Integration Question (1 Viewer)

Peeik · Apr 6, 2012

How do you paste the latex onto your comment if you use codecogs? Do you just copy and paste it like an image or do you write the latex codes and bos will read it from there?

Shadowdude · Apr 6, 2012

tex brackets

[.tex] and [./tex]

remove the full stop

Peeik · Apr 6, 2012

Shadowdude said:
tex brackets

[.tex] and [./tex]

remove the full stop

Thanks Shadow!!

imsleepyz · Apr 8, 2012

Could someone help me out on another question please?

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

Carrotsticks · Apr 8, 2012

Essentially the same as original question, except complete the square at the bottom.

imsleepyz · Apr 8, 2012

Thanks Carrotsticks. One problem though, I'm used to substitution method when splitting up the fractions. But I can't seem to apply substitution or am I not seeing a trick behind it?

SpiralFlex · Apr 8, 2012

imsleepyz said:
Could someone help me out on another question please?

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

$\int \frac{x+1}{\sqrt{-(x^2+x-1)}}\; dx$

$\int \frac{x+1}{\sqrt{-(x^2+x+\frac{1}{4}-\frac{1}{4}-1)}}\; dx$

$\int \frac{x+1}{\sqrt{\frac{5}{4}-(x+\frac{1}{2})^2}}\; dx$

$\int \frac{x}{\sqrt{-x^2-x+1}}\; dx+\int \frac{1}{\sqrt{\frac{5}{4}-(x+\frac{1}{2})^2}}\; dx$

[HR][/HR]

Let's pause!"

I have deliberately left one with an uncompleted square. This is why, in the next step.

[HR][/HR]

Let us focus on the FIRST integral. Well you COULD use substitution, but you should reduce the integral from the form of:

$\int \frac{f'(x)}{2\sqrt{f(x)}}\;dx = \sqrt{f(x)}+C$

The first part of our integration problem is,

$\int \frac{x}{\sqrt{-x^2-x+1}}\; dx$

Our form says, we need the derivative on the top. And a two on the denominator.

First, put derivative on the top!

$\int \frac{-2x-1}{\sqrt{-x^2-x+1}}\; dx$ (NOT FINISHED INTEGRAL! THIS IS A THOUGHT PROCESS)

But this is NO WAY its the same as our previous one. Top compensate, let's see what happens if we negative half it.

$-\frac{1}{2}\int \frac{-2x-1}{\sqrt{-x^2-x+1}}\; dx$ (NOT FINISHED INTEGRAL! THIS IS A THOUGHT PROCESS)

This is still not our integral, but if WE MINUS A HALF! SINCE our resulting integral has an EXCESS of a half, we get,

$-\frac{1}{2}\int \frac{-2x-1}{\sqrt{-x^2-x+1}}\; dx+ -\frac{1}{2}\int\frac{1}{\sqrt{-x^2-x+1}}\; dx$ (Our first integral is done)

Expand it! And see if you get our ORIGINAL ONE!

[HR][/HR]

Second integral, use inverse sine!

(You already know how to do this.)

[HR][/HR]

Thus, we have!

$-\frac{1}{2}\int \frac{-2x-1}{\sqrt{-x^2-x+1}}\; dx-\frac{1}{2}\int \frac{1}{\sqrt{-x^2-x+1}}\;dx+\int \frac{1}{\sqrt{\frac{5}{4}-(x+\frac{1}{2})^2}}\; dx$

$=-\frac{1}{2}\int \frac{-2x-1}{\sqrt{-x^2-x+1}}\; dx+\frac{1}{2}\int \frac{1}{\sqrt{\frac{5}{4}-(x+\frac{1}{2})^2}}\; dx$

$=-\sqrt{-x^2-x+1}+\frac{1}{2}\sin^{-1}(\frac{2x+1}{\sqrt{5}})+C$

imsleepyz · Apr 8, 2012

Wow Spiralflex! Thank you so much for the effort! Okay I guess substitution method does not seem to be reliable for this case, right? I'm going to adopt this new concept of integrating a sqrt function. Thank you!!!

SpiralFlex · Apr 8, 2012

imsleepyz said:
Wow Spiralflex! Thank you so much for the effort! Okay I guess substitution method does not seem to be reliable for this case, right? I'm going to adopt this new concept of integrating a sqrt function. Thank you!!!

No worries. Yes! Try it! It will cut your time in half! If you practice some! You can do it instantly in head! There are some good exercises in the Terry Lee textbook.

Integration Question (1 Viewer)

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