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L e tI1=int(pi/6)^(pi/3)(sinx)/x dx ,I2=...

`L e tI_1=int_(pi/6)^(pi/3)(sinx)/x dx ,I_2=int_(pi/6)^(pi/3)sin(sinx)/(sinx)dx ,I_3=int_(pi/6)^(pi/3)sin(tanx)/(tanx)dx` Then arrange in the decreasing order in which values `I_1,I_2,I_3` lie.

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Verified by Experts

`f(x)=(sinx)/x` is a decreasing function and `(sinx)/xgt0` for all `x` in `(0,pi)`.
Since `sin lt xlt tanx`,
`(sin(sinx))/(sinx)gt(sinx)/xgt(sin(tanx))/(tanx)` for `(pi)/6lt xlt (pi)/3`
`:.I_(2)gtI_(1)gtI_(3)`
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