Question

Consider the integral $I_n={\int }_0^{\frac{\pi }{4}}\frac{\sin (2n-1)x}{\sin x}dx$, 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, $I_n={\int }_0^{\frac{\pi }{4}}\frac{\sin (2n-1)x}{\sin x}dx$
Now, $I_{20}-I_{19}={\int }_0^{\frac{\pi }{4}}\frac{\sin (39x)-\sin (37x)}{\sin x}dx$
$\begin{array}{l}\Rightarrow I_{20}-I_{19}={\int }_0^{\frac{\pi }{4}}\frac{2\sin x\cos (38x)}{\sin x}dx\\ ={\left[\frac{\sin (38x)}{19}\right]}_0^{\frac{\pi }{4}}\\ =\frac{\sin \left(\frac{38\pi }{4}\right)}{19}\\ =\frac{-1}{19}\end{array}$