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प्रश्न
Evaluate :`int_(pi/6)^(pi/3) dx/(1+sqrtcotx)`
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उत्तर
Let `I = ∫_(π/3)^(π/6)1/(1 +sqrt(cotx)dx`
`=int_(pi/6)^(pi/3)1/(1+(sqrt(cosx)/sqrtsinx))dx`
`=int_(pi/6)^(pi/3)1/(sqrtsinx+(sqrt(sinx)+sqrtcosx))dx...(1)`
`now,I=int_(pi/6)^(pi/3)(sqrt(sin(pi/2-x)))/(sqrt(sin(pi/2-x))+sqrt(cos(pi/2-x)))dx .................(int_a^bf(X)dx=int_a^bf(b+a-x)dx)`
`=int_(pi/6)^(pi/3)(sqrt(cosx))/(sqrt(sinx)+sqrt(cosx))dx`
Adding (1) and (2), we get
`2I=int_(pi/6)^(pi/3)(sqrt(sinx)+sqrt(cosx))/(sqrt(sinx)+sqrt(cosx))dx`
`2I=int_(pi/6)^(pi/3)dx`
`2I=[x]_(pi/6)^(pi/3)`
`2I=pi/3-pi/6`
`2I=pi/6`
`I=pi/12`
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