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Question : 78 of 160
Marks:
+1,
-0
Solution:
‌‌ Let ‌I=√tan‌x‌dx‌‌‌‌‌I‌=√cot‌x‌dx⇒2I‌=√2×‌| sin‌x+cos‌x |
| √1−(sin‌x−cos‌x)2 |
‌dx⇒I‌=‌‌‌| d(sin‌x−cos‌x) |
| √1−(sin‌x−cos‌x)2 |
‌=‌[sin‌−1(sin‌x−cos‌x)+C]0π∕2‌=‌(sin‌−1(1)−sin‌−1(−1))‌=‌×‌×2=‌
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