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Question Number 148077 by mathdanisur last updated on 25/Jul/21
∫ ((cos^n x)/(sinx)) dx = ?
$$\int\:\frac{{cos}^{\boldsymbol{{n}}} {x}}{{sinx}}\:{dx}\:=\:? \\ $$
Answered by Olaf_Thorendsen last updated on 26/Jul/21
I_n (x) = ∫((cos^n x)/(sinx)) dx • I_0 (x) = ∫(dx/(sinx)) dx Let t = tan(x/2) I_0 (x) = ∫(1/((2t)/(1+t^2 ))).((2dt)/(1+t^2 )) I_0 (x) = ln∣t∣+C = ln∣tan(x/2)∣+C • I_1 (x) = ∫((cosx)/(sinx)) dx = ln∣sinx∣+C • k ≥ 2 I_k (x)−I_(k−2) (x) = ∫((cos^(k−2) x(cos^2 x−1))/(sinx)) dx I_k (x)−I_(k−2) (x) = −∫sinx.cos^(k−2) x dx I_k (x)−I_(k−2) (x) = (1/(k−1))cos^(k−1) x • If k is even, k = 2p Σ_(p=1) ^n I_(2p) (x)−Σ_(p=1) ^n I_(2p−2) (x) = Σ_(p=1) ^n (1/(2p−1))cos^(2p−1) x I_(2n) (x)−I_0 (x) = Σ_(p=1) ^n ((cos^(2p−1) x)/(2p−1)) I_(2n) (x) = ln∣tan(x/2)∣+ Σ_(p=1) ^n ((cos^(2p−1) x)/(2p−1))+C • If k is odd, k = 2p+1 Σ_(p=1) ^n I_(2p+1) (x)−Σ_(p=1) ^n I_(2p−1) (x) = Σ_(p=1) ^n (1/(2p))cos^(2p) x I_(2n+1) (x)−I_1 (x) = (1/2)Σ_(p=1) ^n ((cos^(2p) x)/p) I_(2n+1) (x) = ln∣sinx∣+(1/2) Σ_(p=1) ^n ((cos^(2p) x)/p)+C
$$\mathrm{I}_{{n}} \left({x}\right)\:=\:\int\frac{\mathrm{cos}^{{n}} {x}}{\mathrm{sin}{x}}\:{dx} \\ $$$$ \\ $$$$\bullet\:\mathrm{I}_{\mathrm{0}} \left({x}\right)\:=\:\int\frac{{dx}}{\mathrm{sin}{x}}\:{dx} \\ $$$$\mathrm{Let}\:{t}\:=\:\mathrm{tan}\frac{{x}}{\mathrm{2}} \\ $$$$\mathrm{I}_{\mathrm{0}} \left({x}\right)\:=\:\int\frac{\mathrm{1}}{\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}.\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\mathrm{I}_{\mathrm{0}} \left({x}\right)\:=\:\mathrm{ln}\mid{t}\mid+\mathrm{C}\:=\:\mathrm{ln}\mid\mathrm{tan}\frac{{x}}{\mathrm{2}}\mid+\mathrm{C} \\ $$$$ \\ $$$$\bullet\:\mathrm{I}_{\mathrm{1}} \left({x}\right)\:=\:\int\frac{\mathrm{cos}{x}}{\mathrm{sin}{x}}\:{dx}\:=\:\mathrm{ln}\mid\mathrm{sin}{x}\mid+\mathrm{C} \\ $$$$ \\ $$$$\bullet\:{k}\:\geqslant\:\mathrm{2} \\ $$$$\mathrm{I}_{{k}} \left({x}\right)−\mathrm{I}_{{k}−\mathrm{2}} \left({x}\right)\:=\:\int\frac{\mathrm{cos}^{{k}−\mathrm{2}} {x}\left(\mathrm{cos}^{\mathrm{2}} {x}−\mathrm{1}\right)}{\mathrm{sin}{x}}\:{dx} \\ $$$$\mathrm{I}_{{k}} \left({x}\right)−\mathrm{I}_{{k}−\mathrm{2}} \left({x}\right)\:=\:−\int\mathrm{sin}{x}.