Therefore: $
I_n = \sum\limits_{k = 0}^{2n} {\int_{ - \pi }^\pi {e^{2 \cdot i \cdot x \cdot \left( {k - n} \right)} dx} } = 2\pi
$

And since: $
\int_{ - \pi }^\pi {\frac{{\sin \left[ {\left( {2n + 1} \right) \cdot x} \right]}}
{{\sin \left( x \right)}}dx} = 4 \cdot \int_0^{\tfrac{\pi }
{2}} {\frac{{\sin \left[ {\left( {2n + 1} \right) \cdot x} \right]}}
{{\sin \left( x \right)}}dx}
$
we have: $
\int_0^{\tfrac{\pi }
{2}} {\frac{{\sin \left[ {\left( {2n + 1} \right) \cdot x} \right]}}
{{\sin \left( x \right)}}dx} = \tfrac{\pi }
{2},\forall n \in \mathbb{N}
$

9. I worked out $I_{n}=\int_{0}^{\pi}\frac{sin^{2}(nx)}{sin^{2}(x)} dx$

if you still care.

For all positive integer values of n.

Let's show that $I_{n}=n{\pi}, \;\ n=1,2,3,....$

We have for $n\geq{2}$:

$D_{n}=I_{n}-I_{n-1}=\int_{0}^{\pi}\frac{(sin^{2}(nx)-sin^{2}((n-1)x)}{sin^{2}(x)}dx$

$=\int_{0}^{\pi}\frac{(sin(nx)-sin((n-1)x))(sin(nx)+sin((n-1)x))}{sin^{2}(x)}dx$

$\int_{0}^{\pi}\frac{2sin(\frac{x}{2})cos(n-\frac{x}{2})2sin(nx-\frac{x}{2})cos(\frac{x}{2})}{sin^{2}(x)}dx$

$=\int_{0}^{\pi}\frac{sin(x)sin((2n-1)x)}{sin^{2}(x)}dx$

$=\int_{0}^{\pi}\frac{sin((2n-1)x)}{sin(x)}dx$

So, $D_{n}=J_{2n-1}, \;\ n\geq{2}$

$J_{m}=\int_{0}^{\pi}\frac{sin(mx)}{sin(x)}dx, \;\ m=0,1,2,..$

Now, for $m\geq{2}$:

$J_{m}-J_{m-2}=\int_{0}^{\pi}\frac{sin(mx)-sin((m-2)x)}{sin(x)}dx$

$=\int_{0}^{\pi}\frac{2sin(x)cos((m-1)x)}{sin(x)}dx$

$=2\int_{0}^{\pi}cos((m-1)x)dx$

$=2\left[\frac{sin((m-1)x)}{m-1}\right]_{0}^{\pi}=0$

So, we have

$J_{m}=J_{m-2}=J_{m-4}=....=\begin{gathered}J_{0}=0, \;\ \text{if m even}\\ J_{1}={\pi}, \;\ \text{if m odd}\end{gathered}$

Therefore, hence and hereftofore:

We get $D_{n}={\pi}, \;\ n\geq{2},$ so that $I_{n}=n{\pi}, \;\ n\geq{1}, \;\ as \;\ I_{1}={\pi}$

WHEW!!

I just noticed. I used Pi instead of Pi/2 all this time. Oh well. Alas.

10. Hi galactus... Isn't it too complicated ?

$\int_0^{\pi /2} \frac{\sin^2 (kx)}{\sin^2(x)} \ dx=k \frac{\pi}{2}$

---------------------
For k=0, $\frac{\sin^2(kx)}{\sin^2(x)}=0 \quad \square.$

Assume that $\int_0^{\pi /2} \frac{\sin^2(kx)}{\sin^2(x)} \ dx = k \frac{\pi}{2}$

From the previous question :

$\int_0^{\pi /2} \frac{\sin^2 [(k+1)x]}{\sin^2(x)} \ dx=\int_0^{\pi /2} \frac{\sin^2 (kx)}{\sin^2(x)} \ dx + \frac{\pi}{2}=\bf (k+1) \frac{\pi}{2} \quad \blacksquare.$

huh?