Welcome to our community

Be a part of something great, join today!

Confused about Substitution

shamieh

Active member
Sep 13, 2013
539
Suppose we have \(\displaystyle \int \frac{4}{x^2 + 4} \)

So I understand the first thing we would so is bring the constant out and do u substitution but what I don't understand is how we can make the substitution u = \(\displaystyle \frac{x}{2}\) when there clearly is no \(\displaystyle \frac{x}{2}\) in the problem. I also understand how to factor out a 4 in the denominator thus getting us this

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

but then I don't understand how we can say u = \(\displaystyle \frac{x}{2}\) then \(\displaystyle du = 1/2dx\)

when \(\displaystyle \frac{x}{2}\) is no where in the problem..What am I missing
 

Guest

Active member
Jan 4, 2014
199
... but then I don't understand how we can say u = \(\displaystyle \frac{x}{2}\) then \(\displaystyle du = 1/2dx\)

when \(\displaystyle \frac{x}{2}\) is no where in the problem..What am I missing
$\displaystyle \frac{x^2}{4} = \left(\frac{x}{2}\right)^2$
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I would instead use the substitution:

\(\displaystyle x=2\tan(\theta)\)

Then you may apply a useful Pythagorean identity...
 

shamieh

Active member
Sep 13, 2013
539
I would instead use the substitution:

\(\displaystyle x=2\tan(\theta)\)

Then you may apply a useful Pythagorean identity...
I don't understand how. Can you show me?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775

shamieh

Active member
Sep 13, 2013
539
First, compute $dx$, and then make the substitution...what do you have now?
\(\displaystyle x = 2tan\theta\)
\(\displaystyle dx = 2sec^2\theta\)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
\(\displaystyle x = 2tan\theta\)
\(\displaystyle dx = 2sec^2\theta\)
Correct, now substitute for $x$ and $dx$ and what do you have?
 

shamieh

Active member
Sep 13, 2013
539
\(\displaystyle \int \frac{4}{2tan^2\theta + 4} 2sec^2\theta\)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
\(\displaystyle \int \frac{4}{2tan^2\theta + 4} 2sec^2\theta\)
Not quite...you want:

\(\displaystyle \int\frac{4}{(2\tan(\theta))^2+4}2\sec^2(\theta)\,d\theta\)

Also, I didn't notice earlier that your computation of $dx$ did not include $d\theta$.

Now, can you simplify this?
 

shamieh

Active member
Sep 13, 2013
539
\(\displaystyle \int \frac{4}{2tan^2\theta + 4} 2sec^2\theta\)

\(\displaystyle \frac{8}{2} \int \frac{sec^2\theta}{tan^2\theta + 4} \, d\theta\)
\(\displaystyle
u = tan^2\theta + 4\)
\(\displaystyle du = sec^2\theta \, d\theta\) ?
 

Guest

Active member
Jan 4, 2014
199
It seems you only squared the tan and not the 2 before it as well.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
\(\displaystyle \frac{8}{2} \int \frac{sec^2\theta}{tan^2\theta + 4} \, d\theta\)
\(\displaystyle
u = tan^2\theta + 4\)
\(\displaystyle du = sec^2\theta \, d\theta\) ?
This is what I suggest...first we have:

\(\displaystyle 2\int\frac{4\sec^2(\theta)}{4\tan^2(\theta)+4}\, d\theta\)

Now, divide each term in the numerator and denominator of the integrand by 4:

\(\displaystyle 2\int\frac{\sec^2(\theta)}{\tan^2(\theta)+1}\, d\theta\)

Now, can you rewrite the denominator of the integrand using a Pythagorean identity?
 

shamieh

Active member
Sep 13, 2013
539
When you say re write it as a pythagorean identity you mean do a substitution right?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
When you say re write it as a pythagorean identity you mean do a substitution right?
Well, it is a substitution of sorts, but not like a typical $u$-substitution. Begin with the well-known Pythagorean identity:

\(\displaystyle \sin^2(\theta)+\cos^2(\theta)=1\)

and divide through by $\cos^2(\theta)$...what do you get?
 

shamieh

Active member
Sep 13, 2013
539
yea but why do i need to do sin + cos when you can make the U substitution for tan and sec?
 

Guest

Active member
Jan 4, 2014
199
yea but why do i need to do sin + cos when you can make the U substitution for tan and sec?
That will take you back to square one. You would exactly be reversing what you did.

Do you know what $\tan^2{x}+1$ is equal to? If no, then perhaps you didn't give MarkFL's question much thought; if yes, then where you have $\tan^2{x}+1$ in the integral replace that with what it equals and there isn't much left.