Wow I would never have thought of it this way! I split it into integrals between -π~0 and 0~π and it turns into (∫[-π, 0] (x + |x|)sinx dx + ∫[-π, 0] (x + |x|)sinx dx), and now since x in the first integral is always negative, you can write |x| as -x, and the same logic allows you to write |x| as x in the second integral. So you end up with ∫[-π, 0] (x - x)sinx dx + ∫[-π, 0] (x + x)sinx dx which is ∫[-π, 0] 0 · sinx dx + ∫[-π, 0] 2x · sinx dx = 0 + 2 · ∫[-π, 0] sin dx = 0 + 2π = 2π
Your writing is so wonderful.❤❤❤❤❤
Wow I would never have thought of it this way! I split it into integrals between -π~0 and 0~π and it turns into (∫[-π, 0] (x + |x|)sinx dx + ∫[-π, 0] (x + |x|)sinx dx), and now since x in the first integral is always negative, you can write |x| as -x, and the same logic allows you to write |x| as x in the second integral. So you end up with ∫[-π, 0] (x - x)sinx dx + ∫[-π, 0] (x + x)sinx dx which is ∫[-π, 0] 0 · sinx dx + ∫[-π, 0] 2x · sinx dx = 0 + 2 · ∫[-π, 0] sin dx = 0 + 2π = 2π
an interesting integral
Wow, just now I realised you could use the d(cosx) notation for IBP 🤯🤯
see many of my other videos, this is a commonly practical way of writing, simplifing integrations
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