# Thread: Evaluating complex indefinite integrals by substitution

1. ## 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.

2. 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

3. 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!)

4. 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
$\int \frac{\,du}{u}{\color{red}\neq}u+C$

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

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

So you should have $-\tfrac15\ln\!\left|u\right|+C$ 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.

5. yeah i realized that right ater i posted and fixed it!

6. 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?

7. Also does anyone wanna take a stab at the first one.

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

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

Since (sec(u))'= sec(u)tan(u), $\int sec(u)tan(u)du$= 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)= $\frac{sin(x)}{cos^2(x)}$. Now, let $u= cos(3\theta)$ so that $dy= -3sin(3\theta)d\theta$ so that $sin(3\theta)d\theta= -\frac{1}{3}du$ and the integral becomes $-\frac{1}{3}\int \frac{1}{u^2} du$