Question Number 64017 by Rio Michael last updated on 12/Jul/19
$${why}\:{can}'{t}\:{we}\:{differentiate}\:{or}\:{intergrate}\:{powers}\:{of}\:{trigonometric} \\ $$$${functions}\:{such}\:{as}\: \\ $$$$\left.\mathrm{1}\left.\right)\:\int{cos}^{\mathrm{2}} {xdx}?\:\:\:\:\mathrm{3}\right)\:\int{tan}^{\mathrm{2}} \mathrm{2}{xdx} \\ $$$$\left.\mathrm{2}\left.\right)\int{sin}^{\mathrm{2}} {xdx}?\:\:\:\:\:\mathrm{4}\right)\:\int{sin}^{\mathrm{10}} {x} \\ $$$${hence}\:{how}\:{do}\:{we}\:{solve}\:{such}\:{problems}.? \\ $$
Commented by kaivan.ahmadi last updated on 12/Jul/19
$${cos}\mathrm{2}{x}=\mathrm{2}{cos}^{\mathrm{2}} {x}−\mathrm{1}\Rightarrow{cos}^{\mathrm{2}} {x}=\frac{\mathrm{1}+{cos}\mathrm{2}{x}}{\mathrm{2}} \\ $$$$\int{cos}^{\mathrm{2}} {xdx}=\int\frac{\mathrm{1}+{cos}\mathrm{2}{x}}{\mathrm{2}}{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int\left(\mathrm{1}+{cos}\mathrm{2}{x}\right){dx}= \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left({x}+\frac{\mathrm{1}}{\mathrm{2}}{sin}\mathrm{2}{x}\right)+{C} \\ $$$$ \\ $$
Commented by kaivan.ahmadi last updated on 12/Jul/19
$${cos}\mathrm{2}{x}=\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} {x}\Rightarrow{sin}^{\mathrm{2}} {x}=\frac{\mathrm{1}−{cos}\mathrm{2}{x}}{\mathrm{2}} \\ $$$$\int{sin}^{\mathrm{2}} {xdx}=\frac{\mathrm{1}}{\mathrm{2}}\int\left(\mathrm{1}−{cos}\mathrm{2}{x}\right){dx}= \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left({x}−\frac{\mathrm{1}}{\mathrm{2}}{sin}\mathrm{2}{x}\right)+{C} \\ $$
Commented by kaivan.ahmadi last updated on 12/Jul/19
$$\int{tan}^{\mathrm{2}} \mathrm{2}{xdx}=\int\left(\mathrm{1}+{tan}^{\mathrm{2}} \mathrm{2}{x}−\mathrm{1}\right){dx}=\frac{\mathrm{1}}{\mathrm{2}}{tan}\mathrm{2}{x}−{x}+{C} \\ $$
Commented by MJS last updated on 12/Jul/19
$$\mathrm{look}\:\mathrm{at}\:\mathrm{qu}.\:\mathrm{64037} \\ $$