Question Number 128408 by bramlexs22 last updated on 07/Jan/21
$$\rho\:=\:\int\:\frac{\mathrm{sin}\:\left(\mathrm{4}{x}\right)}{\mathrm{sin}\:^{\mathrm{4}} \left({x}\right)+\mathrm{cos}\:^{\mathrm{4}} \left({x}\right)}\:{dx}\: \\ $$
Answered by mr W last updated on 07/Jan/21
![=∫((sin (4x))/(1−2 sin^2 x cos^2 x))dx =∫((sin (4x))/(1−(1/2)sin^2 (2x)))dx =∫((sin (4x))/(3+cos (4x)))d(4x) =−∫(1/(3+cos (4x)))d(3+cos (4x)) =−ln [3+cos (4x)]+C](https://www.tinkutara.com/question/Q128409.png)
$$=\int\frac{\mathrm{sin}\:\left(\mathrm{4}{x}\right)}{\mathrm{1}−\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:{x}\:\mathrm{cos}^{\mathrm{2}} \:{x}}{dx} \\ $$$$=\int\frac{\mathrm{sin}\:\left(\mathrm{4}{x}\right)}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}^{\mathrm{2}} \:\left(\mathrm{2}{x}\right)}{dx} \\ $$$$=\int\frac{\mathrm{sin}\:\left(\mathrm{4}{x}\right)}{\mathrm{3}+\mathrm{cos}\:\left(\mathrm{4}{x}\right)}{d}\left(\mathrm{4}{x}\right) \\ $$$$=−\int\frac{\mathrm{1}}{\mathrm{3}+\mathrm{cos}\:\left(\mathrm{4}{x}\right)}{d}\left(\mathrm{3}+\mathrm{cos}\:\left(\mathrm{4}{x}\right)\right) \\ $$$$=−\mathrm{ln}\:\left[\mathrm{3}+\mathrm{cos}\:\left(\mathrm{4}{x}\right)\right]+{C} \\ $$
Commented by bramlexs22 last updated on 07/Jan/21
��✍️