integral((t+3)/sqrt(t^3 +1))dt

[dagger2006]

Not sure how to attempt integrating this with a cubic term in the denominator especially within a square root. As far as I know trig substitution requires square terms, and partial fractions doesn't make this easier. Partial differentiation has been my best guess but haven't been able to make it work. The full question is:


A particle moves along a line so taht its acceleration for t>=0 is given by
a(t)=(t+3)/sqrt(t^3 +1). If the particle's velocity at t=0 is 5, what is the velocity of the particle at t=3.

This is a practice ACT question from the internet. Thanks!

 

[Deleted member 4993]

Not sure how to attempt integrating this with a cubic term in the denominator especially within a square root. As far as I know trig substitution requires square terms, and partial fractions doesn't make this easier. Partial differentiation has been my best guess but haven't been able to make it work. The full question is:


A particle moves along a line so taht its acceleration for t>=0 is given by
a(t)=(t+3)/sqrt(t^3 +1). If the particle's velocity at t=0 is 5, what is the velocity of the particle at t=3.

This is a practice ACT question from the internet. Thanks!

Substitute:

t^3 = tan^2(Θ)

3t^2 dt = 2tan(Θ)*sec^2(Θ) dΘ

dt = 2/3 * [tan(Θ)]^(-1/3) * sec^2(Θ) dΘ ...... continue.... [edited]

 

[tkhunny]

You MAY wish to recall that you can factor t^3 + 1. That may lead nowhere.

 

[dagger2006]

Substitute:

t^3 = tan^2(Θ)

3t^2 dt = 2tan(Θ)*sec^2(Θ) dΘ

dt = 2/3 * [tan(Θ)]^(-1/3) * sec^2(Θ) dΘ ...... continue.... [edited]

I get the trig substitution. My problem is how to resolve the extra t in the numerator with that method. Thanks.

 

[stapel]

Partial differentiation has been my best guess

I get the trig substitution.

In future, kindly please provide that sort of information at the beginning. By your saying that you were trying something else, the helpers had noway of knowing that you'd actually understood that the helper-provided trig-substitution was the way to go.

My problem is how to resolve the extra 5 in the numerator with that method. Also, how do you resolve the t in the numerator?

Please reply showing your work, so we can see what you're talking about. Thank you!

 

[dagger2006]

I meant eliminate the T in the numerator not 5.

t^3=tan^2(theta)

3t^2 dt=2tan(theta) * sec^2(theta) d(theta)

At this point if you substitute this you get:

integral[(t+3)/sqrt(tan^2(theta)+1) dt

The problem I had with trig substitution is no matter what substitution I attempt that (t+3) in the numerator is very difficult to eliminate at my new dt term has t^2.