Question Number 59732 by maxmathsup by imad last updated on 14/May/19
$${find}\:{I}_{{n}} =\int\:\:\frac{{dx}}{{sin}^{{n}} {x}}\:\:{with}\:\:{n}\:{integr}\:{natural}. \\ $$
Answered by tanmay last updated on 14/May/19
![I_n =∫cosec^n x dx let n=even number ∫cosec^(n−2) x cozec^2 xdx ∫(cosec^2 x)^((n−2)/2) cosec^2 xdx ∫(1+cot^2 x)^((n−2)/2) cosec^2 xdx a=cotx da=−cosec^2 xdx ∫(1+a^2 )^((n−2)/2) ×−da =(−1)∫[1+((n−2)/2)C_1 a^2 +((n−2)/2)C_2 a^4 +...+(a^2 )^((n−2)/2) ]da =(−1)[a+((n−2)/2)C_1 ×(a^3 /3)+((n−2)/2)C_2 ×(a^5 /5)+..+(a^(n−1) /(n−1))] now pls put cotx=a] when n=odd i am trying...](https://www.tinkutara.com/question/Q59741.png)
$${I}_{{n}} =\int{cosec}^{{n}} {x}\:{dx} \\ $$$${let}\:{n}={even}\:{number} \\ $$$$\int{cosec}^{{n}−\mathrm{2}} {x}\:{cozec}^{\mathrm{2}} {xdx} \\ $$$$\int\left({cosec}^{\mathrm{2}} {x}\right)^{\frac{{n}−\mathrm{2}}{\mathrm{2}}} {cosec}^{\mathrm{2}} {xdx} \\ $$$$\int\left(\mathrm{1}+{cot}^{\mathrm{2}} {x}\right)^{\frac{{n}−\mathrm{2}}{\mathrm{2}}} {cosec}^{\mathrm{2}} {xdx} \\ $$$${a}={cotx} \\ $$$${da}=−{cosec}^{\mathrm{2}} {xdx} \\ $$$$\int\left(\mathrm{1}+{a}^{\mathrm{2}} \right)^{\frac{{n}−\mathrm{2}}{\mathrm{2}}} ×−{da} \\ $$$$=\left(−\mathrm{1}\right)\int\left[\mathrm{1}+\frac{{n}−\mathrm{2}}{\mathrm{2}}{C}_{\mathrm{1}} {a}^{\mathrm{2}} +\frac{{n}−\mathrm{2}}{\mathrm{2}}{C}_{\mathrm{2}} {a}^{\mathrm{4}} +…+\left({a}^{\mathrm{2}} \right)^{\frac{{n}−\mathrm{2}}{\mathrm{2}}} \right]{da} \\ $$$$=\left(−\mathrm{1}\right)\left[{a}+\frac{{n}−\mathrm{2}}{\mathrm{2}}{C}_{\mathrm{1}} ×\frac{{a}^{\mathrm{3}} }{\mathrm{3}}+\frac{{n}−\mathrm{2}}{\mathrm{2}}{C}_{\mathrm{2}} ×\frac{{a}^{\mathrm{5}} }{\mathrm{5}}+..+\frac{{a}^{{n}−\mathrm{1}} }{{n}−\mathrm{1}}\right] \\ $$$$\left.{now}\:{pls}\:{put}\:{cotx}={a}\right] \\ $$$${when}\:{n}={odd} \\ $$$${i}\:{am}\:{trying}… \\ $$
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
$$\mathrm{trying}\:\mathrm{Weierstrass} \\ $$$${x}=\mathrm{2arctan}\:{t} \\ $$$$\mathrm{sin}\:{x}\:=\frac{\mathrm{2}{t}}{{t}^{\mathrm{2}} +\mathrm{1}};\:{dx}=\frac{\mathrm{2}{dt}}{{t}^{\mathrm{2}} +\mathrm{1}} \\ $$$${I}_{{n}} =\int\frac{{dx}}{\mathrm{sin}^{{n}} \:{x}}=\frac{\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{1}} }\int\frac{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{{n}−\mathrm{1}} }{{t}^{{n}} }{dt}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{1}} }\int\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}−\mathrm{1}}\\{{k}−\mathrm{1}}\end{pmatrix}\frac{{t}^{\mathrm{2}{k}−\mathrm{2}} }{{t}^{{n}} }\right){dt}= \\ $$$$=\frac{\left({n}−\mathrm{1}\right)!}{\mathrm{2}^{{n}−\mathrm{1}} }\int\left(\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{t}^{\mathrm{2}{k}−{n}} }{\left({k}−\mathrm{1}\right)!\left({n}−{k}\right)!}\right){dt} \\ $$$${I}_{\mathrm{1}} =\int\frac{\mathrm{1}}{{t}}{dt} \\ $$$${I}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\int\left(\frac{\mathrm{1}}{{t}^{\mathrm{2}} }+\mathrm{1}\right){dt} \\ $$$${I}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{4}}\int\left(\frac{\mathrm{1}}{{t}^{\mathrm{3}} }+\frac{\mathrm{2}}{{t}}+{t}\right){dt} \\ $$$${I}_{\mathrm{4}} =\frac{\mathrm{1}}{\mathrm{8}}\int\left(\frac{\mathrm{1}}{{t}^{\mathrm{4}} }+\frac{\mathrm{3}}{{t}^{\mathrm{2}} }+\mathrm{3}+{t}^{\mathrm{2}} \right){dt} \\ $$$${I}_{\mathrm{5}} =\frac{\mathrm{1}}{\mathrm{16}}\int\left(\frac{\mathrm{1}}{{t}^{\mathrm{5}} }+\frac{\mathrm{4}}{{t}^{\mathrm{3}} }+\frac{\mathrm{6}}{{t}}+\mathrm{4}{t}+{t}^{\mathrm{3}} \right){dt} \\ $$$${I}_{\mathrm{6}} =\frac{\mathrm{1}}{\mathrm{32}}\int\left(\frac{\mathrm{1}}{{t}^{\mathrm{6}} }+\frac{\mathrm{5}}{{t}^{\mathrm{4}} }+\frac{\mathrm{10}}{{t}^{\mathrm{2}} }+\mathrm{10}+\mathrm{5}{t}^{\mathrm{2}} +{t}^{\mathrm{4}} \right){dt} \\ $$$$… \\ $$
Commented by maxmathsup by imad last updated on 14/May/19
$${thank}\:{you}\:{sir}. \\ $$