\mathrm{cos}^{{k}−\mathrm{2}} {x}\:{dx} \\ $$$$\mathrm{I}_{{k}} \left({x}\right)−\mathrm{I}_{{k}−\mathrm{2}} \left({x}\right)\:=\:\frac{\mathrm{1}}{{k}−\mathrm{1}}\mathrm{cos}^{{k}−\mathrm{1}} {x} \\ $$$$ \\ $$$$\bullet\:\mathrm{If}\:{k}\:\mathrm{is}\:\mathrm{even},\:{k}\:=\:\mathrm{2}{p} \\ $$$$\underset{{p}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{I}_{\mathrm{2}{p}} \left({x}\right)−\underset{{p}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{I}_{\mathrm{2}{p}−\mathrm{2}} \left({x}\right)\:=\:\underset{{p}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{p}−\mathrm{1}}\mathrm{cos}^{\mathrm{2}{p}−\mathrm{1}} {x} \\ $$$$\mathrm{I}_{\mathrm{2}{n}} \left({x}\right)−\mathrm{I}_{\mathrm{0}} \left({x}\right)\:=\:\underset{{p}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{cos}^{\mathrm{2}{p}−\mathrm{1}} {x}}{\mathrm{2}{p}−\mathrm{1}} \\ $$$$\mathrm{I}_{\mathrm{2}{n}} \left({x}\right)\:=\:\mathrm{ln}\mid\mathrm{tan}\frac{{x}}{\mathrm{2}}\mid+\:\underset{{p}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{cos}^{\mathrm{2}{p}−\mathrm{1}} {x}}{\mathrm{2}{p}−\mathrm{1}}+\mathrm{C} \\ $$$$ \\ $$$$\bullet\:\mathrm{If}\:{k}\:\mathrm{is}\:\mathrm{odd},\:{k}\:=\:\mathrm{2}{p}+\mathrm{1} \\ $$$$\underset{{p}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{I}_{\mathrm{2}{p}+\mathrm{1}} \left({x}\right)−\underset{{p}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{I}_{\mathrm{2}{p}−\mathrm{1}} \left({x}\right)\:=\:\underset{{p}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{p}}\mathrm{cos}^{\mathrm{2}{p}} {x} \\ $$$$\mathrm{I}_{\mathrm{2}{n}+\mathrm{1}} \left({x}\right)−\mathrm{I}_{\mathrm{1}} \left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\underset{{p}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{cos}^{\mathrm{2}{p}} {x}}{{p}} \\ $$$$\mathrm{I}_{\mathrm{2}{n}+\mathrm{1}} \left({x}\right)\:=\:\mathrm{ln}\mid\mathrm{sin}{x}\mid+\frac{\mathrm{1}}{\mathrm{2}}\:\underset{{p}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{cos}^{\mathrm{2}{p}} {x}}{{p}}+\mathrm{C} \\ $$
Commented by mathdanisur last updated on 25/Jul/21
Thankyou Sir, answer.?
$${Thankyou}\:{Sir},\:{answer}.? \\ $$
Commented by puissant last updated on 26/Jul/21
prof desole mais I_1 =ln∣sin(x)∣+C car (ln(u))′=((u′)/u) avec u=sin(x)..
$${prof}\:{desole}\:{mais}\:{I}_{\mathrm{1}} ={ln}\mid{sin}\left({x}\right)\mid+{C} \\ $$$${car}\:\left({ln}\left({u}\right)\right)'=\frac{{u}'}{{u}}\:\:{avec}\:\:{u}={sin}\left({x}\right).. \\ $$
Commented by Olaf_Thorendsen last updated on 26/Jul/21
Bien vu ! (mais je ne suis pas prof).
