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Question Number 148798 by abdurehime last updated on 31/Jul/21

Answered by puissant last updated on 31/Jul/21

$$=\frac{\sqrt{2}}{2}(\int \frac{\sin \left(\mathrm{x}\right)}{\sqrt{\sin \left(\mathrm{x}\right)\cos \left(\mathrm{x}\right)}}\mathrm{dx}+\int \frac{\cos \left(\mathrm{x}\right)}{\sqrt{\sin \left(\mathrm{x}\right)\cos \left(\mathrm{x}\right)}}\mathrm{dx})$$$$=\frac{\sqrt{2}}{2}\int \sqrt{\frac{\sin \left(\mathrm{x}\right)}{\cos \left(\mathrm{x}\right)}}\mathrm{dx}+\frac{\sqrt{2}}{2}\int \sqrt{\frac{\cos \left(\mathrm{x}\right)}{\sin \left(\mathrm{x}\right)}}\mathrm{dx}$$$$=\frac{\sqrt{2}}{2}\int (\sqrt{\tan \left(\mathrm{x}\right)}+\sqrt{\mathrm{cotan}\left(\mathrm{x}\right)})\mathrm{dx}$$$$=\frac{\sqrt{2}}{2}\int (\sqrt{\tan \left(\mathrm{x}\right)}+\frac{1}{\sqrt{\tan \left(\mathrm{x}\right)}})\mathrm{dx}$$$$=\frac{\sqrt{2}}{2}\int \frac{1+\tan \left(\mathrm{x}\right)}{\sqrt{\tan \left(\mathrm{x}\right)}}\mathrm{dx}$$$$\mathrm{posons}\mathrm{t}=\sqrt{\tan \left(\mathrm{x}\right)}\Rightarrow \mathrm{dt}=\frac{1+{\tan }^2\left(\mathrm{x}\right)}{2\sqrt{\tan \left(\mathrm{x}\right)}}\mathrm{dx}$$$$\Rightarrow \mathrm{dx}=\frac{2\sqrt{\tan \left(\mathrm{x}\right)}}{1+{\tan }^2\left(\mathrm{x}\right)}\mathrm{dt}$$$$\Rightarrow \mathrm{dx}=\frac{2\mathrm{t}}{1+{\mathrm{t}}^4}\mathrm{dt}$$$$\mathrm{I}=\frac{\sqrt{2}}{2}\int \frac{1+{\mathrm{t}}^2}{\mathrm{t}}\times \frac{2\mathrm{t}}{1+{\mathrm{t}}^4}\mathrm{dt}$$$$=\sqrt{2}\int \frac{1+{\mathrm{t}}^2}{1+{\mathrm{t}}^4}\mathrm{dt}=\sqrt{2}\int \frac{1+\frac{1}{{\mathrm{t}}^2}}{{\mathrm{t}}^2+\frac{1}{{\mathrm{t}}^2}}\mathrm{dt}$$$$=\sqrt{2}\int \frac{1+\frac{1}{{\mathrm{t}}^2}}{{(\mathrm{t}-\frac{1}{\mathrm{t}})}^2+2}\mathrm{dt}$$$$\mathrm{posons}\mathrm{u}=(\mathrm{t}-\frac{1}{\mathrm{t}})\Rightarrow \mathrm{du}=(1+\frac{1}{{\mathrm{t}}^2})\mathrm{dt}$$$$\Rightarrow \mathrm{I}=\sqrt{2}\int \frac{\mathrm{du}}{{\mathrm{u}}^2+{\left(\sqrt{2}\right)}^2}$$$$=\frac{\sqrt{2}}{\sqrt{2}}\arctan \frac{\mathrm{u}}{\sqrt{2}}+\mathrm{C}$$$$=\arctan \left(\frac{\mathrm{t}-\frac{1}{\mathrm{t}}}{\sqrt{2}}\right)+\mathrm{C}$$$$=\arctan \left(\frac{\sqrt{\tan \left(\mathrm{x}\right)}-\frac{1}{\sqrt{\tan \left(\mathrm{x}\right)}}}{\sqrt{2}}\right)+\mathrm{C}$$$$\mathrm{soit}\mathrm{I}=\arctan \left(\frac{\tan \left(\mathrm{x}\right)-1}{\sqrt{2\tan \left(\mathrm{x}\right)}}\right)+\mathrm{C}..$$$${\mathrm{d}}^{\prime }\mathrm{ou}\mathrm{I}=\arctan \left(\frac{\sin \left(\mathrm{x}\right)-\cos \left(\mathrm{x}\right)}{\sqrt{\sin \left(2\mathrm{x}\right)}}\right)+\mathrm{C}$$$$......\mathrm{Le}\mathrm{puissant}....$$

Commented by abdurehime last updated on 31/Jul/21

$$\mathrm{show}\mathrm{me}\mathrm{the}\mathrm{final}\mathrm{answer}\mathrm{step}\mathrm{by}\mathrm{step}$$

Answered by Kamel last updated on 31/Jul/21

$$I=\int \frac{sin\left(x\right)+cos\left(x\right)}{\sqrt{sin\left(2x\right)}}dx\stackrel{t=\sqrt{tg\left(x\right)}}{=}\sqrt{2}\int \frac{1+t^2}{1+t^4}dt$$$$=\sqrt{2}\int \frac{1+\frac{1}{t^2}}{{(t-\frac{1}{t})}^2+2}dt\stackrel{t-\frac{1}{t}=u}{=}\sqrt{2}\int \frac{du}{2+u^2}=Arctg\left(\frac{1}{\sqrt{2}}(\sqrt{tg\left(x\right)}-\frac{1}{\sqrt{tg\left(x\right)}})\right)+c/c\in \mathbb{R}$$$$\therefore \int \frac{sin\left(x\right)+cos\left(x\right)}{\sqrt{sin\left(2x\right)}}dx=Arctg\left(\frac{sin\left(x\right)-cos\left(x\right)}{\sqrt{sin\left(2x\right)}}\right)+c/c\in \mathbb{R}.$$

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