Trig Substitution

razorlead · Postby **razorlead** » Wed Dec 01, 2010 2:20 am

Hello, I have been having trouble with trig substitution. I feel that I understand the basic concepts, but I am struggling to get the same answers as the key. I will use @ symbols instead of thetas

Indefinite Integral

(x^3)sqrt((x^2)+4)dx

With an x= 2tan@
and dx= 2 (sec^2)@ d@

I get to

8(tan^3)@{sqrt((12tan^2)@+8d@

After that I'm stuck

The answer is

(1/5)((x^2)+4)^(5/2)-(4/3)((x^2)+4)^(3/2)+C

Martingale · Postby **Martingale** » Wed Dec 01, 2010 12:30 pm

razorlead wrote:Hello, I have been having trouble with trig substitution. I feel that I understand the basic concepts, but I am struggling to get the same answers as the key. I will use @ symbols instead of thetas

Indefinite Integral

(x^3)sqrt((x^2)+4)dx

With an x= 2tan@
and dx= 2 (sec^2)@ d@

I get to

8(tan^3)@{sqrt((12tan^2)@+8d@

After that I'm stuck

The answer is

(1/5)((x^2)+4)^(5/2)-(4/3)((x^2)+4)^(3/2)+C

$\int x^3\sqrt{x^2+4}dx$

let $x=2\tan(\theta)$ then $dx=2\sec^2(\theta)d\theta$

$\int x^3\sqrt{x^2+4}dx=\int(2\tan(\theta))^3\sqrt{(2\tan(\theta))^2+4}\cdot 2\sec^2(\theta)d\theta=\int 8\tan^3(\theta)\sqrt{4(\tan^2(\theta)+1)}\cdot 2\sec^2(\theta)d\theta$

$=\int 8\tan^3(\theta)\cdot 2\sqrt{\sec^2(\theta)}\cdot 2\sec^2(\theta)d\theta=\int 8\tan^3(\theta)\cdot 2\sec(\theta)\cdot 2\sec^2(\theta)d\theta=\int 32\tan^3(\theta)\sec^3(\theta)d\theta$

...

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