Qualification > Math

Pure math(p3)

<< < (5/9) > >>

twilight:

--- Quote from: ITS MEE... on May 28, 2009, 02:43:25 pm ---can u integrate
sin^3xcosx
please :'(

--- End quote ---
how didya solv it?

ITS MEE...:
well wnever u see an ODD one ie unfamiliar and kind of compound just try to differentiate one of them and if u get the other (regardless of the power or something like a number multiplied or divided by it then u intergrate the one u differentiated
 :-\ hope its a bit clear
Astarmathsandphysics can u explain it a bit more or make it less vague? please

louis:
ii) Find a vector equation for the line of intersection between two planes :
      2x -y - 3z= 7 and x + 2y + 2z=0     
Solution :  The normal vectors of 2 planes are n1= ( 2,-1,-3)    and n2=  (1,2,2)
                    These 2 vectors form a new plane.
                    The normal vector of this new plane is  ( 2,-1,-3)  X   (1,2,2)  =  (4, -7, 5)
                    The normal vector is parallel to the line of intersection of the 2 planes.
                    So the direction vector for the line of intersection of 2 planes is  a ( 4, -7, 5 )

                    Now we have to find the position vector of a point on the line of intersection
                    In the equations of  2x - y -3z= 7 and x + 2y + 2z = 0,there are 3 unknowns.
                    Reduce to 2 equations with 2 unknowns only by putting x=0  or  y=0  or z=o
                    We  choose y=0
                    Solve for x and z in the equations  2x -3Z = 7
                                                                    x +  2Z = 0
                                                     we get        Z   =  -1     x =  2
                     So the vector  equation for the line of intersection of 2 planes is
                                                     r=  2i -k  + a(  4,-7 ,  5 )

twilight:
ooo i think i get wut iu mean ITS ME..
so is it    (1/4)(sin4x) ?

astarmathsandphysics:

--- Quote from: ITS MEE... on May 28, 2009, 02:49:27 pm ---oh i got it no need to solve it but can u integrate this
(1+tanx)^4 sec^2x

--- End quote ---

Substitute u=1+tanx to get u^4dv ans=1/5(1+tanx)^5

Navigation

[0] Message Index

[#] Next page

[*] Previous page

Go to full version