MHB Integrating Symmetric Definite Integrals: A Trick for Evaluating $U_n-U_{n-1}$

[Saitama]

Attempt:
If $\displaystyle U_n=\int_0^{\pi/2} \frac{\sin^2(nx)}{\sin^2x}\,dx$, then find $U_n-U_{n-1}$.

Attempt:
$$U_n=\int_0^{\pi/2} \frac{\sin^2(nx)}{\sin^2x}\,dx$$
$$U_{n-1}=\int_0^{\pi/2} \frac{\sin^2((n-1)x)}{\sin^2x}\,dx$$
$$\Rightarrow U_n-U_{n-1}=\int_0^{\pi/2} \frac{(\sin(nx)-\sin((n-1)x)(\sin(nx)+\sin(n-1)x)}{\sin^2x}\,dx$$
I write the following:
$$\sin(nx)-\sin((n-1)x=2\cos\left(\frac{2n-1}{2}x\right)\sin\left(\frac{x}{2}\right)$$
$$\sin(nx)+\sin((n-1)x=2\sin\left(\frac{2n-1}{2}x\right)\cos\left(\frac{x}{2}\right)$$
The product of above can be written as follows:
$$(\sin(nx)-\sin((n-1)x)(\sin(nx)+\sin((n-1)x)=\sin(x)\sin((2n-1)x)$$
Hence,
$$U_n-U_{n-1}=\int_0^{\pi/2} \frac{\sin((2n-1)x)}{\sin x}\,\,dx$$
I am not sure how to proceed after this. I tried writing $\sin((2n-1)x)=\sin(2nx)\cos(x)-\sin(x)\cos(2nx)$ but that doesn't seem to help. :(

Any help is appreciated. Thanks!

 

[Shobhit]

Let $$T_n=\int_0^{\pi/2}\frac{\sin((2n+1)x)}{\sin x}dx$$

Then $$
\begin{align*}
T_n-T_{n-1} &= \int_0^{\pi/2} \frac{\sin((2n+1)x)-\sin((2n-1)x)}{\sin x}dx \\
&= 2 \int_0^{\pi/2}\cos(2nx)\; dx \\
&= \frac{1}{n}\sin(n\pi) \\
&= 0
\end{align*}
$$

So $$T_n=T_{n-1}=T_{n-2}=\cdots =T_0 = \int_0^{\pi/2}dx=\frac{\pi}{2}$$

Then using the last equation of your post we get $$U_{n}-U_{n-1}=T_{n-1}=\frac{\pi}{2}$$

And from here you can also deduce that $$U_n=\frac{n\pi}{2}$$.

 

[Saitama]

Let $$T_n=\int_0^{\pi/2}\frac{\sin((2n+1)x)}{\sin x}dx$$

Then $$
\begin{align*}
T_n-T_{n-1} &= \int_0^{\pi/2} \frac{\sin((2n+1)x)-\sin((2n-1)x)}{\sin x}dx \\
&= 2 \int_0^{\pi/2}\cos(2nx)\; dx \\
&= \frac{1}{n}\sin(n\pi) \\
&= 0
\end{align*}
$$

So $$T_n=T_{n-1}=T_{n-2}=\cdots =T_0 = \int_0^{\pi/2}dx=\frac{\pi}{2}$$

Thus $$U_{n}-U_{n-1}=T_{n-1}=\frac{\pi}{2}$$

And from here you can also deduce that $$U_n=\frac{n\pi}{2}$$.

Thanks Shobhit! :)

Can I have some insight about how did you think of defining $T_n$? Is there no way to evaluate the definite integral I got or is there any alternative solution?

 

[Shobhit]

Thanks Shobhit! :)

Can I have some insight about how did you think of defining $T_n$? Is there no way to evaluate the definite integral I got?

The integral at the end of your post is actually $T_{n-1}$.

Is there any alternative solution?

I know another solution using the residue theorem but I think that would be too hard for you to understand and probably not appropriate for the calculus section. :)

 

[Saitama]

The integral at the end of your post is actually $T_{n-1}$.

Yes, I know about that but I am more interested to know how did you think of $T_n$? Is this some kind of routine problem?

 

[Shobhit]

Yes, I know about that but I am more interested to know how did you think of $T_n$? Is this some kind of routine problem?

Over the years I have gained a lot of experience in evaluating definite integrals. So that's why I knew which trick can be applied here. The reason behind that particular definition of $T_n$ is the symmetry of the integral.

 

[Saitama]

Over the years I have gained a lot of experience in evaluating definite integrals. So that's why I knew which trick can be applied here. The reason behind that particular definition of $T_n$ is the symmetry of the integral.

Thanks Shobhit! :)