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Question : 64 of 160
Marks:
+1,
-0
Solution:
We have
‌[A‌log‌]∵‌ ⇒A‌log(‌)=∫‌‌dx.....(i)
Let
I=∫‌‌dxLet
x3=sin‌2t3x2‌dx=2sin‌t‌cos‌t I‌=‌‌∫‌| 1×sin‌t‌cos‌t |
| sin‌2t‌cos‌t |
‌dt‌=‌‌∫cosec‌t‌dt=‌‌log|cosec‌t−cot‌t|‌=‌‌log‌|‌|x3∣‌=‌‌log‌|‌|‌=‌‌log‌|‌| (1−√1−x3)2(1+√1−x3) |
| x3(1+√1−x3) |
|.‌=‌‌log‌|‌|. From Eq. (i), we get
A=‌ and
B=−1⇒AB=−‌
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