let
x = 2sinhθ
x^2+4 = 4sinh^2θ+4 = 4cosh^2θ
dx = 2coshθ dθ
Now the integral becomes
∫x^3 / (x^2 + 4)^2 dx
∫8sinh^3θ/16cosh^4θ 2coshθ dθ
∫tanh^3θ dθ
see what you can do with that.
using direct substitution, let
u = x^2+4
du = 2x dx
∫x^3 / (x^2 + 4)^2 dx
(1/2)∫x^2 / (x^2 + 4)^2 2x dx
(1/2)∫(x^2+4)/(x^2 + 4)^2 - 4/(x^2+4)^2 2x dx
(1/2)∫u/u^2 - 4/u^2 du
(1/2)∫1/u - 4/u^2 du
= 1/2(lnu + 4/u)
= 1/2(ln(x^2+4) + 4/(x^2+4))
(i) Evaluate integral [ x^3 / (x^2 + 4)^2 ] using trigonometric substitution.
(ii) Evaluate integral [ x^3 / (x^2 + 4)^2 ] using regular substitution.
(iii) Use a right triangle to check that indeed both answers you obtained in
parts (i) and (ii) are the same.
Thanks!
1 answer
1 answer