If I_{n} = int_{frac{pi}{4}}^{frac{pi}{2}} co | vedclass

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(D) Given $I_{n} = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot^{n} x \, dx = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot^{n-2} x (\csc^{2} x - 1) \, dx$$=

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$\frac{1}{I_{2}+I_{4}}, \frac{1}{I_{3}+I_{5}}, \frac{1}{I_{4}+I_{6}}$ are in $A.P.$

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(D) Given $I_{n} = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot^{n} x \, dx = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot^{n-2} x (\csc^{2} x - 1) \, dx$$= \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot^{n-2} x \csc^{2} x \, dx - \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot^{n-2} x \, dx$$= \left[ -\frac{\cot^{n-1} x}{n-1} \right]_{\frac{\pi}{4}}^{\frac{\pi}{2}} - I_{n-2}$$= \left( 0 - (-\frac{1}{n-1}) \right) - I_{n-2} = \frac{1}{n-1} - I_{n-2}$$\Rightarrow I_{n} + I_{n-2} = \frac{1}{n-1}$For $n=4$,$I_{4} + I_{2} = \frac{1}{3}$For $n=5$,$I_{5} + I_{3} = \frac{1}{4}$For $n=6$,$I_{6} + I_{4} = \frac{1}{5}$Thus,the terms are $\frac{1}{I_{2}+I_{4}} = 3$,$\frac{1}{I_{3}+I_{5}} = 4$,and $\frac{1}{I_{4}+I_{6}} = 5$.Since $3, 4, 5$ are in $A.P.$,the sequence $\frac{1}{I_{2}+I_{4}}, \frac{1}{I_{3}+I_{5}}, \frac{1}{I_{4}+I_{6}}$ is in $A.P.$

If $I_{n} = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot^{n} x \, dx$,then:

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