Asked by mathemagician
Integrate : (x^3)dx/sqrt. of (3x^2 - 5)
Let x=sqrt(5/3)secant theta
dx=sqrt(5/3)secant theta tangent theta
Sqrt. (3x^2-5) = sqrt. (5/3) tangent theta
Let x=sqrt(5/3)secant theta
dx=sqrt(5/3)secant theta tangent theta
Sqrt. (3x^2-5) = sqrt. (5/3) tangent theta
Answers
Answered by
Steve
∫ x^3/√(3x^2-5) dx
x = √(5/3) secθ
3x^2-5 = 5tan^2θ
dx = √(5/3) secθ tanθ dθ
and the integral becomes
∫ √(5/3)^3 sec^3θ/(√5 tanθ) √(5/3) secθ tanθ dθ
= 5/9 √5 ∫sec^4θ dθ
That one's not so hard, eh?
x = √(5/3) secθ
3x^2-5 = 5tan^2θ
dx = √(5/3) secθ tanθ dθ
and the integral becomes
∫ √(5/3)^3 sec^3θ/(√5 tanθ) √(5/3) secθ tanθ dθ
= 5/9 √5 ∫sec^4θ dθ
That one's not so hard, eh?
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