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Question : 76 of 160
Marks:
+1,
-0
Solution:
Consider the expression,
I=e√| 1−sin‌2‌x |
| 1+sin‌2‌x |
dx =e√| cos‌x−sin‌x |
| cos‌x+sin‌x |
dx =e√tan2(−x)dx =e√tan2(−x)dx Further simplify the above,
e[−log(sec(−x))]0 −[−log|sec(−x)|] =e(0+log‌√2)−(−log‌√2+0) =e2‌log‌√2 =elog‌2 =2
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