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Question Number 148077 by mathdanisur last updated on 25/Jul/21

$$\int \frac{{cos}^{\mathit{n}}x}{sinx}dx=?$$

Answered by Olaf_Thorendsen last updated on 26/Jul/21

$${\mathrm{I}}_n\left(x\right)=\int \frac{{\cos }^{n}x}{\sin x}dx$$$$$$$$\bullet {\mathrm{I}}_0\left(x\right)=\int \frac{dx}{\sin x}dx$$$$\mathrm{Let}t=\tan \frac{x}{2}$$$${\mathrm{I}}_0\left(x\right)=\int \frac{1}{\frac{2t}{1+t^2}}.\frac{2dt}{1+t^2}$$$${\mathrm{I}}_0\left(x\right)=\ln \mid t\mid +\mathrm{C}=\ln \mid \tan \frac{x}{2}\mid +\mathrm{C}$$$$$$$$\bullet {\mathrm{I}}_1\left(x\right)=\int \frac{\cos x}{\sin x}dx=\ln \mid \sin x\mid +\mathrm{C}$$$$$$$$\bullet k⩾2$$$${\mathrm{I}}_k\left(x\right)-{\mathrm{I}}_{k-2}\left(x\right)=\int \frac{{\cos }^{k-2}x({\cos }^{2}x-1)}{\sin x}dx$$$${\mathrm{I}}_k\left(x\right)-{\mathrm{I}}_{k-2}\left(x\right)=-\int \sin x.{\cos }^{k-2}xdx$$$${\mathrm{I}}_k\left(x\right)-{\mathrm{I}}_{k-2}\left(x\right)=\frac{1}{k-1}{\cos }^{k-1}x$$$$$$$$\bullet \mathrm{If}k\mathrm{is}\mathrm{even},k=2p$$$$\underset{p=1}{\sum ^n}{\mathrm{I}}_{2p}\left(x\right)-\underset{p=1}{\sum ^n}{\mathrm{I}}_{2p-2}\left(x\right)=\underset{p=1}{\sum ^n}\frac{1}{2p-1}{\cos }^{2p-1}x$$$${\mathrm{I}}_{2n}\left(x\right)-{\mathrm{I}}_0\left(x\right)=\underset{p=1}{\sum ^n}\frac{{\cos }^{2p-1}x}{2p-1}$$$${\mathrm{I}}_{2n}\left(x\right)=\ln \mid \tan \frac{x}{2}\mid +\underset{p=1}{\sum ^n}\frac{{\cos }^{2p-1}x}{2p-1}+\mathrm{C}$$$$$$$$\bullet \mathrm{If}k\mathrm{is}\mathrm{odd},k=2p+1$$$$\underset{p=1}{\sum ^n}{\mathrm{I}}_{2p+1}\left(x\right)-\underset{p=1}{\sum ^n}{\mathrm{I}}_{2p-1}\left(x\right)=\underset{p=1}{\sum ^n}\frac{1}{2p}{\cos }^{2p}x$$$${\mathrm{I}}_{2n+1}\left(x\right)-{\mathrm{I}}_1\left(x\right)=\frac{1}{2}\underset{p=1}{\sum ^n}\frac{{\cos }^{2p}x}{p}$$$${\mathrm{I}}_{2n+1}\left(x\right)=\ln \mid \sin x\mid +\frac{1}{2}\underset{p=1}{\sum ^n}\frac{{\cos }^{2p}x}{p}+\mathrm{C}$$

Commented by mathdanisur last updated on 25/Jul/21

$$ThankyouSir,answer.?$$

Commented by puissant last updated on 26/Jul/21

$$profdesolemais{I}_1=ln\mid sin\left(x\right)\mid +C$$$$car{\left(ln\left(u\right)\right)}^{\prime }=\frac{u^{\prime }}{u}avecu=sin\left(x\right)..$$

Commented by Olaf_Thorendsen last updated on 26/Jul/21

$$\mathrm{Bien}\mathrm{vu}!$$$$\left(\mathrm{mais}\mathrm{je}\mathrm{ne}\mathrm{suis}\mathrm{pas}\mathrm{prof}\right).$$

