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Question : 77 of 160
Marks:
+1,
-0
Solution:
Let,
I=| cos‌x‌d‌x |
| √1+cos‌x‌sin‌x |
Use property
f(x)dx=f(a−x)dx I=| cos(−x) |
| 1+cos(−x)‌sin(−x) |
dx =| cos‌x |
| 1+sin‌x‌cos‌x |
dx Then,
2I=| cos‌x+sin‌x |
| √1+cos‌x‌sin‌x |
×dx =| √2(cos‌x+sin‌x) |
| √2+2‌cos‌x‌sin‌x |
dx =| √2(cos‌x+sin‌x) |
| √3−(sin‌x−cos‌x)2 |
dx Put
sin‌x−cos‌x=t ⇒(cos‌x+sin‌x)dx=dt 2I=dt =2‌dt I=√2[sin−1()]−11 =√2sin−1 ()
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