the integral of sqrt (x^2+8x+6)dx, using trig substitution

Answer 1

If you do the completing of the square, as shown above, you get

∫√(x+4)²-10 dx

If we let u = x+4, then we get du = dx and the integral becomes:

∫√u² - 10 du

If we let u = √10 sec t , then du = √10 tan t sec t dt and this integral becomes

∫ √ 10√(sec² t - 10) * √10 sec t tan t dt

This eventually becomes √10 ∫√(sec²x - 1) √10 tan t sec t dt = 10 ∫ tan t tan t sec t dt

This simplifies to 10∫ tan²t sec t dt

(Trig Identity: tan²t = sec²t - 1)

10∫(sec² t- 1) sec t dt

10∫ sec³t - ∫sec t dt

The first integral is rather tricky and involves integration by parts. I won't go into details since it's a long and gory proof, but if you want to prove it yourself just remember that ∫sec³t dt = ∫sec t sec²t dt. If you choose your u and dv appropriately, set ∫sec³t dt = I, you should get that the first integral is 1/2(sec t tan t + ln | sec t + tan t| ). This implies that:

10(∫ sec³t dt - ∫sec t dt)

= 5 sec t tan t + ln | sec t + tan t | - 5 ln | sec t + tan t|

= 5 sec t tan t - 5 (ln | sec t + tan t |).

Since u = √10 sec t, we get that t = arcsec(u/√10) = arcsec( x+4 /√10):

Substitute that in for t, remembering that sec(arcsec(a)) = a and tan (arcsec(a)) = √(a² - 1)) and after messing with the algebra for a while, you should get that the integral is

(((x+4)/2)*√(x²+8x+6)) - 5 ln |(x+4/√10) + √(1/10)*(x²+8x+6)| + C

Answer 2

The bad news for this problem is that you have to complete the square for the term inside the square root. The reason for this is so that you can get something of the form of either:

1. √(u^2 + a^2) in which case, let tan∅ = u/a
2. √(u^2 - a^2) in which case, let sec∅ = u/a

So, for the inside term

1. x² + 8x + [...] + 6 - [...],
2. what you need to do is to figure out what goes into [...] (hint: (b/2)²)
3. then it will factor out to: (x + ....)² + 6 - [...]

The good news is that once you have completed the square and made the substitution, the rest will be pretty straightforward.

Answer 3

Upon completing of the square :

sqrt( (x+4)^2 - 10)

= sqrt( u^-10) where U = x+4 --> dU = dx

You can then do trig substitution u = sqrt(10) * sec T

so the whole thing will collapse to sqrt(10) tan T