Integration of tan inverse (3x-x^3/1-3x^2)
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/4)(4x-2)/(2x^2–2x+3).dx + integ(5/2)/[2{x^2-x+(3/2)} dx =(3/4)... ... +5/4 . tan inverse of (x-1/2)/(√5/2). +c.
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Hey dear,
● Answer -
3/(1+x^2)
● Explanation -
We have
y = taninv[(3x-x^3)/(1-3x^2)]
Substituting x = tanθ, θ = taninv(x),
y = taninv[(3tanθ-tan^3θ)/(1-3tan^2θ)]
y = taninv[tan(3θ)]
y = 3θ
y = 3taninv(x)
Taking differention on both sides,
dy/dx = d[3taninv(x)]/dx
dy/dx = 3 × 1/(1+x^2)
dy/dx = 3/(1+x^2)
Hope this is helpful...
● Answer -
3/(1+x^2)
● Explanation -
We have
y = taninv[(3x-x^3)/(1-3x^2)]
Substituting x = tanθ, θ = taninv(x),
y = taninv[(3tanθ-tan^3θ)/(1-3tan^2θ)]
y = taninv[tan(3θ)]
y = 3θ
y = 3taninv(x)
Taking differention on both sides,
dy/dx = d[3taninv(x)]/dx
dy/dx = 3 × 1/(1+x^2)
dy/dx = 3/(1+x^2)
Hope this is helpful...
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