Question Number 46101 by Saorey last updated on 21/Oct/18
$$\mathrm{I}=\int\frac{\mathrm{x}^{\mathrm{n}+\mathrm{1}} }{\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{n}} }}\mathrm{dx}=? \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 21/Oct/18
![x^n =tan^2 θ so nx^(n−1) dx=2tanθsec^2 θdθ dx=(2/n)×((tanθsec^2 θ)/([(tanθ)^(2/n) ]^(n−1) ))dθ ∫(([(tanθ)^(2/n) ]^(n+1) )/(secθ))×(2/n)×((tanθsec^2 θ)/([(tanθ)^(2/n) ]^(n−1) ))dθ (2/n)∫tanθsecθ×(tanθ)^(((2n+2)/n)−((2n−2)/n)) dθ (2/n)∫tanθsecθtan^(4/n) θ dθ (2/n)∫tanθsecθ(tan^2 θ)^(2/n) dθ (2/n)∫tanθsecθ(sec^2 θ−1)^(2/n) dθ k=secθ dk=secθtanθdθ (2/n)∫(k^2 −1)^(2/n) dk p=(2/n) p∫(k^2 −1)^p dk=pI_p I_p =∫(k^2 −1)^p dk =(k^2 −1)^p k−∫p(k^2 −1)^(p−1) ×2k×kdk =(k^2 −1)^p k−2p∫(k^2 −1)^(p−1) (k^2 −1+1)dk =k(k^2 −1)^p −2p[∫(k^2 −1)^p +∫(k^2 −1)^(p−1) ]dk =k(k^2 −1)^p −2p×I_p −2pI_(p−1) (2p+1)I_p =k(k^2 −1)^p −2pI_(p−1) I_p =(k/(2p+1))(k^2 −1)^p −((2p)/(2p+1))I_(p−1) pI_p =((kp)/(2p+1))(k^2 −1)^p −((2p^2 )/(2p+2))I_(p−1) (2/n)I_(2/n) =((k((2/n)))/(2((2/n))+1))(k^2 −1)^(2/n) −((2((2/n))^2 )/(2((2/n))+2))I_((2/n)−1) (2/n)I_(2/n) =((2/n)/((4/n)+1))secθ×tan^(4/n) θ−((8/n^2 )/((4/n)+2))I_((2/n)−1) contd...](https://www.tinkutara.com/question/Q46111.png)
$${x}^{{n}} ={tan}^{\mathrm{2}} \theta\:\:\:\:{so}\:{nx}^{{n}−\mathrm{1}} {dx}=\mathrm{2}{tan}\theta{sec}^{\mathrm{2}} \theta{d}\theta \\ $$$${dx}=\frac{\mathrm{2}}{{n}}×\frac{{tan}\theta{sec}^{\mathrm{2}} \theta}{\left[\left({tan}\theta\right)^{\frac{\mathrm{2}}{{n}}} \right]^{{n}−\mathrm{1}} }{d}\theta \\ $$$$\int\frac{\left[\left({tan}\theta\right)^{\frac{\mathrm{2}}{{n}}} \right]^{{n}+\mathrm{1}} }{{sec}\theta}×\frac{\mathrm{2}}{{n}}×\frac{{tan}\theta{sec}^{\mathrm{2}} \theta}{\left[\left({tan}\theta\right)^{\frac{\mathrm{2}}{{n}}} \right]^{{n}−\mathrm{1}} }{d}\theta \\ $$$$\frac{\mathrm{2}}{{n}}\int{tan}\theta{sec}\theta×\left({tan}\theta\right)^{\frac{\mathrm{2}{n}+\mathrm{2}}{{n}}−\frac{\mathrm{2}{n}−\mathrm{2}}{{n}}} {d}\theta \\ $$$$\frac{\mathrm{2}}{{n}}\int{tan}\theta{sec}\theta{tan}^{\frac{\mathrm{4}}{{n}}} \theta\:{d}\theta \\ $$$$\frac{\mathrm{2}}{{n}}\int{tan}\theta{sec}\theta\left({tan}^{\mathrm{2}} \theta\right)^{\frac{\mathrm{2}}{{n}}} {d}\theta \\ $$$$\frac{\mathrm{2}}{{n}}\int{tan}\theta{sec}\theta\left({sec}^{\mathrm{2}} \theta−\mathrm{1}\right)^{\frac{\mathrm{2}}{{n}}} {d}\theta \\ $$$${k}={sec}\theta\:\:\:{dk}={sec}\theta{tan}\theta{d}\theta \\ $$$$\frac{\mathrm{2}}{{n}}\int\left({k}^{\mathrm{2}} −\mathrm{1}\right)^{\frac{\mathrm{2}}{{n}}} {dk} \\ $$$${p}=\frac{\mathrm{2}}{{n}} \\ $$$${p}\int\left({k}^{\mathrm{2}} −\mathrm{1}\right)^{{p}} {dk}={pI}_{{p}} \\ $$$${I}_{{p}} =\int\left({k}^{\mathrm{2}} −\mathrm{1}\right)^{{p}} {dk} \\ $$$$\:\:=\left({k}^{\mathrm{2}} −\mathrm{1}\right)^{{p}} {k}−\int{p}\left({k}^{\mathrm{2}} −\mathrm{1}\right)^{{p}−\mathrm{1}} ×\mathrm{2}{k}×{kdk} \\ $$$$=\left({k}^{\mathrm{2}} −\mathrm{1}\right)^{{p}} {k}−\mathrm{2}{p}\int\left({k}^{\mathrm{2}} −\mathrm{1}\right)^{{p}−\mathrm{1}} \left({k}^{\mathrm{2}} −\mathrm{1}+\mathrm{1}\right){dk} \\ $$$$={k}\left({k}^{\mathrm{2}} −\mathrm{1}\right)^{{p}} −\mathrm{2}{p}\left[\int\left({k}^{\mathrm{2}} −\mathrm{1}\right)^{{p}} +\int\left({k}^{\mathrm{2}} −\mathrm{1}\right)^{{p}−\mathrm{1}} \right]{dk} \\ $$$$={k}\left({k}^{\mathrm{2}} −\mathrm{1}\right)^{{p}} −\mathrm{2}{p}×{I}_{{p}} −\mathrm{2}{pI}_{{p}−\mathrm{1}} \\ $$$$\left(\mathrm{2}{p}+\mathrm{1}\right){I}_{{p}} ={k}\left({k}^{\mathrm{2}} −\mathrm{1}\right)^{{p}} −\mathrm{2}{pI}_{{p}−\mathrm{1}} \\ $$$${I}_{{p}} =\frac{{k}}{\mathrm{2}{p}+\mathrm{1}}\left({k}^{\mathrm{2}} −\mathrm{1}\right)^{{p}} −\frac{\mathrm{2}{p}}{\mathrm{2}{p}+\mathrm{1}}{I}_{{p}−\mathrm{1}} \\ $$$${pI}_{{p}} =\frac{{kp}}{\mathrm{2}{p}+\mathrm{1}}\left({k}^{\mathrm{2}} −\mathrm{1}\right)^{{p}} −\frac{\mathrm{2}{p}^{\mathrm{2}} }{\mathrm{2}{p}+\mathrm{2}}{I}_{{p}−\mathrm{1}} \\ $$$$\frac{\mathrm{2}}{{n}}{I}_{\frac{\mathrm{2}}{{n}}} =\frac{{k}\left(\frac{\mathrm{2}}{{n}}\right)}{\mathrm{2}\left(\frac{\mathrm{2}}{{n}}\right)+\mathrm{1}}\left({k}^{\mathrm{2}} −\mathrm{1}\right)^{\frac{\mathrm{2}}{{n}}} −\frac{\mathrm{2}\left(\frac{\mathrm{2}}{{n}}\right)^{\mathrm{2}} }{\mathrm{2}\left(\frac{\mathrm{2}}{{n}}\right)+\mathrm{2}}{I}_{\frac{\mathrm{2}}{{n}}−\mathrm{1}} \\ $$$$\frac{\mathrm{2}}{{n}}{I}_{\frac{\mathrm{2}}{{n}}} =\frac{\frac{\mathrm{2}}{{n}}}{\frac{\mathrm{4}}{{n}}+\mathrm{1}}{sec}\theta×{tan}^{\frac{\mathrm{4}}{{n}}} \theta−\frac{\frac{\mathrm{8}}{{n}^{\mathrm{2}} }}{\frac{\mathrm{4}}{{n}}+\mathrm{2}}{I}_{\frac{\mathrm{2}}{{n}}−\mathrm{1}} \\ $$$${contd}… \\ $$$$ \\ $$