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Question : 76 of 160
Marks:
+1,
-0
Solution:
I‌=√(x+3)(2−x)‌dx‌⇒√−x2−x+6‌dx‌⇒√−(x2+x−6)‌dx‌⇒√‌−(x+‌)2‌dxPut
u=x+‌⇒du=dxWhen
x=0,u=‌ and
x=2u‌=2+‌=‌I‌=∫‌‌√(‌)2−u2du‌⇒[‌√‌−u2+‌sin‌−1(‌)]‌‌‌⇒[‌√‌−u2+‌sin‌−1(‌)]‌‌‌⇒{‌√‌−‌+‌sin‌−1(1)}−{‌√‌−‌+‌sin‌−1(‌)}‌⇒0+‌⋅‌−‌√6−‌sin‌−1(‌)‌⇒‌−‌−‌sin‌−1(‌)‌⇒‌(‌−sin‌−1(‌))−‌‌⇒‌cos−1(‌)−‌
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