Integral problem

[uberathlete]

Hi everyone. I'm having problems on how to integrate 2/(x^2+4) . I know the answer is arcsin(1/2x) + C, but I can't seem to get to that answer. I tried regular substitution, and integration by parts and they don't seem to work. I'm not so sure if trig substitution would work either. Any help on this would be greatly appreciated. Thanks!

 

[pka]

Surely you mean the answer is \(\displaystyle \L
\arctan \left( {\frac{x}{2}} \right) + C\)

Consider \(\displaystyle \L
\frac{2}{{x^2 + 4}} = \left( {\frac{1}{4}} \right)\frac{2}{{\left( {x/2} \right)^2 + 1}}\), then let \(\displaystyle \L
u = (x/2)\).

The arctangent form is \(\displaystyle \L
\frac{{du}}{{u^2 + 1}}\).

 

[galactus]

It doesn't equal $\displaystyle sin^{-1}(\frac{x}{2})$. I believe you mean

$\displaystyle tan^{-1}(\frac{x}{2})$


Let x=2u and dx=2du

$\displaystyle 2\int{\frac{1}{2^{2}+x^{2}}}dx$

$\displaystyle 2\int{\frac{2du}{2^{2}+2^{2}u^{2}}}$

$\displaystyle 2(\frac{1}{2})\int{\frac{du}{1+u^{2}}}$

$\displaystyle tan^{-1}(u)+C$

Since u=$\displaystyle \frac{x}{2}$

$\displaystyle tan^{-1}(\frac{x}{2})$+C

 

[soroban]

Hello, uberathlete!

Integrate: \(\displaystyle \L\,\int\frac{2}{x^2\,+\,4}\,dx\)

I know the answer is: $\displaystyle \,\arcsin\left(\frac{x}{2}\right)\,+\,C\;$ . . . of course, you mean arctangent

Trig Substitution is the way to go . . .

Let: $\displaystyle x\,=\,2\cdot\tan\theta\;\;\Rightarrow\;\;dx\,=\,2\cdot\sec^2\theta\,d\theta$

And: \(\displaystyle \,x^2\,+\,4\:=\2\tan\theta)^2\,+\,4\:=\:4\tan^2\theta\,+\,4\:=\:4(\tan^2\theta\,+\,1)\:=\:4\cdot\sec^2\theta\)


Substitute: \(\displaystyle \L\:\int\frac{2}{4\cdot\sec^2\theta}\)\(\displaystyle \cdot(2\cdot\sec^2\theta\,d\theta)\;=\;\L\int d\theta \;= \;\theta\,+\,C\)


Back-substitute:$\displaystyle \;x = 2\cdot\tan\theta\;\;\Rightarrow\;\;\tan\theta\,=\,\frac{x}{2}\;\;\Rightarrow\;\;\theta\,=\,\arctan\left(\frac{x}{2}\right)$


\(\displaystyle \text{Answer: }\:\L\arctan\left(\frac{x}{2}\right)\,+\,C\)

 

[uberathlete]

re

Oh my bad. Yeah I meant arctan(1/2x). Dang, I didn't see that. Thank guys.