Question

Consider the integral $I_{n}=\int_{0}^{\frac{\pi}{4}} \frac{\sin (2 n-1) x}{\sin x} d x$, then the value of $I_{20}-I_{19}$ is

• A. $\frac{1}{20}$
• B. $\frac{- 1}{19}$
• C. $\frac{- 1}{25}$
• D. $\frac{1}{19}$

Answer 1

Answer: (B)

We have, $I_{n}=\int_{0}^{\frac{\pi}{4}} \frac{\sin (2 n-1) x}{\sin x} d x$
Now, $I_{20}-I_{19}=\int_{0}^{\frac{\pi}{4}} \frac{\sin (39 x)-\sin (37 x)}{\sin x} d x$
$ \begin{array}{l} \Rightarrow I_{20}-I_{19}=\int_{0}^{\frac{\pi}{4}} \frac{2 \sin x \cos (38 x)}{\sin x} d x \\ =\left[\frac{\sin (38 x)}{19}\right]_{0}^{\frac{\pi}{4}} \\ =\frac{\sin \left(\frac{38 \pi}{4}\right)}{19} \\ =\frac{-1}{19} \end{array} $