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Question Number 105885 by mohammad17 last updated on 01/Aug/20

Answered by bemath last updated on 01/Aug/20

$$\sin \left(x\right)+\cos \left(x\right)=\sqrt{2}\cos (x-\frac{\pi }{4})$$$$\int \frac{1}{\sqrt{2}}\sec (x-\frac{\pi }{4})dx=$$$$\frac{1}{\sqrt{2}}\ln \mid \sec (x-\frac{\pi }{4})+\tan (x-\frac{\pi }{4})\mid +C$$$$\frac{1}{\sqrt{2}}\ln \mid \frac{\sqrt{2}+\sin x-\cos x}{\sin x+\cos x}\mid +C$$

Commented by mohammad17 last updated on 01/Aug/20

$$verythankyousir$$

Commented by bemath last updated on 01/Aug/20

$$\frac{1}{\cos (x-\frac{\pi }{4})}+\frac{\sin (x-\frac{\pi }{4})}{\cos (x-\frac{\pi }{4})}=$$$$\frac{1+\frac{1}{\sqrt{2}}\sin x-\frac{1}{\sqrt{2}}\cos x}{\frac{1}{\sqrt{2}}\cos x+\frac{1}{\sqrt{2}}\sin x}=$$$$\frac{\sqrt{2}+\sin x-\cos x}{\cos x+\sin x}$$

Answered by Dwaipayan Shikari last updated on 01/Aug/20

$$2\int \frac{1}{\frac{2t}{1+t^2}+\frac{1-t^2}{1+t^2}}.\frac{1}{1+t^2}dt(tan\frac{x}{2}=t)\frac{1}{2}{sec}^2\frac{x}{2}=\frac{dt}{dx},{sec}^2\frac{x}{2}=1+t^2$$$$2\int \frac{1}{1+2t-t^2}$$$$-2\int \frac{1}{(t^2-2t+1)-2}=-2\int \frac{1}{{(t-1)}^2-{\left(\sqrt{2}\right)}^2}=-2\frac{1}{2\sqrt{2}}log\left(\frac{t-1-\sqrt{2}}{t-1+\sqrt{2}}\right)+C$$$$=\frac{1}{\sqrt{2}}log\left(\frac{t-1+\sqrt{2}}{t-1-\sqrt{2}}\right)+C=\frac{1}{\sqrt{2}}log\left(\frac{tan\frac{x}{2}-1+\sqrt{2}}{tan\frac{x}{2}-1-\sqrt{2}}\right)+C$$$$$$

Answered by Coronavirus last updated on 01/Aug/20

$$sinx+cosx=\sqrt{2}cos(x-\frac{\pi }{4})$$$$\frac{1}{cosa}=\frac{cosa}{{cos}^{2}a}=\frac{cosa}{1-{sin}^{2}a}=\frac{\frac{d\left(sina\right)}{da}}{1-{\left(sina\right)}^2}=d\left(argth\left(sina\right)\right)/da=\frac{d\left(\frac{1}{2}ln\left(\frac{1+sina}{1-sina}\right)\right)}{da}$$$$yet\int \frac{1}{sin\left(x\right)+cos\left(x\right)}dx=\frac{1}{\sqrt{2}}\int \frac{1}{cos(x-\frac{\pi }{4})}dx$$$$=\frac{1}{\sqrt{2}}argth\left(sin(x-\frac{\pi }{4})\right)+C$$$$=\frac{1}{2\sqrt{2}}ln\left(\frac{1+sin(x-\frac{\pi }{4})}{1-sin(x-\frac{\pi }{4})}\right)+C$$$$so\int \frac{1}{\sin \left(\mathrm{x}\right)+\cos \left(\mathrm{x}\right)}\mathrm{dx}=\frac{1}{2\sqrt{2}}\ln \left(\frac{1+\sin (\mathrm{x}-\frac{\pi }{4})}{1-\sin (\mathrm{x}-\frac{\pi }{4})}\right)+C$$

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