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  • If  <img alt="\mathrm{b_{n}=\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} n x}{\sin x} d x, n \in \mathbb{N}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bb_%7Bn%7D%3D%5Cint_%7

If  \mathrm{b_{n}=\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} n x}{\sin x} d x, n \in \mathbb{N}}, then

Option: 1

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


Option: 2

\mathrm{\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 \mathrm{2}


Option: 3

\mathrm{b_{3}-b_{2},b_{4}-b_{3},b_{5}-b_{4}} are in a G.P


Option: 4

\mathrm{\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 \mathrm{-2}


Answers (1)

best_answer

\mathrm{b_{n}-b_{n-1} =\int_{0}^{\pi / 2} \frac{\cos ^{2} n x-\cos ^{2}(n-1) x}{\sin x}} \\

                    \mathrm{=\int_{0}^{\pi / 2} \frac{\sin (2 n-1) x \sin (-x)}{\sin x}} \\

                   \mathrm{=-\int_{\pi}^{\pi / 2} \sin (2 n-1) x d x }

                   \mathrm{=\frac{1}{2n-1}\left[ \cos \left ( 2n-1 \right )x \right]_{0}^{\frac{\pi}{2}}=-\frac{1}{2n-1}}\\

\mathrm{\frac{1}{b_{3}-b_{2}}=-5,\frac{1}{b_{4}-b_{3}}=-7,\frac{1}{b_{5}-b_{4}}=-9}\\

Common Difference \mathrm{=-2}

Hence the correct answer is option 4

Posted by

vishal kumar

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