Question Number 9581 by dc630248@gmail.com last updated on 18/Dec/16
$$\int\mathrm{sin}^{\mathrm{4}} {x}.\mathrm{cos}^{\mathrm{4}} {xdx} \\ $$
Answered by mrW last updated on 18/Dec/16
![sin^4 x.cos^4 x=(1/(16))(2sinxcosx)^4 =(1/(16))sin^4 (2x) t=2x ∫sin^4 x.cos^4 xdx=(1/(32))∫sin^4 (2x)d(2x) =(1/(32))∫sin^4 tdt=(1/(32))[(3/8)t−((sin(2t))/4)+((sin(4t))/(32))]+C =(1/(1024))[24x−8sin(4x)+sin(8x)]+C](https://www.tinkutara.com/question/Q9582.png)
$$\mathrm{sin}^{\mathrm{4}} {x}.\mathrm{cos}^{\mathrm{4}} {x}=\frac{\mathrm{1}}{\mathrm{16}}\left(\mathrm{2sinxcosx}\right)^{\mathrm{4}} =\frac{\mathrm{1}}{\mathrm{16}}\mathrm{sin}^{\mathrm{4}} \left(\mathrm{2x}\right) \\ $$$$\mathrm{t}=\mathrm{2x} \\ $$$$\int\mathrm{sin}^{\mathrm{4}} {x}.\mathrm{cos}^{\mathrm{4}} {xdx}=\frac{\mathrm{1}}{\mathrm{32}}\int\mathrm{sin}^{\mathrm{4}} \left(\mathrm{2x}\right)\mathrm{d}\left(\mathrm{2x}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{32}}\int\mathrm{sin}^{\mathrm{4}} \mathrm{tdt}=\frac{\mathrm{1}}{\mathrm{32}}\left[\frac{\mathrm{3}}{\mathrm{8}}\mathrm{t}−\frac{\mathrm{sin}\left(\mathrm{2t}\right)}{\mathrm{4}}+\frac{\mathrm{sin}\left(\mathrm{4t}\right)}{\mathrm{32}}\right]+\mathrm{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{1024}}\left[\mathrm{24x}−\mathrm{8sin}\left(\mathrm{4x}\right)+\mathrm{sin}\left(\mathrm{8x}\right)\right]+\mathrm{C} \\ $$