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Question Number 43125 by Raj Singh last updated on 07/Sep/18

Commented by maxmathsup by imad last updated on 07/Sep/18

$$letI=\int \frac{1}{\frac{1}{sinx}+\frac{1}{cosx}}dx\Rightarrow I=\int \frac{cosxsinx}{cosx+sinx}dx$$$$=\frac{1}{2}\int \frac{sin\left(2x\right)}{\sqrt{2}cos(x-\frac{\pi }{4})}dxchangementx-\frac{\pi }{4}=tgive$$$$I=\frac{1}{2\sqrt{2}}\int \frac{sin\left(2(t+\frac{\pi }{4})\right)}{cost}dt=\frac{1}{2\sqrt{2}}\int \frac{cos\left(2t\right)}{cost}dt$$$$=\frac{1}{2\sqrt{2}}\int \frac{2{cos}^{2}t-1}{cost}dt=\frac{1}{2\sqrt{2}}\{\int 2costdt-\int \frac{dt}{cost}\}$$$$=\frac{1}{\sqrt{2}}sin\left(t\right)-\frac{1}{2\sqrt{2}}ln\mid tan(\frac{t}{2}+\frac{\pi }{4})\mid +c$$$$=\frac{1}{\sqrt{2}}sin(x-\frac{\pi }{4})-\frac{1}{2\sqrt{2}}ln\mid tan(\frac{x}{2}-\frac{\pi }{8}+\frac{\pi }{4})+c$$$$=\frac{1}{\sqrt{2}}sin(x-\frac{\pi }{4})-\frac{1}{2\sqrt{2}}ln\mid tan(\frac{x}{2}+\frac{\pi }{8})\mid +c.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 07/Sep/18

$$\int \frac{dx}{\frac{1}{sinx}+\frac{1}{cosx}}$$$$\frac{1}{2}\int \frac{2sinxcosx}{sinx+cosx}dx$$$$\frac{1}{2}\int \frac{1+2sinxcosx-1}{sinx+cosx}dx$$$$\frac{1}{2}\int \frac{{(sinx+cosx)}^2}{sinx+cosx}dx-\frac{1}{2}\int \frac{1}{sinx+cosx}dx$$$$\frac{1}{2}\int sinx+cosxdx-\frac{1}{2}\int \frac{dx}{\sqrt{2}(\frac{1}{\sqrt{2}}sinx+\frac{1}{\sqrt{2}}cosx)}$$$$\frac{1}{2}\int sinx+cosxdx-\frac{1}{2\sqrt{2}}\int \frac{dx}{sin(x+\frac{\mathrm{\Pi }}{4})}$$$$\frac{1}{2}\int sinx+cosxdx-\frac{1}{2\sqrt{2}}\int cosec(\frac{\mathrm{\Pi }}{4}+x)dx$$$$\frac{1}{2}(-cosx+sinx)-\frac{1}{2\sqrt{2}}lntan\left(\frac{x+\frac{\mathrm{\Pi }}{4}}{2}\right)+c$$$$\frac{1}{2}(sinx-cosx)-\frac{1}{2\sqrt{2}}lntan(\frac{x}{2}+\frac{\mathrm{\Pi }}{8})+c$$$$$$$$$$

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