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Question : 78 of 160
Marks:
+1,
-0
Solution:
Let
I=‌‌ . . . (i)
I=‌‌=‌‌ . . . (ii)
From Eqs. (i) and (ii),
⇒2I=‌‌| 1−cos‌x+1+cos‌x |
| (1+cos‌x)(1−cos‌x) |
dx ⇒2I=2‌‌‌dx=‌cosec2xdx ⇒I=−[cot‌x]‌‌=−[cot(‌)−cot(‌)] ⇒I=−[−cot‌−cot‌]=2‌cot‌=2.1 I=2=2‌sin‌
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