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 - <p>2√3 + π<br><!-- [if !supportLineBreakNewLine]--><br><!--[endif]--></p>
Option 2 - <p>3√2 - π</p>
Option 3 - <p>2√3 - π<br><!-- [if !supportLineBreakNewLine]--><br><!--[endif]--></p>
Option 4 - <p>3√2 + π</p>
3 Views|Posted 4 months ago
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1 Answer
A
Answered by
4 months ago
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
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