the integral of sqrt (x^2+6x)dx, using trig substitution

Question

Please answer ASAP with step by step solution. Thank you so much!

Answer 1

I think John M. left out the steps you need. I think these are some of the steps you need.

He is right, first complete the square.

The integrand becomes sqrt[(x+3)² - 9] dx

Then u = = x+3

du=dx

And the integrand becomes sqrt(u² - 9) du

Now set u = 3sec θ

du = 3 sec θ tan θ dθ

and u² = 9 sec² θ

then the integrand becomes 9 tan²θ sec θ dθ.

This should be integrated by parts to get the final result.

Answer 2

sqrt (x^2+6x)dx

∫ √(x² + 6x)dx = ∫ √(x² +6x + 9 - 9)dx =

∫ √(x² +6x + 9 )dx - ∫ √(9)dx =

∫ (x+3)dx - ∫ 3dx = ∫x dx + ∫3dx -∫3dx

∫ xdx = x²/2

Answer 3

Hello Jenny,

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

x² + 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,

x² + 6x = (x² + 6x + 9) - 9

= (x + 3)² - 9

Hence,

∫√(x² + 6x) dx = ∫√[(x+3)² - 9] dx

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

∫√(x² + 6x) dx = ∫√(u² - 9) du

Finally, we can apply trig substitution. When the integrand contains u² - a², 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

∫√(x² + 6x) dx = ∫√(u² - 9) du

= ∫√[(3sec(θ))² - 9] 3sec(θ)tan(θ)dθ

= ∫√[9sec²(θ) - 9] 3sec(θ)tan(θ)dθ

= ∫√[9(sec²(θ)-1)] 3sec(θ)tan(θ)dθ

=9 ∫√[(sec²(θ) - 1] sec(θ)tan(θ)dθ

Recall the trig identity tan²(θ) + 1 = sec²(θ). From this, we have sec²(θ) - 1 = tan²(θ), and we use this result to replace the quantity under the square root in the previous integral.

∫√(x² + 6x) dx =9 ∫√[(sec²(θ) - 1] sec(θ)tan(θ)dθ

= 9 ∫√(tan²(θ)) sec(θ)tan(θ)dθ

=9 ∫tan(θ) sec(θ)tan(θ) dθ

= 9 ∫sec(θ)tan²(θ) dθ

= 9∫(sec³(θ) - sec(θ)) dθ

(where we used sec²(θ) - 1 = tan²(θ) again.)

= 9 ∫sec³(θ)dθ - 9∫ sec(θ)dθ

∫√(x² + 6x) dx= 9 ∫sec³(θ)dθ - 9 Ln|sec(θ)+tan(θ)| (Eq. 1)

It remains to evaluate ∫sec³(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 = sec²(θ) dθ. Please let me know if you would like to see all the details of the integration by parts.] The result is

∫sec³(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

∫√(x² + 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) = [√(u² - 9)]/3. In terms of u, the integral in Eq. 2 becomes

∫√(x^2 + 6x) dx = (1/2)u√(u² - 9) + (9/2)Ln|u/3 + √(u² - 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 + √(x² + 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