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Question : 32 of 160
Marks:
+1,
-0
Solution:
‌dx=I(say)
I=‌| √cos‌x |
| √sin‌x+√cos‌x |
dx...(i)
We have,
f(x)dx=f(a+b−x)dx a+b−x=‌−x I=‌| √cos(−x) |
| √sin(‌−x)+√cos(‌−x) |
dx I=‌| √sin‌x |
| √sin‌x+√cos‌x |
dx...(ii)
Eqs. (i)
+ (ii)
⇒2I =(‌| √cos‌x |
| √sin‌x+√cos‌x |
+‌| √sin‌x |
| √sin‌x+√cos‌x |
)dx =(‌| √sin‌x+√cos‌x |
| √sin‌x+√cos‌x |
)dx =1⋅dx=(x)π/6π/3=(‌−‌) 2I=‌⇒I=‌
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