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Question : 29 of 150
Marks:
+1,
-0
Solution:
Let
I=∫ex()dx =∫ex| (1+2‌sin‌‌cos‌) |
| 2cos2‌ |
dx =∫ex‌sec2‌dx+∫ex‌tan‌‌d‌x =[2ex‌tan‌−∫2ex‌tan‌‌d‌x+∫ex‌tan‌‌d‌x] =ex‌tan‌x‌2−∫ex‌tan‌‌d‌x+∫ex‌tan‌‌d‌x+c =ex‌tan‌+c
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