Methods of integration~ (1 Viewer)

[shinji]

hey, just a lil q on methods of integration.
find the integral of sin²x cos²x dx.

would u make cos²x = 1-sin²x ?

so it becomes integral of {sin²(1-sin²x)}
=> sin²x - sin⁴x
where u let u = sin²x ? lol

 

[word.]

hint: sin(2x) = 2sin(x)cos(x)

 

[Rax]

Pardon my lack of mathematical ability, but where does that hint bust in, unless your converting them using that cos one.

Cant Remember

 

[pLuvia]

int.sin²xcos²xdx
=int.(sinxcosx)²dx
=1/4int.(sin2x)²dx
=1/4int.sin²2xdx
=1/4int.[(1-cos4x)/2]dx
=1/8int.(1-cos4x)dx
=1/8[x-1/4sin4x]dx+C
=x/8-1/32sin4x+C

 

[Rax]

Ah true........Der *smacks head*, And thats why I am not looking too forward to the trial

 

[shinji]

if thre's anyone on atm, can you tell me how pluvia got from

=1/4int.sin²2xdx
to
=1/4int.[(1-cos4x)/2]dx
?
thx

[edit]: nevermind, i got it. hahah