Some integration help.

Printable View

October 27th 2008, 04:09 PM
lord_day

Some integration help.

Hi,
I have two questions I'd like some help with.
Firstly what is $\int cos^3 2x$ dx ?
I tried doing it with trigonometric identities for $cos 2x$ but it became very nasty pretty fast. Any help would be great.

Also, this has been bugging me all day, and I can't see where I am wrong, though I must be... Why is the integral of $\int \tfrac{2}{2x+3} dx = ln(2x+3)$ yet $\int \tfrac{1}{x+1.5} dx = ln(x+1.5)$

Because $\tfrac{2}{2x+3} = \tfrac{1}{x+1.5}$ yet $ln(x+1.5)$ is not the same as $ln(2x+3)$

What am I doing wrong?

Thanks.
October 27th 2008, 04:29 PM
mr fantastic

Quote:

Originally Posted by lord_day

Hi,
I have two questions I'd like some help with.
Firstly what is $\int cos^3 2x$ dx ?
I tried doing it with trigonometric identities for $cos 2x$ but it became very nasty pretty fast. Any help would be great.

[snip]

$\cos^3 (2x) = \cos^2 (2x) \, \cos (2x) = [1 - \sin^2 (2x)] \, \cos (2x)$ .

Now make the substitution $u = \sin (2x)$ .
October 27th 2008, 04:33 PM
mr fantastic

Quote:

Originally Posted by lord_day

[snip]
Also, this has been bugging me all day, and I can't see where I am wrong, though I must be... Why is the integral of $\int \tfrac{2}{2x+3} dx = ln(2x+3)$ yet $\int \tfrac{1}{x+1.5} dx = ln(x+1.5)$

Because $\tfrac{2}{2x+3} = \tfrac{1}{x+1.5}$ yet $ln(x+1.5)$ is not the same as $ln(2x+3)$

What am I doing wrong?

Thanks.

What you're doing wrong is forgetting about the arbitary constant of integration.

$\int \tfrac{2}{2x+3} dx = \ln |2x+3| + C_1 = \ln |2(x + 1.5)| + C_1$ $= \ln (2) + \ln |x + 1.5| + C_1 = \ln |x + 1.5| + C_2$ where $C_2 = \ln (2) + C_1$ is just as arbitrary as $C_1$ .

$\int \tfrac{1}{x+1.5} dx = \ln |x+1.5| + C_2$ .

What do you notice?

BY the way, notice the use of the absolute value brackets NOT round brackets. The answers you gave are therefore also wrong for this reason too.