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Question : 79 of 160
Marks:
+1,
-0
Solution:
I=√1+4sin2(x∕2)+4‌sin(x∕2)dx ∵‌‌(1+2‌sin‌)2=1+4sin2x∕2+4‌sin(x∕2) ∴‌‌I=(1+2‌sin‌x∕2)dx=[x−‌‌cos‌]0π ‌‌=[x−4‌cos(x∕2)]0π=[π−4.0−0+4]=π+4
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