If `int_(0)^(pi//2)(cotx)/(cotx+"cosec "x)dx=m(pi+n)`, them m * n is equal to

Options

A. `-(1)/(2)`
B. 1
C. `(1)/(2)`
D. `-1`

Answer

D

Explanation

Let I `=int_(0)^(pi//2)(cotx)/(cotx+"cosec "x)dx`
`=int_(0)^(pi//2)(cosx/(sinx))/(cosx/(sinx)+(1)/(sinx))dx = int_(0)^(pi//2)(cosx)/(1+cosx)dx`
`=int_(0)^(pi//2)(cosx(1-cosx))/(1-cos^(2)x)dx`
`=int_(0)^(pi//2)(cosx-cos^(2)x)/(sin^(2)x)dx`
`=int_(0)^(pi//2)("cosec "xcotx -cot^(2)x)dx`
`=int_(0)^(pi//2)("cosec" x cot x- " cosec"^(2)x+1) dx`
`=[-"cosec"x+cotx+x]_(0)^(pi//2)`
`=[x+(cosx-1)/(sinx)]_(0)^(pi//2)=[x+((-2"sin"^(2)x/(2)))/(2 "sin"(x)/(2)"cos"(x)/(2))]_(0)^(pi//2)`
`[x-"tan"(x)/(2)]_(0)^(pi//2)=(pi)/(2)-1=(1)/(2)[pi-2]`
`= m [pi +n]` [ given]
On comparing we get m `=(1)/(2)andn=-2`
`:. m*n=-1`
Alternate Solution
Let `I=int_(0)^(pi//2)(cotx)/(cotx+"cosec "x)dx`
`=int_(0)^(pi//2)(cosx/(sinx))/(cosx/(sinx)+(1)/(sinx))`
`int_(0)^(pi//2)("cos"^(2)(x)/(2)-1)/(2 "cos"^(2)(x)/(2))dx `
`[:' costheta=2"cos"^(2)(theta)/(2)-1andcostheta+1=2"cos"^(2)(theta)/(2)]`
`=int_(0)^(pi//2)(1-(1)/(2)"sec"^(2)(x)/(2))dx `
`=[x-"tan"(x)/(2)]_(0)^(pi//2)=(pi
)/(2)-1=(1)/(2)(pi-2)`
Since , I `=m(pi-n)`
`:.(pi-n)=(1)/(2)(pi-2)`
On comparing both sides , we get
`"m"(1)/(2)andn=-2`
Now , mn `=(1)/(2)xx-2=-1`