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Question Number 14014 by tawa tawa last updated on 26/May/17

$$\int {\cos }^{\mathrm{n}}\left(\mathrm{x}\right)\mathrm{dx}$$$$\mathrm{please}\mathrm{i}\mathrm{need}\mathrm{workings}.$$

Answered by mrW1 last updated on 26/May/17

$$I_n=\int {\cos }^{\mathrm{n}}\left(\mathrm{x}\right)\mathrm{dx}=\int {\cos }^{n-1}\left(x\right)d\sin \left(x\right)$$$$using\int udv=uv-\int vdu$$$$=\sin \left(x\right){\cos }^{n-1}\left(x\right)+(n-1)\int {\sin }^2\left(x\right){\cos }^{n-2}\left(x\right)dx$$$$=\sin \left(x\right){\cos }^{n-1}\left(x\right)+(n-1)\int [1-{\cos }^2\left(x\right)]{\cos }^{n-2}\left(x\right)dx$$$$=\sin \left(x\right){\cos }^{n-1}\left(x\right)+(n-1)\int {\cos }^{n-2}\left(x\right)dx-(n-1)\int {\cos }^n\left(x\right)dx$$$$\Rightarrow I_n=\sin \left(x\right){\cos }^{n-1}\left(x\right)+(n-1)\int {\cos }^{n-2}\left(x\right)dx-(n-1)I_n$$$$\Rightarrow {nI}_n=\sin \left(x\right){\cos }^{n-1}\left(x\right)+(n-1)I_{n-2}$$$$\Rightarrow I_n=\frac{1}{n}\sin \left(x\right){\cos }^{n-1}\left(x\right)+\frac{n-1}{n}{I}_{n-2}$$$$\Rightarrow I_{n-2}=\frac{1}{n-2}\sin \left(x\right){\cos }^{n-3}\left(x\right)+\frac{n-3}{n-2}{I}_{n-4}$$$$......ifn=even....$$$$\Rightarrow I_2=\frac{1}{2}\sin \left(x\right)\cos \left(x\right)+\frac{1}{2}{I}_0$$$$\Rightarrow I_0=\int dx=x+C$$$$$$$$......ifn=odd....$$$$\Rightarrow I_3=\frac{1}{3}\sin \left(x\right){\cos }^2\left(x\right)+\frac{2}{3}{I}_1$$$$\Rightarrow I_1=\int \cos \left(x\right)dx=\sin \left(x\right)+C$$

