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Question : 1 of 120
Marks:
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Solution:
Let
I=π∕2‌| √sin‌x |
| √sin‌x+√cos‌x |
dx.....(i)
[∵f(x)dx=f(a+b−x)dx] ⇒‌‌I=π/2‌| √cos‌x |
| √cos‌x+√sin‌x |
dx .....(ii)
On adding Eqs. (i) and (ii), we get
2I=π/2‌| √sin‌x+√cos‌x |
| √sin‌x+√cos‌x |
dx ⇒‌‌2I=π/21dx⇒2I=[x]0 ⇒‌‌2I=‌⇒I=‌
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