Trigonometric substitution

Question

Evaluate the integral:

Integral of (-3x^3) / (sqrt((x^2) + 9) dx, x = 3tan(theta)

Answer 1

One of the key parts to this is the set-up:

Then if we let x = 3tan(θ), it means dx/dθ = 3sec²(θ) or dx = 3sec²(θ) dθ

We can also say that x³ = 27tan³(θ)

Then when you do the substitutions, the integral becomes:

Since tan²(θ) + 1 = sec²(θ) by the Pythagorean Identity we substitute sec2(θ) inside the square root and then, taking the square root, the denominator becomes "sec(θ)" which cancels with the numerator and becomes:

-81∫tan³(θ)sec(θ) dθ

We can now split tan³(θ) into tan²(θ)•tan(θ) and use the variation of of our Pythagorean Identity tan²(θ) = sec²(θ) - 1 as follows:

-81∫tan²(θ)tan(θ)sec(θ) dθ

-81∫(sec²(θ) - 1)tan(θ)sec(θ) dθ

Now, make a second substitution. Let w = sec(θ) which means dw/dθ = sec(θ)tan(θ) or dw = sec(θ)tan(θ) dθ turning the integral into:

-81∫(w² - 1) dw

Integrating, we get -81[1/3w³ - w] + C

Back substituting, w = sec(θ) we get:

-81[1/3sec³(θ) - sec(θ)] + C

In our original trig substitution we had x = 3tan(θ) or tan(θ) = x/3. This means that a triangle could be constructed with opposite side "x" and adjacent side "3". Using the Pythagorean Theorem the hypotenuse is calculated as √(x² + 3²) = √(x² + 9). This means sec(θ) = √(x² + 9)/3 which you can back substitute to get your final answer. Simplify as desired.

Answer 2

x = 3tanu; dx = 3sec²udu;

I = ∫(-3x³)/√(x² + 9)dx = ∫(-3·27tan³u·3sec²du/(3secu) = -81∫tan³usecudu = -81∫sin³u/cos⁴u = 81∫(1 - cos²u)d(cosu)/cos⁴u = 81∫(1/cos⁴u - 1/cos²u)d(cosu) = 81( - 1/(3cos³u) + 1/cosu) + C.

cosu = 3/√9 + x²

Answer 3

The suggested substitution gets you tan³θ sec θ as the integrand.

The tan 1/2 angle substitution doesn't help.

Wolfram Alpha gives a very complicated answer.

Are you certain that you submitted the problem correctly?