Advertisements
Advertisements
प्रश्न
Evaluate:
`int_0^1 sqrt((1 - x)/(1 + x)) * dx`
मूल्यांकन
Advertisements
उत्तर
Let `I = int_0^1 sqrt((1 - x)/(1 + x)) * dx`
Put x = cos θ
dx = − sinθ dθ
When x = 0, cos θ = 0 = cos `pi/(2)` ∴ θ = `pi/(2)`
When x = 1, cos θ = 1 = cos 0 ∴ θ = 0
∴ `I = int_(pi/2)^0 sqrt(( - cos theta)/(1 + cos theta)) * (- sin θ) dθ`
= `int_(pi/2)^0 sqrt((2sin^2(theta//2))/(2cos^2(theta//2)))(- 2sin theta/2 cos theta/2) * dθ`
= `int_(pi/2)^0 (sin(theta//2)/(cos(theta//2)))[- 2sin (theta/2) cos (theta/2)] * dθ`
= `int_(pi/2)^0 - 2sin^2(theta/2) * dθ`
= `- int_(pi/2)^0 (1 - cos θ) * dθ`
= `-[theta - sintheta]_(pi/2)^0`
= `-[(0 - sin0) - (pi/2 - sin pi/2)]`
= `-[0 - pi/2 + 1]`
= `pi/(2) - 1`.
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
