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Question : 11 of 54
Marks:
+1,
-0
Solution:
‌∫−14√‌‌dx‌‌ Let ‌x=‌sin‌θ+‌‌⇒dx=‌‌cos‌θ‌d‌θ‌⇒4−x=4−(‌sin‌θ+‌)‌=‌(1−sin‌θ)‌⇒x+1=‌sin‌θ+‌+1=‌(1+sin‌θ)‌⇒‌‌‌=‌‌⇒‌‌√‌=√‌=‌‌∫√‌‌dx=∫‌‌‌cos‌θ‌d‌θ‌=‌‌∫(1−sin‌θ)dθ‌=‌(θ+cos‌θ)+c‌x=‌sin‌θ+‌⇒sin‌θ=‌‌⇒cos‌θ=√1−sin‌2θ=‌√4+3x−x2‌∫−14√‌‌dx=[‌sin‌−1(‌)+√4+3x−x2]−14‌‌=‌⋅‌−‌(‌)‌‌=‌+‌=‌
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