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Question : 175 of 180
Marks:
+1,
-0
Solution:
‌‌Let ‌I=∫‌| cosec‌x |
| cos2(1+log‌tan‌) |
‌dx‌‌ Put ‌‌‌1+log‌tan‌=t‌⇒‌‌‌⋅ sec2‌⋅‌=dt‌⇒‌‌‌=dt⇒‌=dt‌⇒‌‌cosec‌x‌dx=dt‌‌ So, ‌I=∫‌=∫ sec2t‌dt=tan‌t+c‌=tan(1+log‌tan‌)+c‌
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