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प्रश्न
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))` .
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उत्तर
`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 `
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