मराठी

Evaluate: ∫π6π3dx1+tanx

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प्रश्न

Evaluate: `int_(pi/6)^(pi/3) (dx)/(1 + sqrt(tanx)`

बेरीज
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उत्तर

Let I = `int_(pi/6)^(pi/3) (dx)/(1 + sqrt(tanx)`

= `int_(pi/6)^(pi/3) sqrt(cosx)/(sqrt(sinx) + sqrt(cos x)) dx`  ......(i)

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

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

I = `int_(pi/6)^(pi/3) sqrt(sinx)/(sqrt(cosx) + sqrt(sinx)) dx`  ......(ii)

Adding (i) and (ii), we get

2I = `int_(pi/6)^(pi/3) sqrt(cosx)/(sqrt(sinx) + sqrt(cosx)) dx + int_(pi/6)^(pi/3) sqrt(sinx)/(sqrt(cosx) + sqrt(sinx)) dx`

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

= `[x]_(pi/6)^(pi/3)`

= `pi/3 - pi/6`

= `pi/6`

Hence, I = `int_(pi/6)^(pi/3) (dx)/(1 + sqrt(tanx)) = pi/12`

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