Question Number 36811 by rahul 19 last updated on 05/Jun/18
$$\int\:\frac{\mathrm{sin}\:{x}}{\mathrm{cos}\:^{\mathrm{2}} {x}.\:\sqrt{\mathrm{cos}\:\mathrm{2}{x}}}\:{dx}=\:? \\ $$
Answered by MJS last updated on 05/Jun/18
![∫((sin x)/(cos^2 x (√(cos 2x))))dx=∫((sin x)/(cos^2 x (√(cos^2 x −sin^2 x))))dx= =∫(((tan x)/(sec x))/((1/(sec^2 x))(√((1/(sec^2 x))−((tan^2 x)/(sec^2 x))))))dx= =∫((sec^2 x tan x)/( (√(1−tan^2 x))))dx= [t=tan x → dx=(dt/(sec^2 x))] =∫(t/( (√(1−t^2 ))))dt= [u=1−t^2 → dt=−(du/(2t))] =−(1/2)∫(du/( (√u)))=−(√u)=−(√(1−t^2 ))=−(√(1−tan^2 x))+C](https://www.tinkutara.com/question/Q36813.png)
$$\int\frac{\mathrm{sin}\:{x}}{\mathrm{cos}^{\mathrm{2}} \:{x}\:\sqrt{\mathrm{cos}\:\mathrm{2}{x}}}{dx}=\int\frac{\mathrm{sin}\:{x}}{\mathrm{cos}^{\mathrm{2}} \:{x}\:\sqrt{\mathrm{cos}^{\mathrm{2}} \:{x}\:−\mathrm{sin}^{\mathrm{2}} \:{x}}}{dx}= \\ $$$$=\int\frac{\frac{\mathrm{tan}\:{x}}{\mathrm{sec}\:{x}}}{\frac{\mathrm{1}}{\mathrm{sec}^{\mathrm{2}} \:{x}}\sqrt{\frac{\mathrm{1}}{\mathrm{sec}^{\mathrm{2}} \:{x}}−\frac{\mathrm{tan}^{\mathrm{2}} \:{x}}{\mathrm{sec}^{\mathrm{2}} \:{x}}}}{dx}= \\ $$$$=\int\frac{\mathrm{sec}^{\mathrm{2}} \:{x}\:\mathrm{tan}\:{x}}{\:\sqrt{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:{x}}}{dx}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\left[{t}=\mathrm{tan}\:{x}\:\rightarrow\:{dx}=\frac{{dt}}{\mathrm{sec}^{\mathrm{2}} \:{x}}\right] \\ $$$$=\int\frac{{t}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{dt}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\left[{u}=\mathrm{1}−{t}^{\mathrm{2}} \:\rightarrow\:{dt}=−\frac{{du}}{\mathrm{2}{t}}\right] \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{du}}{\:\sqrt{{u}}}=−\sqrt{{u}}=−\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }=−\sqrt{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:{x}}+{C} \\ $$$$ \\ $$