Question

Prove that `int _a^b f(x) dx = int_a^b f (a + b -x ) dx`  and hence evaluate   `int_(pi/6)^(pi/3) (dx)/(1 + sqrt(tan x))` .   

Answer 1

`int _a^b f(x) dx = int_a^b f (a + b -x ) dx`

Taking L.H.S 

`int _a^b f (x) dx `              ..... ( i )

Let t = a + b - x 

x = a + b - t 

`(dx)/(dt) = 0 + 0 - 1`

⇒ dx = - dt 

changing limits

at x = a  t = a + b - a = b 

x = b  t = a + b - b = a

so integral (i) becomes

` int _b^a f (a + b - t )(- dt )`

using `int_a^b f(x) dx = - int_b^a f (x) dx `

⇒ `int_a^b f ( a + b  -t ) dt`

changing variable

`int _a^b f ( a + b -x ) dx `

L.H.S = R.H.S
Hence proved. 

`I = int _(pi/6)^(pi/3) 1/(1 +sqrt(tan x ))  dx`

`I = int _(pi/6)^(pi/3) sqrt(cos x )/(sqrt(cos x ) + sin x )  dx`        ......( i )

using property

`I = int _(pi/6)^(pi/3) (sqrt(cos (pi/6 + pi/3 -x)))/(cos sqrt(pi/6 + pi/3 - x) +  sqrt(sin (pi/6 + pi/3 - x ))` dx

`I = int _(pi/6)^(pi/3) sqrt(sin x ) /(sqrt (sin x ) + sqrt (cos x) ) dx `         ....... ( ii ) 

Adding (i) & (ii) 

`2I = int _(pi/6)^(pi/3) sqrt(cos x ) /(sqrt(cos x ) + sqrt( sin x ) )  dx  + int_(pi/6)^(pi/3) sqrt( sin x) /( sqrt( sin x ) + sqrt( cos x ) ) dx `

`2I = int _(pi/6)^(pi/3) (sqrt(cos x ) + sqrt( sin x )) /( sqrt ( cos x ) + sqrt( sin x )) dx `

`2I = int _(pi/6)^(pi/3) dx`

`2I = int _(pi/6)^(pi/3) x`

`2I = pi / 3 -  pi / 6 `

`2I = pi /6 `

` I = pi / 12 `