Question Number 212886 by MrGaster last updated on 26/Oct/24
$$\int\frac{\mathrm{sin10}{x}}{\mathrm{sin}\:{x}}{dx}. \\ $$
Answered by Ghisom last updated on 26/Oct/24
![∫((sin 10x)/(sin x))dx= =4∫cos x (cos 8x +cos 4x +(1/2))dx= [t=sin x] =∫(512t^8 −1024t^6 +672t^4 −160t^2 +10)dt= =((512)/9)t^9 −((1024)/7)t^7 +((672)/5)t^5 −((160)/3)t^3 +10t= ... =(2/9)sin 9x +(2/7)sin 7x +(2/5)sin 5x +(2/3)sin 3x +2sin x](https://www.tinkutara.com/question/Q212892.png)
$$\int\frac{\mathrm{sin}\:\mathrm{10}{x}}{\mathrm{sin}\:{x}}{dx}= \\ $$$$=\mathrm{4}\int\mathrm{cos}\:{x}\:\left(\mathrm{cos}\:\mathrm{8}{x}\:+\mathrm{cos}\:\mathrm{4}{x}\:+\frac{\mathrm{1}}{\mathrm{2}}\right){dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{sin}\:{x}\right] \\ $$$$=\int\left(\mathrm{512}{t}^{\mathrm{8}} −\mathrm{1024}{t}^{\mathrm{6}} +\mathrm{672}{t}^{\mathrm{4}} −\mathrm{160}{t}^{\mathrm{2}} +\mathrm{10}\right){dt}= \\ $$$$=\frac{\mathrm{512}}{\mathrm{9}}{t}^{\mathrm{9}} −\frac{\mathrm{1024}}{\mathrm{7}}{t}^{\mathrm{7}} +\frac{\mathrm{672}}{\mathrm{5}}{t}^{\mathrm{5}} −\frac{\mathrm{160}}{\mathrm{3}}{t}^{\mathrm{3}} +\mathrm{10}{t}= \\ $$$$… \\ $$$$=\frac{\mathrm{2}}{\mathrm{9}}\mathrm{sin}\:\mathrm{9}{x}\:+\frac{\mathrm{2}}{\mathrm{7}}\mathrm{sin}\:\mathrm{7}{x}\:+\frac{\mathrm{2}}{\mathrm{5}}\mathrm{sin}\:\mathrm{5}{x}\:+\frac{\mathrm{2}}{\mathrm{3}}\mathrm{sin}\:\mathrm{3}{x}\:+\mathrm{2sin}\:{x} \\ $$