Integration by parts of trig integral of sin(x)^3 * cos(x)^2

[scrum]

I have this integral

Integration of sin(x)^3 * cos(x)^2

When i do integration by parts it seems to turn nasty pretty quickly.

I was just wondering how might i approach this problem. My professor did this in lecture and i thought i understood it but when i try myself the problem just gets bigger and bigger but the integral never falls out,

i am picking u = sin(x)^3 and dv = cos(x)^2

 

[galactus]

Re: Integration by parts of trig,

Don't use parts. It's easier than that.

Use the identity \(\displaystyle sin^{2}(x)=1-cos^{2}(x)\)

\(\displaystyle \int[(1-cos^{2}(x))cos^{2}(x)sin(x)]dx\)

\(\displaystyle \int[(cos^{2}(x)-cos^{4}(x))sin(x)]dx\)

Now, let \(\displaystyle u=cos(x), \;\ du=sin(x)dx\)

 

[PAULK]

I have this integral

Integration of sin(x)^3 * cos(x)^2

When i do integration by parts it seems to turn nasty pretty quickly.

I was just wondering how might i approach this problem. My professor did this in lecture and i thought i understood it but when i try myself the problem just gets bigger and bigger but the integral never falls out,

i am picking u = sin(x)^3 and dv = cos(x)^2

.........................................................


Well, if you MUST use IBP, make sure that the dv part is something easy to integrate. How about:

u = sin^2(x) , du = 2 sin x cos(x)

dv = sin x cos^2(x)dx, v = - cos^3(x)/3, << do a subst of z = cos x

The uv part is - 1/3 sin^2(x) cos^3(x),
The other part is

{
| -1/3 sin x cos^4(x) dx
}

Now let z = cos(x), you get

-1/15 cos^5(x)

Now combine them, to get:
- 1/3 sin^2(x) cos^3(x) - 1/15 cos^5(x)

which you can change to nothing but cosines.

Of course, galactus got it right the first time.