Evaluate using trigonometric substitution

Question

∫(dx)/(((x^2)+9)^3)

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Answer 1

We let x = 3 tan u, so dx = 3 sec^2 u

Then, ∫3 sec^2 u/(9tan^2u+9)^3 =

∫1/243 sec^-4udu = ∫ 1/243 cos^4 u du

cos^4 u = ((cos 2 u + 1)/2)^2 = 1/4 cos^2 2u + 1/2 cos 2u + 1/4 = 1/4 (cos 4u + 1)/2 + 1/2 cos 2u + 1/4 =

1/8 cos 4u + 1/2cos 2u + 3/8

Then, ∫1/243 cos^4u du = 1/243 ∫1/8 cos 4u + 1/2cos 2u + 3/8 du =

1/243 (1/32 sin 4u + 1/4 sin 2u + 3/8 u + C)

As x = 3 tan u, u = arctan(x/3)

Then, you may write the integral as

1/243 (1/32 sin(4 arctan(x/3)) + 1/4 sin (2 arctan(x/3)) + 3/8 arctan(x/3) + C)