# Simple Integrals

1. Jul 17, 2008

### irok

1. The problem statement, all variables and given/known data
Question 1:
Evaluate the indefinite integral.
$$\int \frac{\cos x}{2 \sin x + 6} \, dx$$

Question 2:

Evaluate the indefinite integral.
$$\int \frac{2 \; dx}{x \ln (6 x)}$$
NOTE: The absolute value of x has to be entered as abs(x).

3. The attempt at a solution
Question 1:
Let u = sinx, du = cosx dx
= $$\int \frac{1}{2 u + 6} \, du$$
= $$\frac{1}{2} \int \frac{1}{u + 3} \, du$$
= $$\frac{1}{2} \int (u + 3)^{-1} \, du$$
= $$\frac{1}{2} * [ 1 + 1 ]$$
= 1 + C

Question 2:
$$\int \frac{2 \; dx}{x \ln (6 x)}$$
Let u = ln(6x), du = 1 / 6x
= $$12 \int \frac{1 \; du}{\ln (u)}$$
= $$12 \int \frac{1 \; du}{\ln (u)}$$
= $$12 \int (\ln (u))^{-1}\, du$$
Since inverse of ln is exp
= $$12 \e^(u)$$
= $$12 \e^(ln(6x))$$
= 12 * 6 x = 72 x + C

2. Jul 17, 2008

### rocomath

What is the Integral of $$\int\frac{dx}{x}$$

That's essentially what you have for # 1.

You let $$u=\ln x$$

So what is it that you still have $$\ln x$$ in your Integral? Replace it with "u" completely.

3. Jul 17, 2008

### irok

$$\int \frac{2 \; dx}{x \ln (6 x)}$$
Let u = ln(6x), du = 1 / 6x
= $$12 \int \frac{1}{u} \ du$$
= $$12 [\ln(u)]$$
= $$12 [\ln(ln(6x))]$$ + C

Are there any mistakes?

4. Jul 17, 2008

### rocomath

Good! Did you get the first one too?

5. Jul 17, 2008

### irok

Yep, I got Question #1. I made u = sinx+3 instead of u = sinx.

Thank you rocomath!

One more question, can i simplify ln(ln(x))?

6. Jul 18, 2008

### Defennder

I don't see how to simplify that any further.

7. Jul 20, 2008

### Gib Z

There indeed does seem to be mistakes. If you let u= ln (6x), make sure you use the chain rule to find du.

8. Jul 20, 2008

### HallsofIvy

Staff Emeritus
Even simpler: ln(6x)= ln(x)+ ln(6) and the derivative of ln(6) is 0.