Question:
If $\mathrm{I}_{\mathrm{n}}=\int_{\pi / 4}^{\pi / 2} \cot ^{\mathrm{n}} \mathrm{xdx}$, then:
Correct Option: , 2
Solution:
$I_{n+2}+I_{n}=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot ^{n} x \cdot \cos e c^{2} x d x=\left[\frac{-(\cot x)^{n+1}}{n+1}\right]_{\frac{\pi}{4}}^{\frac{\pi}{2}}$
$I_{n+2}+I_{n}=\frac{1}{n+1}$
$I_{2}+I_{4}=\frac{1}{3}, I_{3}+I_{5}=\frac{1}{4}, I_{4}+I_{6}=\frac{1}{5}$
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