Question-197239

Question Number 197239 by universe last updated on 10/Sep/23

Answered by mahdipoor last updated on 11/Sep/23

⇒^(d/dx) xsin^n x=(((xsin^n x)/n)−(((n−1)xsin^(n−2) cos^2 x)/n)−((sin^(n−1) xcosx)/n)) +(((sin^(n−1) xcosx)/n))+f(n)×(xsin^(n−2) x)⇒ ⇒xsin^(n−2) x(sin^2 x−((sin^2 x)/n)+(((n−1)cos^2 x)/n))= xsin^(n−2) x((((n−1)sin^2 x)/n)+(((n−1)cos^2 x)/n))= xsin^(n−2) x×(((n−1)/n))=xsin^(n−2) x×f(n) ⇒f(n)=((n−1)/n)

$$\overset{\frac{{d}}{{dx}}} {\Rightarrow}{xsin}^{{n}} {x}=\left(\frac{{xsin}^{{n}} {x}}{{n}}−\frac{\left({n}−\mathrm{1}\right){xsin}^{{n}−\mathrm{2}} {cos}^{\mathrm{2}} {x}}{{n}}−\frac{{sin}^{{n}−\mathrm{1}} {xcosx}}{{n}}\right) \\ $$$$+\left(\frac{{sin}^{{n}−\mathrm{1}} {xcosx}}{{n}}\right)+{f}\left({n}\right)×\left({xsin}^{{n}−\mathrm{2}} {x}\right)\Rightarrow \\ $$$$\Rightarrow{xsin}^{{n}−\mathrm{2}} {x}\left({sin}^{\mathrm{2}} {x}−\frac{{sin}^{\mathrm{2}} {x}}{{n}}+\frac{\left({n}−\mathrm{1}\right){cos}^{\mathrm{2}} {x}}{{n}}\right)= \\ $$$${xsin}^{{n}−\mathrm{2}} {x}\left(\frac{\left({n}−\mathrm{1}\right){sin}^{\mathrm{2}} {x}}{{n}}+\frac{\left({n}−\mathrm{1}\right){cos}^{\mathrm{2}} {x}}{{n}}\right)= \\ $$$${xsin}^{{n}−\mathrm{2}} {x}×\left(\frac{{n}−\mathrm{1}}{{n}}\right)={xsin}^{{n}−\mathrm{2}} {x}×{f}\left({n}\right) \\ $$$$\Rightarrow{f}\left({n}\right)=\frac{{n}−\mathrm{1}}{{n}} \\ $$

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Question-197239

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