- Thread starter
- #1

#### Pranav

##### Well-known member

- Nov 4, 2013

- 428

**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!