- #1
Serena_Greene
- 9
- 0
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
∫ 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