integration by subsitution (1 Viewer)

[velox]

cant find my silly mistake, here's the problem


thanks

 

[who_loves_maths]

^

INT[tdt/sqrt(1-t^2)] , let u = 1- t^2, ie. du = -2tdt

I = (-1/2)INT[1/sqrt(u)] = -(1/2)[sqrt(u)/(1/2)] + c = -sqrt(u) + c = -sqrt(1 -t^2) + c

 

[velox]

thanks, what about this one:

Im not sure where the 1/6 came from, i got (1/2)

And this one:

 

[Trev]

du/dx = 2
dx = du/2

1/2 int. {[(u+1)/2].u^-1/2}du
1/4 int. {u^1/2 + u^-1/2}du
1/4*[2/3*u^3/2 + 2u^1/2} + C
Sub u = 2x + 1 back in.

 

[Trev]

Second one;
du/dx = 3x²
du/3 = x²dx
1/3 int. u^1/2du
1/3 [2/3*u^3/2] + C
(2u^3/2)/9 + C
Sub u = 1 + x³ back in.

 

[velox]

Thanks, i keep missing the fractions in the du/dx manipulations =/

 

[velox]

What about this one?

 

[Trev]

d@/dr = 3/(1-r)⁴
du = -dr
@ = -3 int. 1/(u⁴.du
@ = u^-3 + C
@ = 1/(1-r)³ + C

I'm not sure what r(0)=0 means; I have never come across it. (It's to do with working out the constant though, isn't it?)

 

[FinalFantasy]

d@\dr=3\(1-r)^4
@=(1-r)^-3 (3\3)+C=1\(1-r)³+C
@-C=1\(1-r)³
1\(@-C)=(1-r)³
(1\(@-C) )^(1\3)=1-r
r=1-(1\(@-C) )^(1\3)
r(0)=1-1\(-C)^(1\3) =0
1\(-C)^(1\3)=1
1=(-C)^(1\3)
C=-1
.: r=1-1\(@+1)^(1\3)

 

[velox]

d@\dr=3\(1-r)^4
@=(1-r)^-3 (3\3)+C=1\(1-r)³+C
@-C=1\(1-r)³
1\(@-C)=(1-r)³
(1\(@-C) )^(1\3)=1-r
r=1-(1\(@-C) )^(1\3)
r(0)=1-1\(-C)^(1\3) =0
1\(-C)^(1\3)=1
1=(-C)^(1\3)
C=-1
.: r=1-1\(@+1)^(1\3)

thanks, that's correct.

 

[velox]

What about this one...I actually got stuck half way through

 

[FinalFantasy]

I=int. z³sqrt(z²+1) dz from 0 to root 2
let u=z²+1
z=sqrt(u-1)
z³=(u-1)^(3\2)
du\dz=2z
dz=1\2sqrt(u-1) du
I=int. (u-1)^(3\2) u^(1\2) * 1\2sqrt(u-1) du from 1 to 3
=(1\2)int. (u-1) u^(1\2) du from 1 to 3
=(1\2) int. (u^(3\2) - u^(1\2) ) du from 1 to 3
...........