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Question Number 163829 by milandou last updated on 11/Jan/22

$$\int \frac{1}{\cos x}$$

Commented by MJS_new last updated on 11/Jan/22

$$\int x+y=?$$$$\mathrm{maybe}\mathrm{use}\mathrm{proper}\mathrm{syntax}?$$

Answered by Ar Brandon last updated on 11/Jan/22

$$\int \frac{1}{\cos x}dx=\ln (\sec x+\tan x)dx$$$$\mathrm{Let}x=\frac{\pi }{2}-2t\Rightarrow dx=-2dt$$$$\int \frac{dx}{\cos x}=-2\int \frac{1}{\sin 2t}dt=-2\int \frac{{\cos }^{2}t+{\sin }^{2}t}{2\sin t\cos t}dt$$$$=-\int (\frac{\cos t}{\sin t}+\frac{\sin t}{\cos t})dt=\ln \left(\cos t\right)-\ln \left(\sin t\right)+C$$$$=\ln \left(\frac{\cos t}{\sin t}\right)+C=\ln \mid \cot (\frac{\pi }{4}-\frac{x}{2})\mid +C$$

Answered by Ar Brandon last updated on 11/Jan/22

$$\int \frac{1}{\cos x}dx=\int \frac{\cos x}{{\cos }^{2}x}dx=\int \frac{\cos x}{1-{\sin }^{2}x}dx=\int \frac{1}{1-t^2}dt$$$$=\int \frac{1}{(1-t)(1+t)}dt=\frac{1}{2}\int (\frac{1}{1-t}+\frac{1}{1+t})dt=\frac{1}{2}\ln \mid \frac{1+t}{1-t}\mid +C$$$$=\frac{1}{2}\ln \mid \frac{1+\sin x}{1-\sin x}\mid +C=\frac{1}{2}\ln \mid \frac{{(1+\sin x)}^2}{1-{\sin }^{2}x}\mid +C$$$$=\frac{1}{2}\ln \mid \frac{1+\sin x}{\cos x}{\mid }^2+C=\ln \mid \sec x+\tan x\mid +C$$

Answered by stelor last updated on 12/Jan/22

$$I=\int \frac{1}{\cos x}dx$$$$u=sinx$$$$du=cosxdxand\frac{dx}{cosx}=\frac{du}{1-u^2}$$$$I=\int \frac{du}{1-u^2}$$$$I=argth\left(u\right)+c$$$$I=\frac{1}{2}ln\left(\frac{1+u}{1-u}\right)+c$$$$I=\frac{1}{2}ln\left(\frac{1+sinx}{1-sinx}\right)+c$$

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