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If I_n = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot ^n x dx, then
Option: 1 \frac{1}{I_2 +I_4}, \frac{1}{I_3+I_5}, \frac{1}{I_4, I_6} are in G.P
Option: 2 I_2 +I_4 , I_3 +I_5 , I_4 +I_6 are in A.P
Option: 3 I_2 +I_4, (I_3 +I_5)^2 , I_4 +I_6 are in G.P
Option: 4 \frac{1}{I_2+I_4}, \frac{1}{I_3 +I_5}, \frac{1}{I_4 +I_6} are in A.P

Answers (1)

best_answer

Given

\mathrm{I}_{\mathrm{n}}=\int_{\pi / 4}^{\pi / 2} \cot ^{\mathrm{n}} \mathrm{x} \mathrm{d} \mathrm{x}

Let us first solve

\begin{aligned} \\I_n&=\int\cot^nxdx=\int\cot^{n-2}x\cot^2xdx \\&=\int\cot^{n-2}x(\csc^2x-1)dx \\&=\int\cot^{n-2}x\csc^2xdx-\int\cot^{n-2}xdx \\&=\int t^{n-2}dt-I_{n-2} \end{aligned}

\\\text{where, t = }\cot x

I_n=-\frac{\cot^{n-1}x}{n-1}-I_{n-2}

Or

\\I_n\left.=-\frac{\cot ^{\mathrm{n}-1} \mathrm{x}}{\mathrm{n}-1}\right]_{\pi / 4}^{\pi / 2}-\mathrm{I}_{\mathrm{n}-2} \\\mathrm{\;\;\;\;} =\frac{1}{\mathrm{n}-1}-\mathrm{I}_{\mathrm{n}-2}

\\\Rightarrow \mathrm{I}_{\mathrm{n}}+\mathrm{I}_{\mathrm{n}-2}=\frac{1}{\mathrm{n}-1}\\ \Rightarrow \mathrm{I}_{2}+\mathrm{I}_{4}=\frac{1}{3}\\ \mathrm{I}_{3}+\mathrm{I}_{5}=\frac{1}{4}\\ \mathrm{I}_{4}+\mathrm{I}_{6}=\frac{1}{5}\\ \therefore \frac{1}{\mathrm{I}_{2}+\mathrm{I}_{4}}, \frac{1}{\mathrm{I}_{3}+\mathrm{I}_{5}}, \frac{1}{\mathrm{I}_{4}+\mathrm{I}_{6}} \text { are in A.P. }

Posted by

himanshu.meshram

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