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Question Number 59732 by maxmathsup by imad last updated on 14/May/19

$$find{I}_n=\int \frac{dx}{{sin}^{n}x}withnintegrnatural.$$

Answered by tanmay last updated on 14/May/19

$$I_n=\int {cosec}^{n}xdx$$$$letn=evennumber$$$$\int {cosec}^{n-2}x{cozec}^{2}xdx$$$$\int {\left({cosec}^{2}x\right)}^{\frac{n-2}{2}}{cosec}^{2}xdx$$$$\int {(1+{cot}^{2}x)}^{\frac{n-2}{2}}{cosec}^{2}xdx$$$$a=cotx$$$$da=-{cosec}^{2}xdx$$$$\int {(1+a^2)}^{\frac{n-2}{2}}\times -da$$$$=(-1)\int [1+\frac{n-2}{2}{C}_1{a}^2+\frac{n-2}{2}{C}_2{a}^4+...+{\left(a^2\right)}^{\frac{n-2}{2}}]da$$$$=(-1)[a+\frac{n-2}{2}{C}_1\times \frac{a^3}{3}+\frac{n-2}{2}{C}_2\times \frac{a^5}{5}+..+\frac{a^{n-1}}{n-1}]$$$$nowplsputcotx=a]$$$$whenn=odd$$$$iamtrying...$$

Commented by Tawa1 last updated on 14/May/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Answered by MJS last updated on 14/May/19

$$\mathrm{trying}\mathrm{Weierstrass}$$$$x=2\arctan t$$$$\sin x=\frac{2t}{t^2+1};dx=\frac{2dt}{t^2+1}$$$$I_n=\int \frac{dx}{{\sin }^{n}x}=\frac{1}{2^{n-1}}\int \frac{{(t^2+1)}^{n-1}}{t^n}dt=$$$$=\frac{1}{2^{n-1}}\int \left(\underset{k=1}{\sum ^n}\left(\begin{array}{c}n-1\\ k-1\end{array}\right)\frac{t^{2k-2}}{t^n}\right)dt=$$$$=\frac{(n-1)!}{2^{n-1}}\int \left(\frac{1}{t^2}\underset{k=1}{\sum ^n}\frac{t^{2k-n}}{(k-1)!(n-k)!}\right)dt$$$$I_1=\int \frac{1}{t}dt$$$$I_2=\frac{1}{2}\int (\frac{1}{t^2}+1)dt$$$$I_3=\frac{1}{4}\int (\frac{1}{t^3}+\frac{2}{t}+t)dt$$$$I_4=\frac{1}{8}\int (\frac{1}{t^4}+\frac{3}{t^2}+3+t^2)dt$$$$I_5=\frac{1}{16}\int (\frac{1}{t^5}+\frac{4}{t^3}+\frac{6}{t}+4t+t^3)dt$$$$I_6=\frac{1}{32}\int (\frac{1}{t^6}+\frac{5}{t^4}+\frac{10}{t^2}+10+5t^2+t^4)dt$$$$...$$

Commented by maxmathsup by imad last updated on 14/May/19

$$thankyousir.$$

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