Commented by puissant last updated on 26/Jul/21

$$d^{\prime }accordmonsieurolaf..$$

Answered by mathmax by abdo last updated on 25/Jul/21

$${\mathrm{A}}_{\mathrm{n}}=\int \frac{{\cos }^{\mathrm{n}}\mathrm{x}}{\mathrm{sinx}}\mathrm{dx}\Rightarrow {\mathrm{A}}_{2\mathrm{n}}=\int \frac{{\cos }^{2\mathrm{n}}\mathrm{x}}{\mathrm{sinx}}\mathrm{dx}$$$$=\int \frac{{(1-{\sin }^2\mathrm{x})}^{\mathrm{n}}}{\mathrm{sinx}}\mathrm{dx}={(-1)}^{\mathrm{n}}\int \frac{{({\sin }^2\mathrm{x}-1)}^{\mathrm{n}}}{\mathrm{sinx}}\mathrm{dx}$$$$={(-1)}^{\mathrm{n}}\int \frac{\sum _{\mathrm{k}=0}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}{\sin }^{2\mathrm{k}}\mathrm{x}{(-1)}^{\mathrm{n}-\mathrm{k}}}{\mathrm{sinx}}\mathrm{dx}$$$$=\sum _{\mathrm{k}=0}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}{(-1)}^{\mathrm{k}}\int {\sin }^{2\mathrm{k}-1}\mathrm{x}\mathrm{dx}$$$$=\int \frac{\mathrm{dx}}{\mathrm{sinx}}+\sum _{\mathrm{k}=0}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}{(-1)}^{\mathrm{k}}\int {\left(\frac{{\mathrm{e}}^{\mathrm{ix}}-{\mathrm{e}}^{-\mathrm{ix}}}{2}\right)}^{2\mathrm{k}-1}\mathrm{dx}$$$$=\int \frac{\mathrm{dx}}{\mathrm{sinx}}+\sum _{\mathrm{k}=0}^{\mathrm{n}}{(-1)}^{\mathrm{k}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}\frac{1}{2^{2\mathrm{k}-1}}\int \sum _{\mathrm{p}=0}^{2\mathrm{k}-1}{\left({\mathrm{e}}^{\mathrm{ix}}\right)}^{\mathrm{p}}{(-{\mathrm{e}}^{-\mathrm{ix}})}^{2\mathrm{k}-1-\mathrm{p}}\mathrm{dx}$$$$=\int \frac{\mathrm{dx}}{\mathrm{sinx}}+\sum _{\mathrm{k}=0}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{2^{2\mathrm{k}-1}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}(\sum _{\mathrm{p}=0}^{2\mathrm{k}-1}{(-1)}^{2\mathrm{k}-1-\mathrm{p}}\int {\mathrm{e}}^{\mathrm{ipx}}{\mathrm{e}}^{-\mathrm{i}(2\mathrm{k}-1)\mathrm{x}+\mathrm{ipxdx}}$$$$=\int \frac{\mathrm{dx}}{\mathrm{sinx}}+\sum _{\mathrm{k}=0}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{2^{2\mathrm{k}-1}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}(\sum _{\mathrm{p}=0}^{2\mathrm{k}-1}{(-1)}^{\mathrm{p}+1}\int {\mathrm{e}}^{\mathrm{i}(2\mathrm{p}-2\mathrm{k}+1)\mathrm{x}}\mathrm{dx})$$$$=\int \frac{\mathrm{dx}}{\mathrm{sinx}}+\sum _{\mathrm{k}=0}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{2^{2\mathrm{k}-1}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}(\sum _{\mathrm{p}=0}^{2\mathrm{k}-1}\frac{{(-1)}^{\mathrm{p}+1}}{\mathrm{i}(2\mathrm{p}-2\mathrm{k}+1)}{\mathrm{e}}^{\mathrm{i}(2\mathrm{p}-2\mathrm{k}+1)\mathrm{x}}+\lambda )$$$$\mathrm{rest}\mathrm{to}\mathrm{find}{\mathrm{A}}_{2\mathrm{n}+1}....\mathrm{be}\mathrm{continued}....$$

Commented by mathdanisur last updated on 25/Jul/21

$$ThankyouSir,answer.?$$

Commented by mathmax by abdo last updated on 25/Jul/21

$${\mathrm{A}}_{2\mathrm{n}}=\int \frac{\mathrm{dx}}{\mathrm{sinx}}+\sum _{\mathrm{k}=1}^{\mathrm{n}}(....)$$

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