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# 2 More P3 questions.... watch

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1. 1.solve the differential equation

dy/dx = 8y^3 sin^2 x

given that y = 1/2 at x = pi/4

2.1. Use the substitution u = x^3 + 1 to evaluate

x^2/(x^3 + 1)^.5 dx between x =2 and x= 0

2.2 Using integration by parts, or otherwise, find

{ x(3x +1)^-.5 dx

Hence evaluate { x(3x +1)^-.5 dx between x=5 and x = 1.

Thanks.
2. (Original post by jonnymcc2003)
1.solve the differential equation

dy/dx = 8y^3 sin^2 x

given that y = 1/2 at x = pi/4
1. Separate the equation:

dy/8y^3 = sin^2 x dx

Integrate the RHS using cos 2x = 1 - 2sin^2 x and then put your x and y values in to work out the constant.
3. hey johnny, do u have the answers?? thanx
4. (Original post by jonnymcc2003)

2.1. Use the substitution u = x^3 + 1 to evaluate

x^2/(x^3 + 1)^.5 dx between x =2 and x= 0
u=x^3
du/dx = 3x^2
so dx = du/3x^2

x=2, u=8
x=0, u=0

so it now becomes, INT [x^2/(u^3 +1)^0.5](du/3x^2)

the x^2 cancels, so you just get 1/3[INT (u+1)^-0.5 du] , which can be done using p1 methods...
5. (Original post by mockel)
u=x^3
du/dx = 3x^2
so dx = du/3x^2

x=2, u=8
x=0, u=0

so it now becomes, INT [x^2/(u^3 +1)^0.5](du/3x^2)

the x^2 cancels, so you just get 1/3[INT (u+1)^-0.5 du] , which can be done using p1 methods...
hmmm.....i've done it differently :

u = x^3 + 1
du/dx = 3x^2
du = 3x^2 dx

INT [1/3u^5]du
6. (Original post by wonkey)
hmmm.....i've done it differently :

u = x^3 + 1
du/dx = 3x^2
du = 3x^2 dx

INT [1/3u^5]du
oops, misread the substitution....i thought it said u=x^3
my way would still be correct, although for this one, the way you did it is the only way, since it says "use the substitution u= x^3 + 1"
7. (Original post by mockel)
oops, misread the substitution....i thought it said u=x^3
my way would still be correct, although for this one, the way you did it is the only way, since it says "use the substitution u= x^3 + 1"
cool........so i've done what i've donw up to the point where i end up with :

INT ]1/3u^5]du = INT [3u^-5] du.....so i end up with [3u^-4/-4]....is this the point where i substitute 'x^3 +1' back where 'u' appears??
or do i just change the 'x' limits into 'u'? thanx
8. (Original post by wonkey)
cool........so i've done what i've donw up to the point where i end up with :

INT ]1/3u^5]du = INT [3u^-5] du.....so i end up with [3u^-4/-4]....is this the point where i substitute 'x^3 +1' back where 'u' appears??
or do i just change the 'x' limits into 'u'? thanx
if its a definite integral (ie one with limits) change the limits.
if its not a definitee integral simply subsitute and write the expression
9. johnny, i really need some answers pls, its driving me up the wall! lol
10. Can anyone do 2.2 ?

The answers for 1. is 2 tan(x/2) - x + c, and 2.1 is 4/3.
11. i think...

INT x(3x+1)^-5 dx

u = 3x+1
du = 3dx
dx = du/3

x = (u-1)/3

therefore: INT 1/9.(u-1).u^-5 du
which gives: 1/9 INT u^-4-u^-5 du with limits which are u = 16 and u = 4

which gives 1/9[(-1/3u^-3) - (-1/4u^-4)] = 1/9[(-1/3u^-3) + (1/4u^-4)]

put that into ur calculator you 4.615... x 10^-4
12. (Original post by jonnymcc2003)
Can anyone do 2.2 ?

The answers for 1. is 2 tan(x/2) - x + c, and 2.1 is 4/3.

can u show me how to do 1 pls. thx
13. (Original post by wonkey)
can u show me how to do 1 pls. thx
I got 1/(-2 y^2) = 4x - 2 sin2x - Pi
Not sure...

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