Question

`int_0^(π/2)((root(n)(secx))/(root(n)(secx + root(n)("cosec"  x))))dx` is equal to ______.

Options

• `π/2`

• `π/3`

• `π/4`

• `π/6`

Answer 1

`int_0^(π/2)((root(n)(secx))/(root(n)(secx + root(n)("cosec"  x))))dx` is equal to `underlinebb(π/4)`.

Explanation:

Let I = `int_0^(π/2) (root(n)(sec x))/(root(n)(secx) + root(n)("cosec"  x))dx`  ...(i)

= `int_0^(π/2) (root(n)sec(π/2 - x))/(root(n)(sec(π/2 - x)) + root(n)("cosec"  (π/2 - x)))dx`

= `int_0^(π/2) root(n)("cosec"  x)/(root(n)("cosec"  x) + root(n)(sec x))dx`  ...(ii)

After adding equation (i) and (ii), we get

2I = `int_0^(π/2) (root(n)(sec x) + root(n)("cosec"  x))/(root(n)(sec x) + root(n)("cosec"  x))dx`

`\implies` 2I = `int_0^(π/2) dx = [x]_0^(π/2)`

`\implies` 2I = `π/2`

Hence, I = `π/4`