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STD 12 - 7.2 definite integral — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 7.2 definite integral1 Mark
MCQ
If $A_n =$ $\int\limits_0^{\pi /2} \frac{{\sin \,\,(2\,n\,\, - \,\,1)\,\,x}}{{\sin \,\,x}} d x$ ; $B_n =$ $\int\limits_0^{\pi /2}\, {\left( {\frac{{\sin \,\,n\,x}}{{\sin \,\,x}}} \right)^2} d x$ ; for $n \in N$ , then :
  • A
    $A_{n + 1} = A_n$
  • B
    $B_{n + 1} -B_n = A_{n + 1}$
  • C
    $A_{n + 1} -A_n = B_{n + 1}$
  • ✓
    Both $(A)$ and $(B)$
✓

Answer

Correct option: D.
Both $(A)$ and $(B)$
d
Consider $A_{n + 1} -A_n $

$=\int\limits_0^{\pi /2} {\frac{{\sin (2n - 1)x - \sin (2n - 1)x}}{{\sin \,x}}\,}$ 

 $=$ $2\, \cos\, 2\, n x\, d x \,= 0$
$\Rightarrow A_{n + 1} = A_n$
Again consider $B_{n + 1} -B_n$

$=\int\limits_0^{\pi /2}$ $\frac{{\sin 2\,\,(n\, + \,1)\,x\,\, - \,\,{{\sin }^2}\,x}}{{{{\sin }^2}\,x}}$ 

$=$ $\int\limits_0^{\pi /2}$ $\frac{{\sin \,\,(2\,n\,\, + \,\,1)\,\,x}}{{\sin \,\,x}}$ $d x$

$= A_{n + 1}$

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