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Question : 75 of 160
Marks:
+1,
-0
Solution:
I‌=‌‌dx‌=‌| 2sin‌‌cos‌ |
| 2cos2‌+2sin‌‌‌cos‌ |
‌dx‌=‌‌dx‌=‌‌‌| sin‌+cos‌+sin‌−cos‌ |
| cos‌+sin‌‌ |
‌dx‌=‌‌(1+‌| sin‌−cos‌ |
| cos‌+sin‌‌ |
)‌dx‌=‌[x+{−2‌ln(cos‌+sin‌‌)}]0‌‌=‌[x−2‌ln(cos‌+sin‌‌)]0‌‌=‌[(‌−2‌ln‌√2)−0]‌=‌−ln‌√2=‌−‌‌ln‌2
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