# Evaluating complex indefinite integrals by substitution

• Feb 13th 2009, 01:17 PM
fattydq
Evaluating complex indefinite integrals by substitution
I've uploaded a picture with two problems I'm currently stuck on, I got the first four problems of this type correct, because they were just simple trig functions and whatnot, but now that quotients and complex trig stuff is coming into play I'm getting stuck. For example, on the first one, I don't know what I'd use as u since one term is always contained in another, if that makes sense...

And with the second I just don't understand what to do with that quotient that's there.
• Feb 13th 2009, 02:00 PM
cm3pyro
for the second one:
the integral of 1/(9-5x)dx

first, let u = 9-5x
then, du = -5
so, -1/5du = dx

substitute

(-1/5) times integral of 1/u du

which is easy to integrate

(-1/5) times ln(u) + c

substitute back for u

(-1/5)(ln(5x-9)) + C

edit: forgot the + C
• Feb 13th 2009, 02:03 PM
fattydq
Quote:

Originally Posted by cm3pyro
for the second one:
the integral of 1/(9-5x)dx

first, let u = 9-5x
then, du = -5
so, -1/5du = dx

substitute

(-1/5) times integral of 1/u du

which is the same as

(-1/5) times integral of u^-1 du
which is really easy to integrate

-1/5u + C

edit: forgot the + C

Nope that's incorrect. That was my first response as well (and I just tried it again to make sure!)
• Feb 13th 2009, 02:09 PM
Chris L T521
Quote:

Originally Posted by cm3pyro
for the second one:
the integral of 1/(9-5x)dx

first, let u = 9-5x
then, du = -5
so, -1/5du = dx

substitute

(-1/5) times integral of 1/u du

which is the same as

(-1/5) times integral of u^-1 du
which is really easy to integrate

-1/5u + C

edit: forgot the + C

$\displaystyle \int \frac{\,du}{u}{\color{red}\neq}u+C$

You can't use power rule here, because you will end up dividing by zero!!

You must know that $\displaystyle \int\frac{\,du}{u}=\color{red}\ln\!\left|u\right|+ C$!

So you should have $\displaystyle -\tfrac15\ln\!\left|u\right|+C$ as the "solution". I say "solution" because its not the solution we're looking for. You need to back substitute the value we defined u to be.
• Feb 13th 2009, 02:13 PM
cm3pyro
yeah i realized that right ater i posted and fixed it!
• Feb 13th 2009, 02:15 PM
fattydq
Ok but why does du = -5 lead to -1/5du = dx? In other words I don't understand where the -1/5 comes from in the response. I understand that du is -5 and why it is but why would that mean we make that the denominator?
• Feb 13th 2009, 02:22 PM
fattydq
Also does anyone wanna take a stab at the first one.
• Feb 13th 2009, 02:54 PM
HallsofIvy
Quote:

Originally Posted by fattydq
Ok but why does du = -5 lead to -1/5du = dx? In other words I don't understand where the -1/5 comes from in the response. I understand that du is -5 and why it is but why would that mean we make that the denominator?

It doesn't. In fact, you can't have "du= -5" because du is a differential and -5 isn't. But if you have du= -5 dx, then you can divide both sides by -5!

For the other problem, $\displaystyle \int sec(3\theta)tan(3\theta)d\theta$, the substitution $\displaystyle u= 3\theta$, so that $\displaystyle du= 3 d\theta$ and $\displaystyle \frac{1}{3}du= d\theta$ and the integral becomes
$\displaystyle \frac{1}{3} sec(u)tan(u)du$

Now, you should have learned in "differential" calculus that $\displaystyle (sec(x))'= \left(\frac{1}{cos(x)}\right)'= \frac{sin(x)}{cos^2(x)}= \frac{1}{cos(x)}\frac{sin(x)}{cos(x)}= sec(x)tan(x)$

Since (sec(u))'= sec(u)tan(u), $\displaystyle \int sec(u)tan(u)du$= what?

Another way to do that, if you don't like remembering a lot of derivatives and integrals of trig functions, is to reduce everything to sine and cosine (what I would be inclined to do!). sec(x)= 1/cos(x) and tan(x)= sin(x)/cos(x) so sec(x)tan(x)= $\displaystyle \frac{sin(x)}{cos^2(x)}$. Now, let $\displaystyle u= cos(3\theta)$ so that $\displaystyle dy= -3sin(3\theta)d\theta$ so that $\displaystyle sin(3\theta)d\theta= -\frac{1}{3}du$ and the integral becomes $\displaystyle -\frac{1}{3}\int \frac{1}{u^2} du$