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Question Number 97041 by bemath last updated on 06/Jun/20

$$\int \sin {}^8\left(x\right)\cos {}^8\left(x\right)dx=?$$

Answered by john santu last updated on 06/Jun/20

$$\Rightarrow \sin {}^{8}x.\cos {}^{8}x=\frac{\sin {}^8\left(2x\right)}{2^8}$$$$\sin \left(2x\right)=\frac{e^{2ix}-e^{-2ix}}{2i}$$$$\sin {}^{8}x.\cos {}^{8}x=\frac{{(e^{2ix}-e^{-2ix})}^8}{2^{16}}$$$$\mathrm{I}=\frac{1}{2^{15}}\int (\cos 16x+8\cos 12x+56\cos 4x+35)dx$$$$\mathrm{I}=\frac{1}{2^{15}}(\frac{\sin 16x}{16}+\frac{8\sin 12x}{12}+\frac{56\sin 4x}{4}+35x)+c$$$$$$

Answered by Sourav mridha last updated on 06/Jun/20

$$\mathit{l}\mathit{e}\mathit{t}\mathit{s}\mathit{i}\mathit{n}\mathit{x}=\mathit{m}$$$$\int {(1-{\mathit{m}}^2)}^7{\mathit{m}}^8\mathit{d}\mathit{m}$$$$=\int [\underset{r=0}{\sum ^7}{\stackrel{7}{\mathit{C}}}_{\mathit{r}}{\left(1\right)}^{7-\mathit{r}}.{(-{\mathit{m}}^2)}^{\mathit{r}}.].{\mathit{m}}^8\mathit{d}\mathit{m}$$$$=\underset{\mathit{r}=0}{\sum ^7}{(-1)}^{\mathit{r}}{\stackrel{7}{\mathit{C}}}_{\mathit{r}}[\int {\mathit{m}}^{2\mathit{r}+8}\mathit{d}\mathit{m}]$$$$=\underset{\mathit{r}=0}{\sum ^7}{(-1)}^{\mathit{r}}{\stackrel{7}{\mathit{C}}}_{\mathit{r}}\frac{{\left(\mathit{s}\mathit{i}\mathit{n}\left(\mathit{x}\right)\right)}^{2\mathit{r}+9}}{2\mathit{r}+9}.+\mathit{k}$$

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