standard integral (1 Viewer)

[dawso]

ok, the bos standard table has

int 1 / sqrt a^2 - x^2

however....many books have this as a f(x) instead of x, which is quite clear, but then have f'(x) on the top.......but this wouldnt work with the x^2 as f'(x) would be 2x, not 1..... (see cambridge p149 btw..)

anyway....can this new standard integral as given here be assumed??

 

[dawso]

ok......well steele, how would u do this q:

int (sin x) / sqrt (2-cos^2 (x))

 

[Jumbo Cactuar]

I think you are confused;

∫f'(x)dx/sqrt(a² - f(x)²)

consider f(x) = x , f'(x) = 1;

∫dx/sqrt(a² - x²)

I don't mean to be condecending btw. It is just the use of a substitution. My uni tables use x rather than f(x).

Or if you prefer;
I = ∫f'(x)dx/sqrt(a² - f(x)²)

let u=f(x)
u'=f'(x)
du=f'(x)dx

I = ∫du/sqrt(a² - u²)
= arcsin(u/a) + C
= arcsin(f(x)/a) + C

 

[Jumbo Cactuar]

ok......well steele, how would u do this q:

int (sin x) / sqrt (2-cos^2 (x))

I = ∫ (sin x) dx / sqrt (2-cos²(x))

let;
u = cos x
du = - sin x dx

I = - ∫ du / sqrt (2-u²)
= arccos(cos(x)/rt2) + C