# Trig Sub Problems:

• Feb 14th 2010, 11:19 PM
XFactah416
Trig Sub Problems:
Hi guys. I have two problems that revolve around trig sub and just want to make sure that I got these right.

1) integral of cos (sqrt(x))^2 / (sqrt(x))

I got (sqrt) x + (1/2)sin(2(sqrt(x)) + C.

2) integral of (x^3) / sqrt (x^2 + 4), with limits from 0 to 2.

For this one, I'm a little confused on how to set this one up. Which trig identity do I use that will successfully satisfy this beginning?
• Feb 14th 2010, 11:25 PM
Jhevon
Quote:

Originally Posted by XFactah416
Hi guys. I have two problems that revolve around trig sub and just want to make sure that I got these right.

1) integral of cos (sqrt(x))^2 / (sqrt(x))

I got (sqrt) x + (1/2)sin(2(sqrt(x)) + C.

2) integral of (x^3) / sqrt (x^2 + 4), with limits from 0 to 2.

For this one, I'm a little confused on how to set this one up. Which trig identity do I use that will successfully satisfy this beginning?

Personally i would not use trig sub for either of these. a substitution of $u = \sqrt x$ works nicely for the first (is the square on the cosine? if so, you are correct) and a substitution of $u^2 = x^2 + 4$ works wonders for the second. if you insist on doing trig sub for this, you would use $x = 2 \tan \theta$, the first does not have a required form to apply trig sub.
• Feb 14th 2010, 11:54 PM
XFactah416
Quote:

Originally Posted by Jhevon
Personally i would not use trig sub for either of these. a substitution of $u = \sqrt x$ works nicely for the first (is the square on the cosine? if so, you are correct) and a substitution of $u^2 = x^2 + 4$ works wonders for the second. if you insist on doing trig sub for this, you would use $x = 2 \tan \theta$, the first does not have a required form to apply trig sub.

Yeah, the square is on the cosine for the first one.

Thanks a lot!
• Feb 15th 2010, 08:27 PM
XFactah416
Sorry, hate to bring this topic back up, but I would like to know the answer to #2.

It seems like after I do a substitiution of x = 2 tan (theta), my limits of integration go to 0 to -4.37. Then, after some algebra, and another substitution (this time of u = sec (theta), my limits of integration is now 1 to -2.98.
• Feb 15th 2010, 11:12 PM
ione
Quote:

Originally Posted by XFactah416
Sorry, hate to bring this topic back up, but I would like to know the answer to #2.

It seems like after I do a substitiution of x = 2 tan (theta), my limits of integration go to 0 to -4.37. Then, after some algebra, and another substitution (this time of u = sec (theta), my limits of integration is now 1 to -2.98.

$\int_0^2\frac{x^3}{\sqrt{x^2+4}}\,dx$

$=\int_0^{\frac{\pi}{4}}\frac{8tan^3{\theta}}{2sec{ \theta}}2sec^2{\theta}\,d\theta$

$=\int_0^{\frac{\pi}{4}}8tan^2{\theta}sec{\theta}ta n{\theta}\,d\theta$

$=\int_0^{\frac{\pi}{4}}8(sec^2{\theta}-1)sec{\theta}tan{\theta}\,d\theta$

$=8\left[\frac{sec^3{\theta}}{3}-sec{\theta}\right]_0^{\frac{\pi}{4}}$
• Feb 16th 2010, 12:41 AM
Prove It
Quote:

Originally Posted by XFactah416
2) integral of (x^3) / sqrt (x^2 + 4), with limits from 0 to 2.

Hyperbolic Substitution is better for this integral.

Let $x = 2\sinh{t}$ so that $dx = 2\cosh{t}\,dt$.

So the integral becomes

$\int{\frac{x^3}{\sqrt{x^2 + 4}}\,dx} = \int{\frac{(2\sinh{t})^3}{\sqrt{(2\sinh{t})^2 + 4}}\,2\cosh{t}\,dt}$

$= \int{\frac{16\sinh^3{t}\cosh{t}}{\sqrt{4\sinh^2{t} + 4}}\,dt}$

$= \int{\frac{16\sinh^3{t}\cosh{t}}{\sqrt{4(\sinh^2{t } + 1)}}\,dt}$

$= \int{\frac{16\sinh^3{t}\cosh{t}}{2\sqrt{\cosh^2{t} }}\,dt}$

$= \int{\frac{8\sinh^3{t}\cosh{t}}{\cosh{t}}\,dt}$

$= 8\int{\sinh^3{t}\,dt}$

$= 8\int{\sinh{t}\sinh^2{t}\,dt}$

$= 8\int{\sinh{t}(\cosh^2{t} - 1)\,dt}$

Now let $u = \cosh{t}$ so that $\frac{du}{dt} = \sinh{t}$, so that the integral becomes

$= 8\int{(u^2 - 1)\,\frac{du}{dt}\,dt}$

$= 8\int{u^2 - 1\,du}$

$= 8\left[\frac{1}{3}u^3 - u\right] + C$

$= 8\left[\frac{1}{3}\cosh^3{t} - \cosh{t}\right] + C$

$= \frac{8}{3}\cosh^3{t} - 8\cosh{t} + C$.

And since $x = 2\sinh{t}$

$\frac{x}{2} = \sinh{t}$

$\frac{x^2}{4} = \sinh^2{t}$

$\frac{x^2}{4} = \cosh^2{t} - 1$

$1 + \frac{x^2}{4} = \cosh^2{t}$

$\cosh{t} = \sqrt{1 + \frac{x^2}{4}}$.

$\frac{8}{3}\sqrt{\left(1 + \frac{x^2}{4}\right)^2} - 8\sqrt{1 + \frac{x^2}{4}} + C$.