Is this a valid method in Trig Substitution to skip steps?

Asked May 08 '22 at 10:35

Active May 08 '22 at 10:35

Viewed 27 times

I've been practicing for a test tomorrow, and with the past few Trig Sub questions I have done involving $\sqrt{x^2 + a^2}$, $sec(u)$ always equals the square root in the expression.

Is this true for all values of $a$, or just $a = 1$? And if this is true, can I use this to skip the step of drawing the diagram?

asked May 08 '22 at 10:35

Andog

1

In your opinion, why does $x=\sec u$ work? – José Carlos Santos May 08 '22 at 10:43
@JoséCarlosSantos I don't understand your question. Nowhere does $x = secu$. However $x = tanu$ because of the trig identity $tan^2u + 1 = sec^2u$. – Andog May 08 '22 at 11:00
Sorry; what I meant to ask was if you knew why you get $\sec u$. And indeed $\tan^2u+1=\sec^2u$. But, if $a\ne1$, you don't have $\tan^2u+a=\sec^2u$, right?! – José Carlos Santos May 08 '22 at 11:02
Well, I know the identity extends to $tan^2u + a^2 = asec^2u$. – Andog May 08 '22 at 11:04
If $a=2$ and $u=\frac\pi4$, then $\tan^2u+a^2=5$, whereas $a\sec^2u=4$. So, that equality doesn't hold in general. – José Carlos Santos May 08 '22 at 11:08
Ah, I just realized I forgot the restrictions that $\frac{-pi}{2} \lt u \lt \frac{pi}{2}$, but I understand the example you gave. – Andog May 08 '22 at 11:16
Since $\tan^2 u + 1 = \sec^2 u$, $a^2\tan^2 u + a^2 = a^2\sec^2 u$ – Joe May 08 '22 at 11:38

Is this a valid method in Trig Substitution to skip steps?

0 Answers0