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Question : 149 of 180
Marks:
+1,
-0
Solution:
We have,
I=∫0dt =∫0| 2‌sin‌t‌cos‌t |
| sin4t+cos4t |
dt =∫0| 2‌tan‌t‌sec2‌t |
| (tan2t)2+1 |
dt Put
tan2t=x (2‌tan‌t‌sec2‌t)dt=dx ∴I=∫0−dx=[tan−1x]0−=
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