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