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Question : 91 of 160
Marks:
+1,
-0
Solution:
Let
I=∫(√+√)dx Put
x=a‌cos‌2‌θ⇒dx=−2a‌sin‌2‌θ‌d‌θ ∴=−∫(√| a+a‌cos‌2‌θ |
| a−a‌cos‌2‌θ |
+√| a−a‌cos‌2‌θ |
| a+a‌cos‌2‌θ |
) ×2a‌sin‌2‌θ‌d‌θ =−∫(√+√) 2a‌sin‌2‌θ‌d‌θ =−2a‌∫(+)‌sin‌2‌θ‌d‌θ =−2a‌∫‌sin‌2‌θ‌d‌θ =−4a‌∫dθ=−4aθ+C1 =−2acos−1()+C1 =−2a[−sin−1()]+C1 =2asin−1()+C1−πa =2asin−1()+C, where
C=C1−πa
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