Having big problem with integration work (1st year tertiary level)

[capt. crunch]

I am currently working on revision and the following question came up:
Find the integral of:
$$\int$$ x^2 $$\sqrt{x-5}$$dx

now the way I went about solving this was to let u = x-5 and du = dx so that for the first step i got:
$$\int$$ x^2 U^1/2 du. I felt wrong right from the get go and had a look at the worked solution and they came up with:
Let u =$$\sqrt{x-5}$$
then they got
2$$\int$$u^2 (U^2 +5)^2 du
I do not understand where this comes from?

EDIT i figured it out:
u = $$\sqrt{x-5}$$
x= u^2+5 therefore x^2 = (u^2+5)^2
dx = 2U du

This gives me all the values who's origins i was unsure of.

 

[capt. crunch]

for a 2nd question:
$$\int$$sin^7xcos^3xdx

i worked it down to the following using reduction eqn and also substitution:
= - cos^4x/40 -1/10 sin^6xcos^4x - 3/40 sin^4xcos^4x - 1/20sin^2xcos^4x + c

should I simplify this further and where should i begin if i do?

 

[tiny-tim]

welcome to pf!

hi capt. crunch! welcome to pf!

(have an integral: ∫ and a square-root: √ and try using the X² icon just above the Reply box )

hint: cos³x = cosx - cosx sin²x

 

[capt. crunch]

hi capt. crunch! welcome to pf!

(have an integral: ∫ and a square-root: √ and try using the X² icon just above the Reply box )

hint: cos³x = cosx - cosx sin²x

Thanks, so if I use what you gave above, how do i treat the sin⁷x? i thought i had to reduce that with the reduction equations?

 

[tiny-tim]

there won't be a sin⁷x

 

[capt. crunch]

i got it... thanks dude you are a hero!