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Question : 42 of 89
Marks:
+1,
-0
Solution:
‌‌ Let ‌I=∫‌‌dx‌=∫‌| 1 |
| (sin‌x+cos‌x)(sin‌2x+cos2x−sin‌x‌cos‌x) |
‌dx‌=∫‌| sin‌x+cos‌x |
| (sin‌x+cos‌x)2(1−sin‌x‌cos‌x) |
‌dx‌=∫‌| sin‌x+cos‌x |
| (1+2sin‌x‌cos‌x)(1−sin‌x‌cos‌x) |
‌dx‌=∫‌Put
sin‌x−cos‌x=t(cos‌x+sin‌x)‌dx=dt‌1−2sin‌x‌cos‌x=t2‌2sin‌x‌cos‌x=1−t2 ‌=∫‌=−2‌∫‌‌=−‌‌∫(‌−‌)‌dt‌=−‌[‌‌ln(‌)−tan−1t]‌=−‌‌ln(‌| sin‌x−cos‌x−√2 |
| sin‌x−cos‌x+√2 |
)+‌tan−1(sin‌x−cos‌x)+C‌=‌‌ln‌|‌| sin‌x−cos+√2 |
| sin‌x−cos‌x−√2 |
|+‌tan−1(sin‌x−cos‌x)+C‌∵A=‌,B=‌‌ and ‌t=sin‌x−cos‌x‌∵(‌,t)=(2√2,sin‌x−cos‌x)
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