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Question Number 63566 by aliesam last updated on 05/Jul/19

prove that    ∫sin^n (x) dx , p∈n , p≥2 =− (1/n)cos(x) sin^(n−1) (x) + (p−1)∫sin^(n−2) (x) dx

$${prove}\:{that} \\ $$$$ \\ $$$$\int{sin}^{{n}} \left({x}\right)\:{dx}\:,\:{p}\in{n}\:,\:{p}\geqslant\mathrm{2}\:=−\:\frac{\mathrm{1}}{{n}}{cos}\left({x}\right)\:{sin}^{{n}−\mathrm{1}} \left({x}\right)\:+\:\left({p}−\mathrm{1}\right)\int{sin}^{{n}−\mathrm{2}} \left({x}\right)\:{dx} \\ $$

Commented by aliesam last updated on 05/Jul/19

thats right it was a typo

$${thats}\:{right}\:{it}\:{was}\:{a}\:{typo} \\ $$

Commented by MJS last updated on 05/Jul/19

it′s not true. the formula is for (n≥2)∈N  ∫sin^n  x dx=−(1/n)cos x sin^(n−1)  x +((n−1)/n)∫sin^(n−2)  x dx  ⇒ p−1=((n−1)/n) ⇔ p=2−(1/n) ⇒ p∉N∧p<2

$$\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{true}.\:\mathrm{the}\:\mathrm{formula}\:\mathrm{is}\:\mathrm{for}\:\left({n}\geqslant\mathrm{2}\right)\in\mathbb{N} \\ $$$$\int\mathrm{sin}^{{n}} \:{x}\:{dx}=−\frac{\mathrm{1}}{{n}}\mathrm{cos}\:{x}\:\mathrm{sin}^{{n}−\mathrm{1}} \:{x}\:+\frac{{n}−\mathrm{1}}{{n}}\int\mathrm{sin}^{{n}−\mathrm{2}} \:{x}\:{dx} \\ $$$$\Rightarrow\:{p}−\mathrm{1}=\frac{{n}−\mathrm{1}}{{n}}\:\Leftrightarrow\:{p}=\mathrm{2}−\frac{\mathrm{1}}{{n}}\:\Rightarrow\:{p}\notin\mathbb{N}\wedge{p}<\mathrm{2} \\ $$

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