Help in sin^2x and cos^2x :S (1 Viewer)

[addikaye03]

its not sin 3x its sin^3 x



substitutions is slow for this, u shudn't need it

sin(2x+x)=Sin2xcosx+cos2xsinx
sin3x=2sinxcosx.cosx+(1-2sin^2x).sinx
sin3x=2sinxcos^2x+sinx-2sin^3x
sin3x=2sinx(1-sin^2x)+sinx-2sin^3x
sin3x=2sinx-2sin^3x+sinx-2sin^3x
sin3x=3sinx-4sin^3x

therefore sin^3x=1/4(3sinx-sin3x)

Int. sin^3x dx= 1/4 Int (3sinx-sin3x)
=1/4(-3cosx+1/3cos3x)
=-3/4cosx+1/12cos3x or/ 1/12cos3x-3/4cosx for neatness

 

[kwabon]

its not sin 3x its sin^3 x



substitutions is slow for this, u shudn't need it

really???!!! wow i thought u sub is the quickest way to integrate cos^2 x sinx, could u please tell me wat way would u use. thanks.

 

[tommykins]

sin(2x+x)=Sin2xcosx+cos2xsinx
sin3x=2sinxcosx.cosx+(1-2sin^2x).sinx
sin3x=2sinxcos^2x+sinx-2sin^3x
sin3x=2sinx(1-sin^2x)+sinx-2sin^3x
sin3x=2sinx-2sin^3x+sinx-2sin^3x
sin3x=3sinx-4sin^3x

therefore sin^3x=1/4(3sinx-sin3x)

Int. sin^3x dx= 1/4 Int (3sinx-sin3x)
=1/4(-3cosx+1/3cos3x)
=-3/4cosx+1/12cos3x or/ 1/12cos3x-3/4cosx for neatness

yuck solution

 

[M@ster P]

its sin^3x not sin3x

 

[addikaye03]

yuck solution

but a solution never the less, looking back on the post it is pretty awful... Btw, hows engineering going so far? PM me your msn, talk further

 

[addikaye03]

its sin^3x not sin3x

mate, we all understand that it isn't sin3x... read my solution and figure it out yourself

 

[M@ster P]

mate, we all understand that it isn't sin3x... read my solution and figure it out yourself

my bad

 

[addikaye03]

my bad

haha its all good man