Reduction Formula and Integration by Parts

[Serena_Greene]

I was sick and missed class

∫ sin^2x dx = x/2 – sin2x/4 + C
(I see that there is a trig identity in the answer as sin2x = 2sinxcosx)

What I tried (copying the example in the book)
u = sinx du = sinxcosx dx (why is du not cosx?)
v = -cos x dv = sinx dx (why is this not just dx?)

∫ sin^2x dx = -cosxsinx + 1 ∫ sinx cos^2x dx (okay why did I add +1 and where did the funky ∫ come from) I see that cos^2x = 1 – sin^2x.

So now (from the example in the book but why?)
∫ sin^2x dx = -cosxsinx + 1 ∫ sinx dx – 1 ∫ sin^2x dx

2 ∫ sin^2x dx = -cosxsinx + 1 ∫ sinx dx (what happened to the -1?)

∫ sin^2x dx = -½cosxsinx + 1/2 ∫ sinx dx

Which I thought should give

-½cosxsinx – ½cosx + C

Nothing like the above answer……. Can someone help?

-Serena

 

[jacobrhcp]

I don't want to read trough the whole thing, but I think it's easier to use the trig identity from the start:

sin^2 x = 1/2 - 1/2 cos(2x)

edit: this way you'll get to the answer easily too :)

 

[Vid]

u = sin(x)
du = cos)dx
v = -cos(x)
dv = sin(x)dx

Int[sin^2(x)] = -sin(x)cos(x) + Int[cos^2(x)]
Int[sin^2(x)] = -1/2sin(2x) + Int[1] - Int[sin^2x]
2Int[sin^2(x)] = -1/2sin(2x) + x
Int[sin^2(x)] = x/2 - 1/4sin(2x)