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Question : 97 of 120
Marks:
+1,
-0
Solution:
Let
I=‌‌| 2sin‌x |
| 2sin‌x+2cos‌x |
dx.(i)
⇒‌‌I=‌‌| 2sin(−x) |
| 2sin(‌−x)+2cos(‌−x) |
⇒‌‌I=‌‌| 2cos‌x |
| 2cos‌x+2sin‌x |
....(ii)
By adding Eqs. (i) and (ii), we get
2I=‌‌| 2sin‌x+2cos‌x |
| 2sin‌x+2cos‌x |
dx ⇒‌‌2I=‌1⋅dx=[x]0 ⇒‌‌2I=‌ ⇒‌‌I=‌
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