Question

If the integral $I_n={\int }_0^{\frac{\pi }{2}}\frac{sin\left(2n-1\right)x}{sinx}dx$ , 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)

$I_{20}-I_{19}={\int }_0^{\frac{\pi }{2}}\frac{sin\left(39x\right)-sin\left(37x\right)}{sinx}dx$
$\Rightarrow I_{20}-I_{19}={\int }_0^{\frac{\pi }{2}}\frac{2sinxcos\left(38x\right)}{sinx}dx$
$\Rightarrow I_{20}-I_{19}={\int }_0^{\frac{\pi }{2}}2cos\left(38x\right)dx$
$$\begin{gathered} =\left[\frac{sin\left(38x\right)}{19}\right]\frac{\pi }{2} \\ 0 \end{gathered}$$
$=\frac{sin\left(\frac{38\pi }{2}\right)}{19}$
$=0$
$\therefore I_{20}=I_{19}$
Hence, ${\left[I_{20}\right]}^3-{\left[I_{19}\right]}^3=0$