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Question : 96 of 120
Marks:
+1,
-0
Solution:
Let
I=‌‌dx ...(i)
I=| (+‌−x) |
| 1+sin(‌+‌−x) |
=‌‌ ....(ii)
[∵f(x)dx=f(a+b−x)dx] By adding Eqs. (i) and (ii), we get
2I=‌‌ ⇒‌‌2I=π‌‌‌dx ⇒‌‌2I=π[tan‌x−sec‌x]π/43π/4 ⇒‌‌2I=π[−1−(−√2)−(1−√2)]
⇒‌‌2J=π[−1+√2−1+√2] ⇒‌‌2I=π[−2+2√2] ∴‌‌I=π[√2−1] =‌(√2+1) =‌
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