Commented by tawa tawa last updated on 26/May/17

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

Commented by mrW1 last updated on 27/May/17

$$Ifnisevenweget:$$$$I_n=\int {\cos }^n\left(x\right)dx$$$$=\frac{1}{n}\sin \left(x\right){\cos }^{n-1}\left(x\right)$$$$+\frac{n-1}{n}\cdot \frac{1}{n-2}\sin \left(x\right){\cos }^{n-3}\left(x\right)$$$$+\frac{n-1}{n}\cdot \frac{n-3}{n-2}\cdot \frac{1}{n-4}\sin \left(x\right){\cos }^{n-5}\left(x\right)$$$$+......$$$$+\frac{n-1}{n}\cdot \frac{n-3}{n-2}\cdot \frac{n-5}{n-4}\cdot \cdot \cdot \frac{5}{6}\cdot \frac{1}{4}\sin \left(x\right){\cos }^3\left(x\right)$$$$+\frac{n-1}{n}\cdot \frac{n-3}{n-2}\cdot \frac{n-5}{n-4}\cdot \cdot \cdot \frac{5}{6}\cdot \frac{3}{4}\cdot \frac{1}{2}\sin \left(x\right)\cos \left(x\right)$$$$+\frac{n-1}{n}\cdot \frac{n-3}{n-2}\cdot \frac{n-5}{n-4}\cdot \cdot \cdot \frac{5}{6}\cdot \frac{3}{4}\cdot \frac{1}{2}x+C$$$$=\underset{\mathit{k}=1}{\mathbf{\sum }^{\frac{\mathit{n}}{2}}}\frac{\stackrel{=1ifk=1}{(\mathit{n}-1)(\mathit{n}-3)\cdot \cdot \cdot (\mathit{n}-2\mathit{k}+3)}}{\mathit{n}(\mathit{n}-2)\cdot \cdot \cdot (\mathit{n}-2\mathit{k}+2)}\cdot \mathbf{sin}\left(\mathit{x}\right){\mathbf{cos}}^{\mathit{n}-2\mathit{k}+1}\left(\mathit{x}\right)$$$$+\frac{(\mathit{n}-1)!!}{\mathit{n}!!}\mathit{x}+\mathit{C}$$$$$$$$ifnisoddweget:$$$$I_n=\int {\cos }^n\left(x\right)dx$$$$=\frac{1}{n}\sin \left(x\right){\cos }^{n-1}\left(x\right)$$$$+\frac{n-1}{n}\cdot \frac{1}{n-2}\sin \left(x\right){\cos }^{n-3}\left(x\right)$$$$+\frac{n-1}{n}\cdot \frac{n-3}{n-2}\cdot \frac{1}{n-4}\sin \left(x\right){\cos }^{n-5}\left(x\right)$$$$+......$$$$+\frac{n-1}{n}\cdot \frac{n-3}{n-2}\cdot \frac{n-5}{n-4}\cdot \cdot \cdot \frac{4}{5}\cdot \frac{1}{3}\sin \left(x\right){\cos }^2\left(x\right)$$$$+\frac{n-1}{n}\cdot \frac{n-3}{n-2}\cdot \frac{n-5}{n-4}\cdot \cdot \cdot \frac{4}{5}\cdot \frac{2}{3}\cdot \frac{1}{1}\sin \left(x\right){\cos }^0\left(x\right)+C$$$$=\underset{\mathit{k}=1}{\mathbf{\sum }^{\frac{\mathit{n}+1}{2}}}\frac{\stackrel{=1ifk=1}{(\mathit{n}-1)(\mathit{n}-3)\cdot \cdot \cdot (\mathit{n}-2\mathit{k}+1)}}{\mathit{n}(\mathit{n}-2)\cdot \cdot \cdot (\mathit{n}-2\mathit{k}+2)}\cdot \mathbf{sin}\left(\mathit{x}\right){\mathbf{cos}}^{\mathit{n}-2\mathit{k}+1}\left(\mathit{x}\right)+\mathit{C}$$

Commented by tawa tawa last updated on 27/May/17

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

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 27/May/17

$$I_n=\int {cos}^{2}x.{cos}^{n-2}xdx=$$$$=\int {cos}^{n-2}xdx-\int {sin}^{2}x.{cos}^{n-2}xdx=$$$$=I_{n-2}-[\frac{-1}{n-1}sinx.{cos}^{n-1}x+\frac{1}{n-1}\int cosx.{cos}^{n-1}xdx=$$$$=I_{n-2}+\frac{1}{n-1}sinx.{cos}^{n-1}x-\frac{1}{n-1}{I}_n+\mathit{C}.$$$$\Rightarrow I_n+\frac{1}{n-1}{I}_n=I_{n-2}+\frac{1}{n-1}sinx.{cos}^{n-1}x+\mathit{C}\Rightarrow$$$$\frac{n}{n-1}{I}_n=I_{n-2}+\frac{1}{n-1}sinx.{cos}^{n-1}x+C$$$$\Rightarrow I_n=\frac{n-1}{n}.I_{n-2}+\frac{1}{n}sinx.{cos}^{n-1}x+C.$$$$0)I_0=\int dx=x+C.$$$$1)I_1=\int cosxdx=sinx+\mathit{C}$$$$2)I_2=\int {cos}^{2}xdx=\frac{1}{2}x+\frac{1}{4}sin2x+\mathit{C}$$$$3)n>2\Rightarrow formulagivenabove.$$$$example:I_3$$$$I_3=\int {cos}^{3}xdx=\int cosx(1-{sin}^{2}x)dx=$$$$=sinx-\frac{1}{3}{sin}^{3}x+\mathit{C}$$$$I_3=\frac{3-1}{3}{I}_1+\frac{1}{3}sinx.{cos}^{2}x=\frac{2}{3}{I}_1+\frac{1}{3}sinx.{cos}^{2}x=$$$$=\frac{2}{3}sinx+\frac{1}{3}sinx(1-{sin}^{2}x)=sinx-\frac{1}{3}{sin}^{3}x+C$$

Commented by tawa tawa last updated on 27/May/17

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

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