- #1

Serena_Greene

- 9

- 0

∫ 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