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Question : 77 of 160
Marks:
+1,
-0
Solution:
∫‌dx=∫‌| √cot‌x |
| 2‌sin‌x‌cos‌x |
dx =∫‌‌=‌‌∫(cot‌x)−‌cosec2xdx Let
cot‌x=t⇒−cosec2xdx=dt ‌‌=‌‌∫t−1∕2(−dt)=‌(‌)+C ‌‌=−t1∕2+C=−√t+C=−√cotx+C
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