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Question : 76 of 160
Marks:
+1,
-0
Solution:
We have,
‌ =‌ =‌| (1+tan2) |
| 4+4tan2‌+10‌tan‌ |
dx =‌‌‌| sec2 |
| tan2‌+‌‌tan‌+1 |
dx let
tan‌=t ⇒‌‌‌‌sec2‌(‌)dx=dt ⇒sec2‌dx=2dt at
x=0,t=0 and
‌‌x=‌,t=1=‌−‌dt =‌‌‌ =‌‌‌=‌‌‌ =‌[log|‌|]01 =‌[log(‌)−log(‌)] =‌[log(‌)−log(‌)]=‌[log(‌4)] =‌‌log‌2
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