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Question : 113 of 150
Marks:
+1,
-0
Solution:
(cos‌x+|cos‌x|)dx =cos‌x‌d‌x+[|cos‌x|]dx =[sin‌x]02π+2‌|cos‌x|dx =sin‌2‌π+sin‌0+2[cos‌x‌d‌x −cos‌x‌d‌x] =0+0+2[cos‌x‌d‌x −cos‌x‌d‌x] =2[[sin‌x]0π∕2−[sin‌x]π∕2π] =2[(sin‌π∕2−sin‌0)−(sin‌π−sin‌π∕2)] =2[(1−0)−(0−1)] =2[1+1]=4
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