If b_{n} = int_{0}^{frac{pi}{2}} frac{cos^{2} | vedclass

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(D) Consider the difference $b_{n+1} - b_{n} = \int_{0}^{\frac{\pi}{2}} \frac{\cos^{2}(n+1)x - \cos^{2}nx}{\sin x} dx$.Using the identity $\cos^{2}A -

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$\frac{1}{b_{3}-b_{2}}, \frac{1}{b_{4}-b_{3}}, \frac{1}{b_{5}-b_{4}}$ are in an $A.P.$ with common difference $-2$

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(D) Consider the difference $b_{n+1} - b_{n} = \int_{0}^{\frac{\pi}{2}} \frac{\cos^{2}(n+1)x - \cos^{2}nx}{\sin x} dx$.Using the identity $\cos^{2}A - \cos^{2}B = -\sin(A+B)\sin(A-B)$,we get:$b_{n+1} - b_{n} = \int_{0}^{\frac{\pi}{2}} \frac{-\sin(2n+1)x \sin x}{\sin x} dx = -\int_{0}^{\frac{\pi}{2}} \sin(2n+1)x dx$.Evaluating the integral:$b_{n+1} - b_{n} = \left[ \frac{\cos(2n+1)x}{2n+1} \right]_{0}^{\frac{\pi}{2}} = \frac{\cos((2n+1)\frac{\pi}{2}) - \cos(0)}{2n+1} = \frac{0 - 1}{2n+1} = -\frac{1}{2n+1}$.Now,let $a_{n} = b_{n+1} - b_{n} = -\frac{1}{2n+1}$.Then $\frac{1}{a_{n}} = -(2n+1)$.For $n=2, 3, 4$,we have $\frac{1}{a_{2}} = -5, \frac{1}{a_{3}} = -7, \frac{1}{a_{4}} = -9$.These terms are in an $A.P.$ with common difference $-7 - (-5) = -2$ and $-9 - (-7) = -2$.Thus,$\frac{1}{b_{3}-b_{2}}, \frac{1}{b_{4}-b_{3}}, \frac{1}{b_{5}-b_{4}}$ are in an $A.P.$ with common difference $-2$.

If $b_{n} = \int_{0}^{\frac{\pi}{2}} \frac{\cos^{2} nx}{\sin x} dx$,$n \in N$,then

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