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B
ln|√2−1|
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C
ln|2−√3|
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D
ln∣∣∣1√2−1∣∣∣
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Solution
The correct option is Aln∣∣∣√2−12−√3∣∣∣ Let I=π4∫π6cosecxdx I=[ln|cosecx−cotx|]π4π6[∵∫cosecxdx=ln|cosecx−cotx|+c]=ln∣∣∣cosecπ4−cotπ4∣∣∣−ln∣∣∣cosecπ6−cotπ6∣∣∣=ln∣∣√2−1∣∣−ln∣∣2−√3∣∣=ln∣∣∣√2−12−√3∣∣∣