$$\mathrm{Bien}\:\mathrm{vu}\:! \\ $$$$\left(\mathrm{mais}\:\mathrm{je}\:\mathrm{ne}\:\mathrm{suis}\:\mathrm{pas}\:\mathrm{prof}\right). \\ $$
Commented by puissant last updated on 26/Jul/21
d′accord monsieur olaf..
$${d}'{accord}\:{monsieur}\:{olaf}.. \\ $$
Answered by mathmax by abdo last updated on 25/Jul/21
A_n =∫ ((cos^n x)/(sinx))dx ⇒A_(2n) =∫ ((cos^(2n) x)/(sinx))dx =∫ (((1−sin^2 x)^n )/(sinx))dx =(−1)^n ∫ (((sin^2 x−1)^n )/(sinx))dx =(−1)^n ∫ ((Σ_(k=0) ^n C_n ^k sin^(2k) x(−1)^(n−k) )/(sinx))dx =Σ_(k=0) ^n C_n ^k (−1)^k ∫sin^(2k−1) x dx =∫ (dx/(sinx)) +Σ_(k=0) ^n C_n ^k (−1)^k ∫ (((e^(ix) −e^(−ix) )/2))^(2k−1) dx =∫ (dx/(sinx)) +Σ_(k=0) ^n (−1)^k C_n ^k (1/2^(2k−1) )∫ Σ_(p=0) ^(2k−1) (e^(ix) )^p (−e^(−ix) )^(2k−1−p ) dx =∫ (dx/(sinx)) +Σ_(k=0) ^n (((−1)^k )/2^(2k−1) )C_n ^k (Σ_(p=0) ^(2k−1) (−1)^(2k−1−p) ∫ e^(ipx) e^(−i(2k−1)x+ipxdx) =∫ (dx/(sinx)) +Σ_(k=0) ^n (((−1)^k )/2^(2k−1) )C_n ^k (Σ_(p=0) ^(2k−1) (−1)^(p+1) ∫ e^(i(2p−2k+1)x) dx) =∫ (dx/(sinx))+Σ_(k=0) ^n (((−1)^k )/2^(2k−1) )C_n ^k (Σ_(p=0) ^(2k−1) (((−1)^(p+1) )/(i(2p−2k+1)))e^(i(2p−2k+1)x) +λ) rest to find A_(2n+1) ....be continued....
$$\mathrm{A}_{\mathrm{n}} =\int\:\:\frac{\mathrm{cos}^{\mathrm{n}} \mathrm{x}}{\mathrm{sinx}}\mathrm{dx}\:\Rightarrow\mathrm{A}_{\mathrm{2n}} =\int\:\:\frac{\mathrm{cos}^{\mathrm{2n}} \mathrm{x}}{\mathrm{sinx}}\mathrm{dx} \\ $$$$=\int\:\:\frac{\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \mathrm{x}\right)^{\mathrm{n}} }{\mathrm{sinx}}\mathrm{dx}\:=\left(−\mathrm{1}\right)^{\mathrm{n}} \:\int\:\:\frac{\left(\mathrm{sin}^{\mathrm{2}} \mathrm{x}−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{sinx}}\mathrm{dx} \\ $$$$=\left(−\mathrm{1}\right)^{\mathrm{n}} \:\int\:\:\frac{\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}} \:\mathrm{C}_{\mathrm{n}} ^{\mathrm{k}} \:\mathrm{sin}^{\mathrm{2k}} \mathrm{x}\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{k}} }{\mathrm{sinx}}\mathrm{dx} \\ $$$$=\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}} \:\mathrm{C}_{\mathrm{n}} ^{\mathrm{k}} \:\left(−\mathrm{1}\right)^{\mathrm{k}} \:\:\int\mathrm{sin}^{\mathrm{2k}−\mathrm{1}} \:\mathrm{x}\:\mathrm{dx} \\ $$$$=\int\:\frac{\mathrm{dx}}{\mathrm{sinx}}\:+\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}} \:\mathrm{C}_{\mathrm{n}} ^{\mathrm{k}} \left(−\mathrm{1}\right)^{\mathrm{k}} \:\:\int\:\:\left(\frac{\mathrm{e}^{\mathrm{ix}} −\mathrm{e}^{−\mathrm{ix}} }{\mathrm{2}}\right)^{\mathrm{2k}−\mathrm{1}} \mathrm{dx} \\ $$$$=\int\:\:\frac{\mathrm{dx}}{\mathrm{sinx}}\:+\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}} \:\left(−\mathrm{1}\right)^{\mathrm{k}} \:\mathrm{C}_{\mathrm{n}} ^{\mathrm{k}} \:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2k}−\mathrm{1}} }\int\:\sum_{\mathrm{p}=\mathrm{0}} ^{\mathrm{2k}−\mathrm{1}} \:\left(\mathrm{e}^{\mathrm{ix}} \right)^{\mathrm{p}} \:\left(−\mathrm{e}^{−\mathrm{ix}} \right)^{\mathrm{2k}−\mathrm{1}−\mathrm{p}\:} \mathrm{dx} \\ $$$$=\int\:\frac{\mathrm{dx}}{\mathrm{sinx}}\:+\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{k}} }{\mathrm{2}^{\mathrm{2k}−\mathrm{1}} }\mathrm{C}_{\mathrm{n}} ^{\mathrm{k}} \:\left(\sum_{\mathrm{p}=\mathrm{0}} ^{\mathrm{2k}−\mathrm{1}} \left(−\mathrm{1}\right)^{\mathrm{2k}−\mathrm{1}−\mathrm{p}} \:\int\:\:\mathrm{e}^{\mathrm{ipx}} \:\mathrm{e}^{−\mathrm{i}\left(\mathrm{2k}−\mathrm{1}\right)\mathrm{x}+\mathrm{ipxdx}} \right. \\ $$$$=\int\:\frac{\mathrm{dx}}{\mathrm{sinx}}\:+\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{k}} }{\mathrm{2}^{\mathrm{2k}−\mathrm{1}} }\mathrm{C}_{\mathrm{n}} ^{\mathrm{k}} \left(\sum_{\mathrm{p}=\mathrm{0}} ^{\mathrm{2k}−\mathrm{1}} \left(−\mathrm{1}\right)^{\mathrm{p}+\mathrm{1}} \int\:\:\mathrm{e}^{\mathrm{i}\left(\mathrm{2p}−\mathrm{2k}+\mathrm{1}\right)\mathrm{x}} \mathrm{dx}\right) \\ $$$$=\int\:\frac{\mathrm{dx}}{\mathrm{sinx}}+\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{k}} }{\mathrm{2}^{\mathrm{2k}−\mathrm{1}} }\mathrm{C}_{\mathrm{n}} ^{\mathrm{k}} \:\left(\sum_{\mathrm{p}=\mathrm{0}} ^{\mathrm{2k}−\mathrm{1}} \frac{\left(−\mathrm{1}\right)^{\mathrm{p}+\mathrm{1}} }{\mathrm{i}\left(\mathrm{2p}−\mathrm{2k}+\mathrm{1}\right)}\mathrm{e}^{\mathrm{i}\left(\mathrm{2p}−\mathrm{2k}+\mathrm{1}\right)\mathrm{x}} +\lambda\right) \\ $$$$\mathrm{rest}\:\mathrm{to}\:\mathrm{find}\:\mathrm{A}_{\mathrm{2n}+\mathrm{1}} ….\mathrm{be}\:\mathrm{continued}…. \\ $$
Commented by mathdanisur last updated on 25/Jul/21
Thankyou Sir, answer.?
$${Thankyou}\:{Sir},\:{answer}.? \\ $$
Commented by mathmax by abdo last updated on 25/Jul/21
A_(2n) =∫ (dx/(sinx))+Σ_(k=1) ^n (....)
$$\mathrm{A}_{\mathrm{2n}} =\int\:\frac{\mathrm{dx}}{\mathrm{sinx}}+\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \left(….\right) \\ $$

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