If the value of the integral ∫(from 0 to 1/2) x²/(1-x²)³/² dx is k/6, then k is equal to:
If the value of the integral ∫(from 0 to 1/2) x²/(1-x²)³/² dx is k/6, then k is equal to:
Option 1 -
2√3 + π
Option 2 -
3√2 - π
Option 3 -
2√3 - π
Option 4 -
3√2 + π
-
1 Answer
-
Correct Option - 3
Detailed Solution:k/6 = ∫? ^ (π/6) (x²)/ (1-x²)³/² dx x = sinθ dx = cosθdθ
=> k/6 = ∫? ^ (π/6) (sin²θ)/ (1-sin²θ)³/² * cosθdθ
=> k/6 = ∫? ^ (π/6) (sin²θ)/ (cos³θ) * cosθdθ
=> k/6 = ∫? ^ (π/6) tan²θdθ = ∫? ^ (π/6) (sec²θ-1)dθ
=> k/6 = [tanθ - θ]? ^ (π/6) = (1/√3 - π/6) = (2√3-π)/6
=> k = 2√3 - π
Similar Questions for you
(put )
[Applying by parts]
Taking an Exam? Selecting a College?
Get authentic answers from experts, students and alumni that you won't find anywhere else
Sign Up on ShikshaOn Shiksha, get access to
- 65k Colleges
- 1.2k Exams
- 688k Reviews
- 1800k Answers