Hello Jenny,

To carry out the integral using trig substitution, we will need to complete the square on the expression

x^{2} + 6x first. To do this we will add (and subtract) 1/2 of the coefficient of x. That is, we will add and subtract

[1/2(6)]^{2} =[3]^{2} = 9.

Thus,

x^{2} + 6x = (x^{2} + 6x + 9) - 9

= (x + 3)^{2} - 9

Hence,

∫√(x^{2} + 6x) dx = ∫√[(x+3)^{2} - 9] dx

Now make the substitution u = x + 3. Then du = dx, so the above integral may be written

∫√(x^{2} + 6x) dx = ∫√(u^{2} - 9) du

Finally, we can apply trig substitution. When the integrand contains u^{2} - a^{2}, the appropriate trig substitution is u = asec(θ). Thus, we set

u = 3sec(θ),

and we obtain

du = 3sec(θ)tan(θ)dθ.

Substituting these results into the integral, we have

∫√(x^{2} + 6x) dx = ∫√(u^{2} - 9) du

= ∫√[(3sec(θ))^{2} - 9] 3sec(θ)tan(θ)dθ

= ∫√[9sec^{2}(θ) - 9] 3sec(θ)tan(θ)dθ

= ∫√[9(sec^{2}(θ)-1)] 3sec(θ)tan(θ)dθ

=9 ∫√[(sec^{2}(θ) - 1] sec(θ)tan(θ)dθ

Recall the trig identity tan^{2}(θ) + 1 = sec^{2}(θ). From this, we have sec^{2}(θ) - 1 = tan^{2}(θ), and we use this result to replace the quantity under the square root in the previous integral.

∫√(x^{2} + 6x) dx =9 ∫√[(sec^{2}(θ) - 1] sec(θ)tan(θ)dθ

= 9 ∫√(tan^{2}(θ)) sec(θ)tan(θ)dθ

=9 ∫tan(θ) sec(θ)tan(θ) dθ

= 9 ∫sec(θ)tan^{2}(θ) dθ

= 9∫(sec^{3}(θ) - sec(θ)) dθ

(where we used sec^{2}(θ) - 1 = tan^{2}(θ) again.)

= 9 ∫sec^{3}(θ)dθ - 9∫ sec(θ)dθ

∫√(x^{2} + 6x) dx= 9 ∫sec^{3}(θ)dθ - 9 Ln|sec(θ)+tan(θ)| (Eq. 1)

It remains to evaluate ∫sec^{3}(θ)dθ. This can be accomplished using integration by parts, and in the interest of space I will omit some of the details. [Hint: Use U = sec(θ) and dV = sec^{2}(θ) dθ. Please let me know if you would like to see all the details of the integration by parts.] The result is

∫sec^{3}(θ)dθ = (1/2)sec(θ)tan(θ) + (1/2)Ln|sec(θ)+tan(θ)| + C.

Substituting this result into the previous integral (Eq.1), and combining terms, we obtain

∫√(x^{2} + 6x) dx = (9/2)sec(θ)tan(θ) - (9/2)Ln|sec(θ)+tan(θ)| + C. (Eq. 2)

Now recalling the original trig substitution u = 3tan(θ), we have sec(θ) = u/3. Using a right-triangle diagram to represent the substitution (or trig identities), we also note that tan(θ) = [√(u^{2} - 9)]/3. In terms of u, the integral in Eq. 2 becomes

∫√(x^2 + 6x) dx = (1/2)u√(u^{2} - 9) + (9/2)Ln|u/3 + √(u^{2} - 9)/3| + C

Now use u = x + 3 and obtain

∫√(x^2 + 6x) dx = (1/2)(x+3)√((x+3)^{2} - 9) + (9/2)Ln|(x+3)/3 + √((x+3)^{2} - 9)/3| + C

With some simplification, this can be rewritten as

∫√(x^2 + 6x) dx = (1/2)(x+3)√(x+6x) + (9/2)Ln|x+3 + √(x^{2} + 6x)| + C

Note that in writing the above -(9/2)Ln(3) has been absorbed into the constant of integration C. Hope this helps! Please let me know if you would like any further explanation.

William

John M.

02/29/20