Question

`int _(0)^(pi //2)((root(n)(secx))/(root(n)(secx)+root(n)("cosec"x)))dx=`
A. `pi/2`
B. `pi/3`
C. `pi/4`
D. `pi/6`

Answer 1

Correct Answer - C
Let `l=int_(0)^((pi)/(2))(root(n)(sec x))/(root(n)(sec x)+root(n)("cosec" x))dx` …(i)
`= int_(0)^((pi)/(2))(root(n)(sec((pi)/(2)-x)))/(root(n)(sec((pi)/(2)-x))+root(n)("cosec "((pi)/(2)-x)))dx`
`= int_(0)^((pi)/(2))(root(n)("cosec" x))/(root(n)("cosec" x)+root(n)(sec x))dx` ...(ii)
On adding Eq. (i) and (ii), we get
`2l+int_(0)^((pi)/(2))(root(n)(sec x)+root(n)("cosec " x))/(root(n)(sec x)+root(n)("cosec "x))dx`
`rArr 2l = int_(0)^((pi)/(2))dx=[x]_(0)^(pi//2)`
`rArr 2l=(pi)/(2) rArr l = (pi)/(4)`