Integration by Trig Substitution

[MathHawk]

I've attempted solutions at this in two different manners and found myself stuck both ways. I'll show you the way that seems to make progress. The other way involves not factoring the junk under the radical in the denominator.

Homework Statement

\int^{1}_{0} \frac{3x^2 -1}{\sqrt{x-x^3}} dx

Homework Equations

For case \sqrt{a^2-x^2}, use substitution a*sin\theta.

The Attempt at a Solution

\int^{1}_{0} \frac{3x^2 -1}{\sqrt{x-x^3}} dx

= \int^{1}_{0} \frac{3x^2 -1}{\sqrt{1-x^2}\sqrt{x}} dx

Substitute: x = sin\theta, dx = cos\theta d\theta, bounds of 0 - 1 remain the same as sin 0 = 0 and sin 1 = 1.

\int^{1}_{0} \frac{3sin^2\theta -1}{\sqrt{cos^2\theta}\sqrt{sin\theta}} cos\theta d\theta


= \int^{1}_{0} \frac{3sin^2\theta -1}{cos\theta\sqrt{sin\theta}} d\theta



At this point I no longer know where to go. Factoring the numerator appeared to bear no fruit, no u substitutions make any sense to me at the moment.

 

[Mark44]

Trig substitution is the wrong way to go. A simple substitution will work here: u = x - x³. Note that the numerator is very close to du.

 

[MathHawk]

Oh my gosh, thank you. Made that way too hard on myself ;). I even thought about substituting using the 1-sin^2x inside the radical (instead of changing it to cos^2x) and it never occurred to me to look at the original question. Your help is very appreciated.

 

[MathHawk]

This changes my bounds to 0 and 0. Can I stop here and claim that the answer is 0?

 

[rock.freak667]

This changes my bounds to 0 and 0. Can I stop here and claim that the answer is 0?

Yes you can

EDIT: Actually, on second thought for

\frac{3x^2-1}{\sqrt{x-x^3}}=0 \Rightarrow x= \pm \frac{1}{\sqrt{3}}


Meaning that between 1 and 0 the curve passes through the x-axis, you would probably need to change from 1 to 0 to

\int ^{1} _{\frac{1}{\sqrt{3}}} \frac{3x^2-1}{\sqrt{x-x^3}} dx + \left | \int ^{\frac{1}{\sqrt{3}}} _{0} \frac{3x^2-1}{\sqrt{x-x^3}}dx \right |

 

[Mark44]

You can change the limits of integration as you were thinking, or you can leave them unchanged, integrate using the substitution, and then undo the substitution, and finally evaluate. When I do it this way, I remind myself that these are x limits by writing x = <whatever> for the lower limit of integration.

 

[Mark44]

You should also note that this is an improper integral. The integrand is undefined at both limits of integration.