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Question : 74 of 160
Marks:
+1,
-0
Solution:
‌I=∫‌‌‌| cos‌x−sin‌x |
| sin‌2x |
‌dx‌I=∫√2‌+‌‌Put
sin‌x+cos‌x=t(cos‌x−sin‌x)‌dx=dt1+sin‌2x=t2 ‌=[‌‌ln(‌)]√2‌‌=‌[ln(‌)−ln(‌)]‌=‌[ln‌−ln(‌)]‌=‌[ln‌−ln(3−2√2)]‌=‌[ln‌+ln(3+2√2)]‌=‌[ln(‌)+ln(3+2√2)]‌=‌‌ln[‌]
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