Does Apostol ever introduce Trig Substitutions ?

[process91]

Does Apostol ever introduce "Trig Substitutions"?

I took Calculus before, but I am going over Apostol's Calculus Vol. 1 book. In section 5.7 he introduces Integration by Substitution, but never really discusses what was commonly referred to as "Trig Substitutions" in my Calc classes. For instance, #16 in section 5.8 is

$\int(x^2+1)^{\frac{-3}{2}}dx$

Now I know (from my previous class) that I can solve this by letting $x=tan\theta$, but Apostol never introduced this notion. More importantly, to properly solve it in terms of $x$ I will need to use $arctan$, which isn't introduced until section 6. Just wondering if there's a way to solve it without using trig substitutions.

 

[Dustinsfl]

I took Calculus before, but I am going over Apostol's Calculus Vol. 1 book. In section 5.7 he introduces Integration by Substitution, but never really discusses what was commonly referred to as "Trig Substitutions" in my Calc classes. For instance, #16 in section 5.8 is

$\int(x^2+1)^{\frac{-3}{2}}dx$

Now I know (from my previous class) that I can solve this by letting $x=tan\theta$, but Apostol never introduced this notion. More importantly, to properly solve it in terms of $x$ I will need to use $arctan$, which isn't introduced until section 6. Just wondering if there's a way to solve it without using trig substitutions.

Trig substitutions tend to come after u-substitution. Have you looked in other chapters?

 

[process91]

Yes, section 5.7 is "u-substitution", and section 5.8 contains the exercises which pertain to it. He does several examples of the typical u-substitution methods, but then comes question 16. I skimmed section 6.21, where he introduces the inverse trig functions, but there's mostly integration of the inverse trig functions, not using them as substitution.

 

[process91]

Got it. He probably wants this solution at this stage:

Let $u=\frac{x}{\sqrt{x^2+1}}$, then $du=(x^2+1)^\frac{-3}{2}dx$
$\int(x^2+1)^\frac{-3}{2}dx=\int du = u + C = \frac{x}{\sqrt{x^2+1}} + C$

I suppose that's useful (to be able to recognize that setting u to that value will yield a desirable result), but I hope he does hit trig substitutions at some point.