Question Number 193852 by SaRahAli last updated on 21/Jun/23
Answered by cortano12 last updated on 21/Jun/23
![(1/( (√2))) ∫ ((sin 2x)/(cos (x−(π/4))))= (1/( (√2))) ∫ ((cos 2u)/(cos u)) du = (1/( (√2))) ∫ ((2cos^2 u−1)/(cos u)) du = (1/( (√2))) [∫ 2cos u du−∫sec u du ] = (1/( (√2))) [ 2sin u−ln ∣sec u+tan u∣ +c = (1/( (√2))) [ 2sin (x−(π/4))−ln ∣((1+sin (x−(π/4)))/(cos (x−(π/4))))∣ ] + c](https://www.tinkutara.com/question/Q193854.png)
$$\:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\int\:\frac{\mathrm{sin}\:\mathrm{2x}}{\mathrm{cos}\:\left(\mathrm{x}−\frac{\pi}{\mathrm{4}}\right)}=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\int\:\frac{\mathrm{cos}\:\mathrm{2u}}{\mathrm{cos}\:\mathrm{u}}\:\mathrm{du} \\ $$$$\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\int\:\frac{\mathrm{2cos}\:^{\mathrm{2}} \mathrm{u}−\mathrm{1}}{\mathrm{cos}\:\mathrm{u}}\:\mathrm{du} \\ $$$$\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\left[\int\:\mathrm{2cos}\:\mathrm{u}\:\mathrm{du}−\int\mathrm{sec}\:\mathrm{u}\:\mathrm{du}\:\right] \\ $$$$\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\left[\:\mathrm{2sin}\:\mathrm{u}−\mathrm{ln}\:\mid\mathrm{sec}\:\mathrm{u}+\mathrm{tan}\:\mathrm{u}\mid\:+\mathrm{c}\:\right. \\ $$$$\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\left[\:\mathrm{2sin}\:\left(\mathrm{x}−\frac{\pi}{\mathrm{4}}\right)−\mathrm{ln}\:\mid\frac{\mathrm{1}+\mathrm{sin}\:\left(\mathrm{x}−\frac{\pi}{\mathrm{4}}\right)}{\mathrm{cos}\:\left(\mathrm{x}−\frac{\pi}{\mathrm{4}}\right)}\mid\:\right]\:+\:\mathrm{c}\: \\ $$
Answered by witcher3 last updated on 21/Jun/23
![sin(2x)=(1/2)[(sin(x)+cos(x))^2 −(sin(x)−cos(x))^2 ]](https://www.tinkutara.com/question/Q193855.png)
$$\mathrm{sin}\left(\mathrm{2x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left[\left(\mathrm{sin}\left(\mathrm{x}\right)+\mathrm{cos}\left(\mathrm{x}\right)\right)^{\mathrm{2}} −\left(\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{x}}\right)−\boldsymbol{\mathrm{cos}}\left(\boldsymbol{\mathrm{x}}\right)\right)^{\mathrm{2}} \right] \\ $$
Answered by MM42 last updated on 21/Jun/23
$$\int\:\frac{\mathrm{1}+{sin}\mathrm{2}{x}−\mathrm{1}}{{sinx}+{cosx}}{dx}=\int\:\frac{\left({sinx}+{cosx}\right)^{\mathrm{2}} −\mathrm{1}}{{sinx}+{cosx}}{dx} \\ $$$$=\int\left({sinx}+{cosx}\right){dx}−\int\frac{{dx}}{\:\sqrt{\mathrm{2}}{sin}\left({x}+\frac{\pi}{\mathrm{4}}\right)} \\ $$$$=−{cosx}+{sinx}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{ln}\left({tan}\left(\frac{{x}}{\mathrm{2}}+\frac{\pi}{\mathrm{8}}\right)\right)+{c} \\ $$$$ \\ $$