Question

If the integral $I_{n}=\displaystyle \int _{0}^{\frac{\pi }{2}}\frac{sin \left(2 n - 1\right) x}{sin ⁡ x}dx$ , then the value of $\left[I_{20}\right]^{3}-\left[I_{19}\right]^{3}$ is

• A. $400$
• B. $200$
• C. $361$
• D. $0$

Answer 1

Answer: (D)

$I_{20} - I_{19} = \displaystyle \int _{0}^{\frac{\pi }{2}} \frac{sin \left(39 x\right) - sin \left(37 x\right)}{sin x} d x$
$\Rightarrow I_{20}-I_{19}=\displaystyle \int _{0}^{\frac{\pi }{2}}\frac{2 sin x cos \left(38 x\right)}{sin x}dx$
$\Rightarrow I_{20} - I_{19} = \displaystyle \int _{0}^{\frac{\pi }{2}} 2 cos \left(38 x\right) d x$
$=\left[\frac{sin \left(38 x\right)}{19}\right]\frac{\pi }{2} \\ 0$
$=\frac{sin \left(\frac{38 \pi }{2}\right)}{19}$
$=0$
$\therefore I_{20}=I_{19}$
Hence, $\left[I_{20}\right]^{3}-\left[I_{19}\right]^{3}=0$