Integral Calculus Question 150
Question: $ A_{n}=\int\limits_0^{\pi /2}{\frac{\sin (2n-1)x}{\sin x}}dx;B_{n}=\int\limits_0^{\pi /2}{{{( \frac{\sin nx}{\sin x} )}^{2}}dx;} $ For $ n\in N, $ then
Options:
A) $ {A_{n+1}}=A_{n},{B_{n+1}}-B_{n}={A_{n+1}} $
B) $ {B_{n+1}}=B_{n} $
C) $ {A_{n+1}}-A_{n}={B_{n+1}} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
[a] We have $ {A_{n+1}}-A_{n}=\int\limits_0^{\frac{\pi }{2}}{\frac{\sin (2n+1)x-\sin (2n-1)x}{\sin x}dx} $ $ =\int\limits_0^{\frac{\pi }{2}}{\frac{2\cos 2nx\sin x}{\sin x}dx=2\int\limits_0^{\frac{\pi }{2}}{\cos 2n\pi dx=0}} $ Again $ {B_{n+1}}-B_{n}=\int\limits_0^{\frac{\pi }{2}}{\frac{{{\sin }^{2}}(n+1)x-{{\sin }^{2}}nx}{{{\sin }^{2}}x}dx} $ $ =\int\limits_0^{\frac{\pi }{2}}{\frac{\sin (2n+1)x\sin x}{{{\sin }^{2}}x}dx={A_{n+1}}} $
BETA
