Advertisements
Advertisements
प्रश्न
Find:
`int1/x sqrt((x + a)/(x - a)) dx`
Advertisements
उत्तर
`int1/x sqrt((x + a)/(x - a)) dx`
`int1/x sqrt((x + a)/(x - a) xx (x - a)/(x - a)) dx`
`intsqrt((x^2 - a^2))/(x(x - a)) dx`
Let x = a sec θ
dx = a sec θ tan θ dθ
= `int(sqrt(a^2(sec^2 θ - 1)) xx a sec θ tan θ dθ)/(a sec θ xx a(sec θ- 1))`
= `int(a tan θ xx a sec θ tan θ dθ)/(a xx a sec θ(sec θ - 1))`
= `int (tan^2 θ dθ)/(sec θ - 1)`
= `int(sec^2 θ - 1^2)/(sec θ - 1) dθ`
= `int((sec θ - 1)(sec θ + 1))/(sec θ - 1) dθ`
= `int(sec θ + 1) dθ`
= log |sec θ + tan θ| + θ + c ...`[{:(∵ x = a sec θ"," θ = sec^-1 x/a),(sec θ = x/a"," tan θ = sqrt(x^2/a^2 - 1)):}]`
I = `log|x/a + sqrt(x^2/a^2 - 1)| + sec^-1 x/a + c_1`
= `log|(x + sqrt(x^2 - a^2))/a| + sec^-1 x/a + c_1`
= `log |x + sqrt(x^2 - a^2)| + sec^-1 x/a + c_1 - loga`
= `log|x + sqrt(x^2 - a^2)| + sec^-1 x/a + c`
Where C = C1 − log a
