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Question Number 17600 by mondodotto@gmail.com last updated on 08/Jul/17

Answered by alex041103 last updated on 08/Jul/17

$$\mathrm{We}\mathrm{use}\mathrm{the}\mathrm{following}\mathrm{trig}\mathrm{sub}:$$$$x=tan\left(\theta \right)\Rightarrow dx={sec}^2\left(\theta \right)d\theta$$$$\mathrm{We}\mathrm{know}\mathrm{that}1+{tan}^2\theta ={sec}^2\theta$$$$\Rightarrow \frac{dx}{{(1+x^2)}^n}=\frac{{sec}^2\theta d\theta }{{sec}^{2n}\theta }=\frac{d\theta }{{sec}^{2(n-1)}\theta }$$$$\mathrm{But}\frac{1}{\sec \mathrm{x}}=\cos \mathrm{x}$$$$\Rightarrow \frac{d\theta }{{sec}^{2(n-1)}\theta }={cos}^{2(n-1)}\theta d\theta$$$$\mathrm{Now}\mathrm{we}\mathrm{change}\mathrm{the}\mathrm{limits}\mathrm{of}\mathrm{integration}:$$$$x=tan\theta \Rightarrow \theta ={tan}^{-1}x$$$$\Rightarrow {\mathrm{I}}_n=\underset{{tan}^{-1}\left(0\right)}{\stackrel{{tan}^{-1}\left(1\right)}{\int }}{cos}^{2(n-1)}\theta d\theta$$$$=\underset{0}{\stackrel{\pi /4}{\int }}{cos}^{2(n-1)}\theta d\theta$$$$\Rightarrow {\mathrm{I}}_{n+1}=\underset{0}{\stackrel{\pi /4}{\int }}{cos}^{2n}\theta d\theta$$$$\mathrm{Now}\mathrm{we}\mathrm{use}\mathrm{integration}\mathrm{by}\mathrm{parts}:$$$${\mathrm{I}}_{n+1}=\underset{0}{\stackrel{\pi /4}{\int }}{cos}^{2n-1}\theta \cos \theta d\theta$$$$u={cos}^{2n-1}\theta \Rightarrow du=(2n-1){cos}^{2(n-1)}\theta (-sin\theta )$$$$dv=cos\theta d\theta \Rightarrow v=sin\theta$$$$\text{Missing left or extra right}$$$$\Rightarrow {\mathrm{I}}_{n+1}=\underset{0}{\stackrel{\pi /4}{\int }}{cos}^{2n-1}\theta \cos \theta d\theta =$$$$\text{Missing left or extra right}$$$$\text{Missing left or extra right}$$$$\text{Missing left or extra right}$$$$={\left(\frac{1}{\sqrt{2}}\right)}^{2n}=\frac{1}{2^n}=2^{-n}$$$$\mathrm{Let}\mathrm{substitute}{sin}^2\theta =1-{cos}^2\theta$$$${\mathrm{I}}_{n+1}=2^{-n}+(2n-1)\underset{0}{\stackrel{\pi /4}{\int }}(1-{cos}^2\theta ){cos}^{2(n-1)}\theta d\theta =$$$$=2^{-n}+(2n-1)[\underset{0}{\stackrel{\pi /4}{\int }}{cos}^{2(n-1)}\theta d\theta -\underset{0}{\stackrel{\pi /4}{\int }}{cos}^{2n}\theta d\theta ]$$$$\mathrm{But}$$$$\underset{0}{\stackrel{\pi /4}{\int }}{cos}^{2(n-1)}\theta d\theta ={\mathrm{I}}_n\mathrm{and}$$$$\underset{0}{\stackrel{\pi /4}{\int }}{cos}^{2n}\theta d\theta ={\mathrm{I}}_{n+1}$$$$\Rightarrow {\mathrm{I}}_{n+1}=2^{-n}+(2n-1)({\mathrm{I}}_n-{\mathrm{I}}_{n+1})$$$${\mathrm{I}}_{n+1}=2^{-n}+(2n-1){\mathrm{I}}_n-(2n-1){\mathrm{I}}_{n+1}$$$$\Rightarrow 2n{\mathrm{I}}_{n+1}=2^{-n}+(2n-1){\mathrm{I}}_n$$$$Anyquestions?$$

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