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Question : 162 of 180
Marks:
+1,
-0
Solution:
We have,
‌I=‌| cos‌x |
| (cos‌+sin‌‌)3 |
‌dx=‌| cos2−sin‌2 |
| (cos‌+sin‌‌)3 |
‌dx‌‌=‌| cos‌−sin‌ |
| (cos‌+sin‌‌)2 |
‌dxPut
cos‌+sin‌‌=t ⇒‌(−sin‌‌+cos‌)‌dx=dtAlso, when
x=0⇒t=1When
x=‌⇒t=√2∴‌‌I‌=‌‌=2[−‌]1√2=2[−‌+1]=(2−√2)
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