Trigonometric Substitution Integration

Question

3.) ∫(dx)/((x)*(9+25x^2)^(1/2))

Answer 1

Let x = (3/5)tan z. Then z = tan^-1(5x/3), dx = 3/5sec²z dz and we have:

∫dx/(x(9+25x²)^1/2) = (3/5) ∫sec²z dz/[(3/5)tan z(9 + 9tan²z)^1/2]

= (1/3) ∫sec²z dz/(tan z sec z)

= (1/3) ∫sec z dz/(sin z sec z)

= (1/3) ∫ csc z dz = -(1/3) ln |csc z + cot z| + C

= -(1/3) ln |csc(tan^-1(5x/3)) + cot (tan^-1(5x/3))| + C

= -(1/3) ln|sqrt(25x² + 9)/(5x) + 3/(5x)| + C

Answer 2

In the denominator, factor out a 9 from inside the radical making it √9(1 + 25/9x²) then take the square root of the 9 to get 3√(1 + 25/9x²). Now make the substitution 5/3x = tan(θ) which makes 25/9x² = tan²(θ). It also means x = 3/5tan(θ) so that dx/dθ = 3/5sec²(θ) or dx = 3/5sec²(θ)dθ.

If you sub into the integral, the radical (now √1 + tan²(θ)) now becomes √sec²(θ) = sec(θ). Since there is also an "x" in the denominator that becomes 3/5tan(θ).

Cancelling and turning into sine and cosine simplifies this down to a csc(θ) integration which is available as a standard in a table of integrals [-ln(cot(θ) +csc(θ))]. You can then make your back substitutions to get back to the "x world". Remember that since your sub was tan(θ) = 5x/3 that you can set up a triangle where the opposite side is 5x and the adjacent side is 3 then use the Pythagorean Theorem to solve for the hypotenuse. Then, you can convert cot(θ) into a ratio of adjacent/opposite and csc(θ) as a ratio of hypotenuse/opposite to get your final answer.