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Question Number 114072 by Lordose last updated on 17/Sep/20

$$\int \frac{1}{\mathbf{sinx}+\mathbf{cosx}}\mathbf{dx}$$

Answered by Olaf last updated on 17/Sep/20

$$x=\frac{\pi }{4}-u$$$$\sin x+\cos x=$$$$(\sin \frac{\pi }{4}\cos u-\sin u\cos \frac{\pi }{4})+(\cos \frac{\pi }{4}\cos u+\sin \frac{\pi }{4}\sin u)$$$$=\frac{\sqrt{2}}{2}(\cos u-\sin u)+\frac{\sqrt{2}}{2}(\cos u+\sin u)$$$$=\sqrt{2}\cos u$$$$-\frac{1}{\sqrt{2}}\int \frac{du}{\cos u}=-\frac{1}{\sqrt{2}}\int \frac{du}{\frac{1-t^2}{1+t^2}}$$$$\mathrm{with}t=\tan \frac{u}{2}$$$$dt=\frac{1}{2}(1+{\tan }^2\frac{u}{2})du$$$$du=\frac{2dt}{1+t^2}$$$$-\frac{1}{\sqrt{2}}\int \frac{1+t^2}{1-t^2}.\frac{2dt}{1+t^2}$$$$=-\sqrt{2}\int \frac{dt}{1-t^2}=-\frac{1}{\sqrt{2}}\int (\frac{1}{1-t}+\frac{1}{1+t})dt$$$$=\frac{1}{\sqrt{2}}\ln \mid \frac{1-t}{1+t}\mid =\frac{1}{\sqrt{2}}\ln \mid \frac{1-\tan \frac{u}{2}}{1+\tan \frac{u}{2}}\mid$$$$\frac{1}{\sqrt{2}}\ln \mid \frac{1-\tan (\frac{\pi }{8}-\frac{x}{2})}{1+\tan (\frac{\pi }{8}-\frac{x}{2})}\mid ...$$

Answered by MJS_new last updated on 17/Sep/20

$$\int \frac{dx}{\sin x+\cos x}=$$$$[t=x+\frac{\pi }{4}\to dx=dt]$$$$=\frac{\sqrt{2}}{2}\int \csc tdt=\frac{\sqrt{2}}{2}\ln \mid \tan \frac{t}{2}\mid =$$$$=\frac{\sqrt{2}}{2}\ln \mid \tan (\frac{x}{2}+\frac{\pi }{8})\mid =$$$$=\frac{\sqrt{2}}{2}\ln \mid \frac{\sqrt{2}+2\sin x}{\sqrt{2}+2\cos x}\mid +C$$

Answered by Dwaipayan Shikari last updated on 17/Sep/20

$$\int \frac{1}{sinx+cosx}dx$$$$=\sqrt{2}\int \frac{1}{cos(\frac{\pi }{4}-x)}dx$$$$=\sqrt{2}\int sec(\frac{\pi }{4}-x)dx$$$$=-\sqrt{2}\int secudu=-\sqrt{2}log(secu+tanu)$$$$=-\sqrt{2}log(sec(\frac{\pi }{4}-x)+tan(\frac{\pi }{4}-x))$$

Commented by MJS_new last updated on 17/Sep/20

$$\mathrm{no}.$$

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