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Question Number 36811 by rahul 19 last updated on 05/Jun/18

∫ ((sin x)/(cos^2 x. (√(cos 2x)))) dx= ?

$$\int\:\frac{\mathrm{sin}\:{x}}{\mathrm{cos}\:^{\mathrm{2}} {x}.\:\sqrt{\mathrm{cos}\:\mathrm{2}{x}}}\:{dx}=\:? \\ $$

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

$$\int\frac{\mathrm{sin}\:{x}}{\mathrm{cos}^{\mathrm{2}} \:{x}\:\sqrt{\mathrm{cos}\:\mathrm{2}{x}}}{dx}=\int\frac{\mathrm{sin}\:{x}}{\mathrm{cos}^{\mathrm{2}} \:{x}\:\sqrt{\mathrm{cos}^{\mathrm{2}} \:{x}\:−\mathrm{sin}^{\mathrm{2}} \:{x}}}{dx}= \\ $$$$=\int\frac{\frac{\mathrm{tan}\:{x}}{\mathrm{sec}\:{x}}}{\frac{\mathrm{1}}{\mathrm{sec}^{\mathrm{2}} \:{x}}\sqrt{\frac{\mathrm{1}}{\mathrm{sec}^{\mathrm{2}} \:{x}}−\frac{\mathrm{tan}^{\mathrm{2}} \:{x}}{\mathrm{sec}^{\mathrm{2}} \:{x}}}}{dx}= \\ $$$$=\int\frac{\mathrm{sec}^{\mathrm{2}} \:{x}\:\mathrm{tan}\:{x}}{\sqrt{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:{x}}}{dx}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\left[{t}=\mathrm{tan}\:{x}\:\rightarrow\:{dx}=\frac{{dt}}{\mathrm{sec}^{\mathrm{2}} \:{x}}\right] \\ $$$$=\int\frac{{t}}{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{dt}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\left[{u}=\mathrm{1}−{t}^{\mathrm{2}} \:\rightarrow\:{dt}=−\frac{{du}}{\mathrm{2}{t}}\right] \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{du}}{\sqrt{{u}}}=−\sqrt{{u}}=−\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }=−\sqrt{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:{x}}+{C} \\ $$$$ \\ $$

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