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Question

0π/211+cot x dx

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Solution

Let I=0π211+cotxdx ... (i)= 0π211+cotπ2-xdx Using 0afxdx=0afa-xdx=0π211+tanxdx ... (ii) Adding (i) and (ii)2I=0π211+cotx+11+tanx dx =0π21+tanx+ 1+cotx1+cotx1+tanx dx =0π22+tanx+ cotx1+tanx +cotx + tanx cotxdx =0π22+tanx+ cotx2+tanx+ cotx dx =0π2dx = x0π2=π2Hence , I=π4

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