Question about integrals using trig sub

A

angelina135

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Hi,

I'm learning trig substitution right now, and there's something I'm a bit unclear about. If you could help me, I would really appreciate it.

Basically, if you had a radical like this:
[MATH]\sqrt{2x^2 + 9}[/MATH]Where the [MATH]x^2[/MATH] term has a constant coefficient, is there any reason you can't use the substitution[MATH] \sqrt{\text{coefficient}} * x = [/MATH] ... instead of dividing the expression inside the radical by the coefficient and using x = ...?
For example, with the expression above, is there any reason why I couldn't use [MATH]x\sqrt{2}=3tan(\theta)[/MATH] and substitute in [MATH]3tan(\theta)[/MATH] for [MATH]x\sqrt{2}[/MATH] instead of using [MATH]x=3tan(\theta)/\sqrt{2}[/MATH]?

Regardless of which substitution I use, the expression seems to simplify to [MATH]3*|sec(\theta)|[/MATH]. Am I right in thinking I can use either substitution in this sort of situation?

Thank you so much,
Angelina
 
angelina135 said:
Hi,

I'm learning trig substitution right now, and there's something I'm a bit unclear about. If you could help me, I would really appreciate it.

Basically, if you had a radical like this:
[MATH]\sqrt{2x^2 + 9}[/MATH]Where the [MATH]x^2[/MATH] term has a constant coefficient, is there any reason you can't use the substitution[MATH] \sqrt{\text{coefficient}} * x = [/MATH] ... instead of dividing the expression inside the radical by the coefficient and using x = ...?
For example, with the expression above, is there any reason why I couldn't use [MATH]x\sqrt{2}=3tan(\theta)[/MATH] and substitute in [MATH]3tan(\theta)[/MATH] for [MATH]x\sqrt{2}[/MATH] instead of using [MATH]x=3tan(\theta)/\sqrt{2}[/MATH]?

Regardless of which substitution I use, the expression seems to simplify to [MATH]3*|sec(\theta)|[/MATH]. Am I right in thinking I can use either substitution in this sort of situation?

Thank you so much,
Angelina
Click to expand...
as lex said

[MATH]\sqrt{2}x = 3\tan \theta[/MATH]
is the same as

[MATH]x = \frac{3\tan \theta}{\sqrt{2}}[/MATH]
 
angelina135 said:
Hi,

I'm learning trig substitution right now, and there's something I'm a bit unclear about. If you could help me, I would really appreciate it.

Basically, if you had a radical like this:
[MATH]\sqrt{2x^2 + 9}[/MATH]Where the [MATH]x^2[/MATH] term has a constant coefficient, is there any reason you can't use the substitution[MATH] \sqrt{\text{coefficient}} * x = [/MATH] ... instead of dividing the expression inside the radical by the coefficient and using x = ...?
For example, with the expression above, is there any reason why I couldn't use [MATH]x\sqrt{2}=3tan(\theta)[/MATH] and substitute in [MATH]3tan(\theta)[/MATH] for [MATH]x\sqrt{2}[/MATH] instead of using [MATH]x=3tan(\theta)/\sqrt{2}[/MATH]?

Regardless of which substitution I use, the expression seems to simplify to [MATH]3*|sec(\theta)|[/MATH]. Am I right in thinking I can use either substitution in this sort of situation?

Thank you so much,
Angelina
Click to expand...
do not forget to apply substitution for dx also!
 
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