मराठी

Evaluate :∫π/3 π/6 dx/(1+√cotx)

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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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2013-2014 (March) Delhi Set 1

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