Need some help on a couple integrations

• Dec 2nd 2006, 02:18 PM
DaAzNJRiCh
Need some help on a couple integrations
You dont need to do them all but one would be nice. I want to see how to do it rather than just get answers. Thanks.

Integral of [(e^x + cos(x)]^2 dx
I tried to expand it and then got confused with integration by parts but I think you use that

Integral of e^2x ln (1+e^x) dx
I always hated ln

Integral of (12-4x-x^2)^.5 dx
I tried to factor it and got it to [-(x+6)(x-2)]^.5

Integral of cos^4 (x) sin^3 (x) dx
Should i change the cos^4 (x) to like (1-sin(x))^2 or something?

Integral of [(4-x^2)^.5]/ x dx
• Dec 2nd 2006, 02:40 PM
ThePerfectHacker
Quote:

Originally Posted by DaAzNJRiCh

Integral of e^2x ln (1+e^x) dx
I always hated ln

I assume,
$\displaystyle \int e^{2x}\ln (1+e^x)dx$
You can write it as,
$\displaystyle \int [(1+e^x)-1] \ln (1+e^x) e^x dx$
Then we can express this as,
$\displaystyle (f\circ g) \cdot g'$
Where,
$\displaystyle g=1+e^x$
$\displaystyle f=(x-1)\ln x$
Thus we need to find,
$\displaystyle F=\int (x-1)\ln x dx$
Which you can do with integration by parts.
$\displaystyle F\circ g+C$
• Dec 2nd 2006, 02:45 PM
CaptainBlack
Quote:

Originally Posted by DaAzNJRiCh
You dont need to do them all but one would be nice. I want to see how to do it rather than just get answers. Thanks.

Integral of [(e^x + cos(x)]^2 dx
I tried to expand it and then got confused with integration by parts but I think you use that

$\displaystyle \int [e^x+\cos(x)]^2\ dx=\int e^{2x}+2e^x \cos(x)+(\cos(x))^2\ dx=$$\displaystyle \int e^{2x}\ dx +2\int e^x \cos(x)\ dx+\int (\cos(x))^2\ dx$

The first and last integrals on the right are standard integrals and should pose
no problem. Now consider the middle integral on the right:

$\displaystyle 2\int e^x \cos(x)\ dx=2\ Re \left[ \int e^x e^{ix}\ dx \right]=2\ Re \left[ \int e^{x(1+i)}\ dx \right]$

Which is fairly simple to deal with.

RonL
• Dec 2nd 2006, 02:46 PM
ThePerfectHacker
Quote:

Originally Posted by DaAzNJRiCh

Integral of (12-4x-x^2)^.5 dx
I tried to factor it and got it to [-(x+6)(x-2)]^.5

Okay you have,
$\displaystyle \int \sqrt{12-4x-x^2} dx$
You factored, but you should have completed the square.
$\displaystyle \int \sqrt{-(x^2+4x-12)}dx$
Thus,
$\displaystyle \int \sqrt{-(x^2+4x+4-12-4)}dx$
Thus,
$\displaystyle \int \sqrt{16-(x+2)^2}dx$
Use a linear subsitution,
$\displaystyle u=x+2$
Thus,
$\displaystyle \int \sqrt{4^2-u^2} du$
This is a standard inverse sine function substitution.
• Dec 2nd 2006, 02:53 PM
ThePerfectHacker
Quote:

Originally Posted by DaAzNJRiCh

Integral of cos^4 (x) sin^3 (x) dx
Should i change the cos^4 (x) to like (1-sin(x))^2 or something?

Almost!
$\displaystyle \int \cos^4 x\sin^2 x\sin x dx$
$\displaystyle \int \cos^4 x(1-\cos^2 x)\sin xdx$
Let,
$\displaystyle u=\sin x$
Thus,
$\displaystyle \int u^4(1-u^2)du$
The rest is trivial.