Question

If $S=\sum _{n=1}^{10}{I}_n,$ where $I_n=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{\sin (2n+1)x}{\sin x}dx$ then the value of $S$ is

• A. $5\pi$
• B. $4\pi$
• C. $10\pi$
• D. $12\pi$

Answer 1

The correct option is A $5\pi$

$I_n=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{\sin (2n+1)x}{\sin x}dx$
$I_{n-1}=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{\sin \left(2\right(n-1)+1)x}{\sin x}dx$
​​​​$I_n-I_{n-1}=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{\sin (2n+1)x-\sin (2n-1)x}{\sin x}dx$
$\Rightarrow I_n-I_{n-1}=2\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{\cos 2nx\sin x}{\sin x}dx$
$\Rightarrow I_n-I_{n-1}=\frac{2}{2n}[\sin 2nx{]}_0^{\frac{\pi }{2}}=0$
$\therefore I_n=I_{n-1}=......=I_0$
$\Rightarrow I_0=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{\sin x}{\sin x}dx=\frac{\pi }{2}$
$\therefore S=10\times \frac{\pi }{2}=5\pi$
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