solve it: integrate it sqr(tanx) dx
- Thread starter fredie
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If you're in a calculus class and "don't know anything about" radicals, trigonometric functions, or integrals, then I'm afraid you need months of intensive, hands-on, in-person tutoring (or else a year or two of classes at college). There is, I'm sorry to say, nothing we can do for you here.fredie said:
My best wishes to you.
Eliz.
This is a booger of an integral for someone with no clue. First, maybe work on your language skills. 'abt dis'?.
One recourse could be to let \(\displaystyle \L\\u=tan(x), \;\ tan^{-1}(u)=x, \;\ dx=\frac{1}{1+u^{2}}du\)
Then you have:
\(\displaystyle \L\\\int\frac{\sqrt{u}}{1+u^{2}}du\)
That's a start. Hope it helps.
What I would do is run this through some technology.
I ran this thorugh Maple and she gave me:
\(\displaystyle \L\\\frac{1}{2}\frac{\sqrt{tan(x)}cos(x)\sqrt{2}cos^{-1}(cos(x)-sin(x))}{\sqrt{cos(x)sin(x)}}-\frac{1}{2}\sqrt{2}ln(cos(x)+\sqrt{2tan(x)}cos(x)+sin(x))\)
WHEW!!. Good luck deriving that by hand.
One recourse could be to let \(\displaystyle \L\\u=tan(x), \;\ tan^{-1}(u)=x, \;\ dx=\frac{1}{1+u^{2}}du\)
Then you have:
\(\displaystyle \L\\\int\frac{\sqrt{u}}{1+u^{2}}du\)
That's a start. Hope it helps.
What I would do is run this through some technology.
I ran this thorugh Maple and she gave me:
\(\displaystyle \L\\\frac{1}{2}\frac{\sqrt{tan(x)}cos(x)\sqrt{2}cos^{-1}(cos(x)-sin(x))}{\sqrt{cos(x)sin(x)}}-\frac{1}{2}\sqrt{2}ln(cos(x)+\sqrt{2tan(x)}cos(x)+sin(x))\)
WHEW!!. Good luck deriving that by hand.
Start off by letting \(\displaystyle u^2 = tan(x)\), so \(\displaystyle dx = \frac{2u du}{sec^2(x)}\\), therefore: \(\displaystyle dx = \frac{2u du}{u^4 + 1}\\). So the integral becomes:
\(\displaystyle 2 \int \frac{u^2 du}{u^4 + 1}\\\)
Now, if we let \(\displaystyle m = u^2\), the integrand becomes:
\(\displaystyle \frac{m}{m^2+1}\ = m \frac{1}{m^2+1}\ = m [ 1 - m^2 + m^4 - m^6 + m^8 + ... ]\)
Therefore, our integral is:
\(\displaystyle 2 \int u^2 [ 1 - u^4 + u^8 - u^12 + u^16 + ... ] du \ = 2 \int [ u^2 - u^6 + u^10 - u^14 + u^18 + ... ] du \\)
Somebody please correct me if I'm wrong if these are not valid steps. I'd then integrate term-by-term, back-substitute, and attempt to express the series integral in closed form.
\(\displaystyle 2 \int \frac{u^2 du}{u^4 + 1}\\\)
Now, if we let \(\displaystyle m = u^2\), the integrand becomes:
\(\displaystyle \frac{m}{m^2+1}\ = m \frac{1}{m^2+1}\ = m [ 1 - m^2 + m^4 - m^6 + m^8 + ... ]\)
Therefore, our integral is:
\(\displaystyle 2 \int u^2 [ 1 - u^4 + u^8 - u^12 + u^16 + ... ] du \ = 2 \int [ u^2 - u^6 + u^10 - u^14 + u^18 + ... ] du \\)
Somebody please correct me if I'm wrong if these are not valid steps. I'd then integrate term-by-term, back-substitute, and attempt to express the series integral in closed form.
You're thinking on the same lines as I, morson.
By making another substitution in the form I posted, you can get to the form you have:
With partial fractions:
\(\displaystyle \L\\\frac{u^{2}}{1+u^{4}}=\frac{\sqrt{2}u}{4(u^{2}-\sqrt{2}u+1)}-\frac{\sqrt{2}u}{4(u^{2}+\sqrt{2}u+1)}\)
It ain't pretty no matter how you look at it.
\(\displaystyle \L\\\frac{\sqrt{2}u}{4(u^{2}-\sqrt{2}u+1)}=\frac{\sqrt{2}}{4}\left[\frac{2u-\sqrt{2}}{2(u^{2}-\sqrt{2}u+1)}+\frac{\sqrt{2}}{2(u^{2}-\sqrt{2}u+1)}\right]\)
If you let \(\displaystyle \L\\w=u^{2}-\sqrt{2}u+1, \;\ dw=(2u-\sqrt{2})du\)
\(\displaystyle \L\\\frac{1}{2}\int\frac{1}{w}dw=\frac{1}{2}ln(w)=\frac{1}{2}ln(u^{2}-\sqrt{2}u+1)=\frac{1}{2}ln(tan(x)-\sqrt{2tan(x)}+1)\)
For the other part, let \(\displaystyle \L\\w=\frac{2u-\sqrt{2}}{2}, \;\ dw=du\)
\(\displaystyle \L\\\sqrt{2}\int\frac{1}{1+2w^{2}}dw=tan^{-1}(\sqrt{2}w)=tan^{-1}(\sqrt{2}(\frac{2u-\sqrt{2}}{2}))=tan^{-1}\left(\sqrt{2}\left(\frac{2\sqrt{tan(x)}-\sqrt{2}}{2}\right)\right)\)
There's half of it.
Continue in this way and you can hammer it into shape. Soroban or Skeeter will probably be along with some slick show off method. :wink:
By making another substitution in the form I posted, you can get to the form you have:
With partial fractions:
\(\displaystyle \L\\\frac{u^{2}}{1+u^{4}}=\frac{\sqrt{2}u}{4(u^{2}-\sqrt{2}u+1)}-\frac{\sqrt{2}u}{4(u^{2}+\sqrt{2}u+1)}\)
It ain't pretty no matter how you look at it.
\(\displaystyle \L\\\frac{\sqrt{2}u}{4(u^{2}-\sqrt{2}u+1)}=\frac{\sqrt{2}}{4}\left[\frac{2u-\sqrt{2}}{2(u^{2}-\sqrt{2}u+1)}+\frac{\sqrt{2}}{2(u^{2}-\sqrt{2}u+1)}\right]\)
If you let \(\displaystyle \L\\w=u^{2}-\sqrt{2}u+1, \;\ dw=(2u-\sqrt{2})du\)
\(\displaystyle \L\\\frac{1}{2}\int\frac{1}{w}dw=\frac{1}{2}ln(w)=\frac{1}{2}ln(u^{2}-\sqrt{2}u+1)=\frac{1}{2}ln(tan(x)-\sqrt{2tan(x)}+1)\)
For the other part, let \(\displaystyle \L\\w=\frac{2u-\sqrt{2}}{2}, \;\ dw=du\)
\(\displaystyle \L\\\sqrt{2}\int\frac{1}{1+2w^{2}}dw=tan^{-1}(\sqrt{2}w)=tan^{-1}(\sqrt{2}(\frac{2u-\sqrt{2}}{2}))=tan^{-1}\left(\sqrt{2}\left(\frac{2\sqrt{tan(x)}-\sqrt{2}}{2}\right)\right)\)
There's half of it.
Continue in this way and you can hammer it into shape. Soroban or Skeeter will probably be along with some slick show off method. :wink: