How do we evaluate ∫1+x2(1−x2)1+x4−−−−−√dx? ∫ 1 + � 2 ( 1 − � 2 ) 1 + � 4 d � ?
𝙰𝙽𝚂𝚆𝙴𝚁 = One possible way to evaluate the integral ∫(1+x^2(1−x^2)1+x^4)^(-1/2)dx is by using trigonometric substitution. Let x = tanθ, so that dx = sec^2θdθ and 1 + x^2 = 1 + tan^2θ = sec^2θ 1 - x^2 = 1 - tan^2θ = cos^2θ 1 + x^4 = 1 + tan^4θ = sec^4θ Substituting these expressions in the integral yields: ∫(1+x^2(1−x^2)1+x^4)^(-1/2)dx = ∫(sec^2θcos^2θsec^4θ)^(-1/2)sec^2θdθ = ∫(cos^2θ)^(1/2)(sec^4θ)^(1/2)sec^2θdθ = ∫cosθsec^3θdθ We can evaluate this integral using integration by substitution: Let u = secθ, so that du = secθtanθdθ Substituting u and du, we have: ∫cosθsec^3θdθ = ∫du/u^3 = -1/(2u^2) + C Substituting back u = secθ, we get: ∫cosθsec^3θdθ = -1/(2sec^2θ) + C = -1/(2+2x^2(1-x^2)^(-1/2)) + C = -1/(2+2tan^2θsecθ) + C = -1/(2+x√(1-x^2)) + C Therefore, the solution to the integral is: ∫(1+x^2(1−x^2)1+x^4)^(-1/2)dx = -1/(2+x√(1-x^2)) + C.
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