Integration: Trig Substitution

[ChaoticLlama]

\(\displaystyle \L\int {\frac{{\sqrt {1 + 4x^2 } }}{{x^4 }}} dx\)

I've been trying \(\displaystyle x = 2\tan \theta\) without much success. Is this the proper way to go about it?

thanks.

 

[galactus]

Try \(\displaystyle \L\\\frac{tan(x)}{2}\)

You could also try \(\displaystyle \L\\u^{2}=4+\frac{1}{x^{2}}\)

 

[ChaoticLlama]

I'm running into some trouble with the trig substitution still.

After using the correct one you gave to me, I ended up with:
\(\displaystyle \L\ 8\int {\frac{{\sec ^3 \theta }}{{\tan ^4 \theta }}d\theta } = 8\int {\csc ^3 \theta \cot \theta d\theta } = \frac{{ - 8}}{3}\csc ^3 \theta+C\)

And when making the equation in terms of x again:
\(\displaystyle \L\left( {\frac{{ - 8}}{3}} \right)\frac{{\sqrt {1 + 4x^2 } }}{{2x}} + C\)

but it's not the right answer.

 

[pka]

Using a computer algebra I get \(\displaystyle \L
\frac{{ - \left[ {\sqrt {1 + 4x^2 } } \right]^3 }}{{3x^3 }}\).
If you differentiate that you may see how it works.

Here is a usual site to check answers.
http://integrals.wolfram.com/index.jsp

 

[galactus]

Do you have to use trig sub?.

The other suggestion:

\(\displaystyle \L\\u^{2}=4+\frac{1}{x^{2}}\)

\(\displaystyle \L\\2udu=\frac{-2}{x^{3}}dx\)

\(\displaystyle \L\\-udu=\frac{1}{x^{3}}dx\)

Therefore, \(\displaystyle \L\\x^{2}=\frac{1}{u^{2}-4}\)

You have:

\(\displaystyle \L\\-\int\frac{\sqrt{1+4(\frac{1}{u^{2}-4})}u}{\sqrt{\frac{1}{u^{2}-4}}}du\)

This looks much worse than it is. Believe it or not, it whittles down to:

\(\displaystyle \L\\-\int{u^{2}}du\)

Now, it's easy as...........well, you know.

 

[soroban]

Hello, ChaoticLlama!

\(\displaystyle \L\int {\frac{{\sqrt {1 + 4x^2 } }}{{x^4 }}} dx\)

I've been trying \(\displaystyle x\,=\,2\tan \theta\) without much success.

Of course not!

Then \(\displaystyle \,1\,+\,4x^2\:=\:1\,+\,16\tan^2\theta\) . . . and what good is that?


You should have begun with: \(\displaystyle \,4x^2\,=\,\tan^2\theta\;\;\Rightarrow\;\;x\,=\,\frac{1}{2}\tan\theta\;\;\Rightarrow\;\;dx\,=\,\frac{1}{2}\sec^2\theta\ d\theta\)

Substitute: \(\displaystyle \L\,\int\frac{\sec\theta}{\frac{1}{16}\tan^4\theta}\cdot\left(\frac{1}{2}\sec^2\theta\ d\theta\right) \;= \;8\int\frac{\sec^3\theta}{\tan^4\theta}\ d\theta\)


Multiply top and bottom by \(\displaystyle \cos^4\theta:\)

\(\displaystyle \L\;\;8\int\frac{\cos\theta}{\sin^4\theta}\ d\theta \;= \;8\int(\sin\theta)^{-4}\cos\theta\ d\theta\)

Let \(\displaystyle u\,=\,\sin\theta\) . . . and you're on your way!

 

[galactus]

I'm running into some trouble with the trig substitution still.

After using the correct one you gave to me, I ended up with:
\(\displaystyle \L\ 8\int {\frac{{\sec ^3 \theta }}{{\tan ^4 \theta }}d\theta } = 8\int {\csc ^3 \theta \cot \theta d\theta } = \frac{{ - 8}}{3}\csc ^3 \theta+C\)

And when making the equation in terms of x again:
\(\displaystyle \L\left( {\frac{{ - 8}}{3}} \right)\frac{{\sqrt {1 + 4x^2 } }}{{2x}} + C\)

but it's not the right answer.

You're correct, you just forgot to cube:

\(\displaystyle \L\\-\frac{8}{3}csc^{3}({\theta})d{\theta}\)

\(\displaystyle \L\\csc({\theta})=\frac{\sqrt{1+4x^{2}}}{2x}\)

Sub:

\(\displaystyle \L\\-\frac{8}{3}\left(\frac{\sqrt{1+4x^{2}}}{2x}\right)^{3}\)

Which, as you can see, equals:

\(\displaystyle \L\\-\frac{8}{3}\frac{(1+4x^{2})^{\frac{3}{2}}}{8x^{3}}\)

Cancel 8's:

\(\displaystyle \L\\-\frac{(1+4x^{2})^{\frac{3}{2}}}{3x^{3}}\)

 

[ChaoticLlama]

Thanks for all your help. Missing that cube was really getting on my nerves.