Some integration help.

Hi,
I have two questions I'd like some help with.
Firstly what is $\displaystyle \int cos^3 2x$ dx ?
I tried doing it with trigonometric identities for $\displaystyle cos 2x$ 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 $\displaystyle \int \tfrac{2}{2x+3} dx = ln(2x+3) $ yet $\displaystyle \int \tfrac{1}{x+1.5} dx = ln(x+1.5) $

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

What am I doing wrong?

Thanks.

Quote:

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Originally Posted by lord_day

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

[snip]

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

Now make the substitution $\displaystyle u = \sin (2x)$.

Quote:

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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 $\displaystyle \int \tfrac{2}{2x+3} dx = ln(2x+3) $ yet $\displaystyle \int \tfrac{1}{x+1.5} dx = ln(x+1.5) $

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

What am I doing wrong?

Thanks.

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

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

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

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.