Solve Simple Integrals Homework Statement

[irok]

Homework Statement

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).

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

 

[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.

 

[irok]

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

\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?

 

[rocomath]

\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?

Good! Did you get the first one too?

 

[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))?

 

[Defennder]

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

 

[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.

 

[HallsofIvy]

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