Evaluating complex indefinite integrals by substitution

**fattydq** · February 13th 2009, 01:17 PM

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.

**cm3pyro** · February 13th 2009, 02:00 PM

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

so the answer is:

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

edit: forgot the + C

**fattydq** · February 13th 2009, 02:03 PM

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

so the answer is:

-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!)

**Chris L T521** · February 13th 2009, 02:09 PM

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

so the answer is:

-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.

**cm3pyro** · February 13th 2009, 02:13 PM

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

**fattydq** · February 13th 2009, 02:15 PM

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?

**fattydq** · February 13th 2009, 02:22 PM

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

**HallsofIvy** · February 13th 2009, 02:54 PM

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 $-\frac{1}{3}\int \frac{1}{u^2} du$

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