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Question Number 38397 by Fawomath last updated on 25/Jun/18

$$evaluate\int secxdx$$

Commented by maxmathsup by imad last updated on 25/Jun/18

$$letI=\int \frac{dx}{cosx}changementtan\left(\frac{x}{2}\right)=tgive$$$$I=\int \frac{1}{\frac{1-t^2}{1+t^2}}\frac{2dt}{1+t^2}=\int \frac{2dt}{1-t^2}=\int (\frac{1}{1+t}+\frac{1}{1-t})dt$$$$=ln\mid \frac{1+t}{1-t}\mid +c=ln\mid \frac{1+tan\left(\frac{x}{2}\right)}{1-tan\left(\frac{x}{2}\right)}\mid +c=ln\mid tan(\frac{\pi }{4}+\frac{x}{2})\mid +c$$

Commented by Cheyboy last updated on 25/Jun/18

$$\int \mathit{s}\mathit{e}\mathit{c}\mathit{x}\mathit{d}\mathit{x}$$$$\int (\mathit{s}\mathit{e}\mathit{c}\mathit{x}.\frac{\mathit{s}\mathit{e}\mathit{c}\mathit{x}+\mathit{t}\mathit{a}\mathit{n}\mathit{x}}{\mathit{s}\mathit{e}\mathit{c}\mathit{x}+\mathit{t}\mathit{a}\mathit{n}\mathit{x}})\mathit{d}\mathit{x}$$$$\int \left(\frac{{\mathit{s}\mathit{e}\mathit{c}}^2\mathit{x}+\mathit{s}\mathit{e}\mathit{c}\mathit{x}\mathit{t}\mathit{a}\mathit{n}\mathit{x}}{\mathit{s}\mathit{e}\mathit{c}\mathit{x}+\mathit{t}\mathit{a}\mathit{n}\mathit{x}}\right)\mathit{d}\mathit{x}$$$$\mathit{l}\mathit{e}\mathit{t}\mathit{u}=\mathit{s}\mathit{e}\mathit{c}\mathit{x}+\mathit{t}\mathit{a}\mathit{n}\mathit{x}$$$$\mathit{d}\mathit{u}=(\mathit{s}\mathit{e}\mathit{c}\mathit{x}\mathit{t}\mathit{a}\mathit{n}\mathit{x}+{\mathit{s}\mathit{e}\mathit{c}}^2\mathit{x})\mathit{d}\mathit{x}$$$$\int \frac{1}{\mathit{u}}\mathit{d}\mathit{u}$$$$\ln \mid \mathit{u}\mid +C$$$$\ln \mid \mathit{s}\mathit{e}\mathit{c}\mathit{x}+\mathit{t}\mathit{a}\mathit{n}\mathit{x}\mid +C$$

Answered by tanmay.chaudhury50@gmail.com last updated on 25/Jun/18

$$\int \frac{dx}{cosx}$$$$=\int \frac{cosx}{{cos}^{2}x}dx$$$$=\int \frac{cosx}{1-{sin}^{2}x}dx$$$$t=sinxdt=cosxdx$$$$\int \frac{dt}{1-t^2}=\frac{1}{2}\int \frac{1+t+1-t}{(1+t)(1-t)}dt$$$$\frac{1}{2}\int \frac{dt}{1-t}+\frac{1}{2}\int \frac{dt}{1+t}$$$$$$$$=-\frac{1}{2}ln\mid 1-t\mid +\frac{1}{2}ln\mid 1+t\mid$$$$=\frac{1}{2}ln\mid \frac{1+t}{1-t}\mid +c$$$$=\frac{1}{2}ln\mid \frac{1+sinx}{1-sinx}\mid +c$$$$=\frac{1}{2}ln\mid \frac{{(1+sinx)}^2}{1-{sin}^{2}x}\mid +c$$$$=\frac{1}{2}ln\mid \frac{{(1+sinx)}^2}{{cos}^{2}x}\mid +c$$$$=ln\mid \frac{1}{cosx}+\frac{sinx}{cosx}\mid +c$$$$=ln\mid secx+tanx\mid +c$$$$$$$$$$

Answered by malwaan last updated on 25/Jun/18

$$\int \frac{\mathrm{secx}(\mathrm{secx}+\mathrm{tanx})}{\mathrm{secx}+\mathrm{tanx}}$$$$=\int \frac{{\sec }^2\mathrm{x}+\mathrm{secx}.\mathrm{tanx}}{\mathrm{tanx}+\mathrm{secx}}$$$$=\ln \mid \mathrm{tanx}+\mathrm{secx}\mid +\mathrm{c}$$

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