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Question Number 164612 by bounhome last updated on 19/Jan/22

$$solve:$$$$1.\int \frac{1}{sinx}dx$$$$2.\int \frac{1}{cosx}dx$$

Answered by Ar Brandon last updated on 19/Jan/22

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

Answered by Ar Brandon last updated on 19/Jan/22

$$\int \frac{1}{\cos x}dx=Q163829$$

Answered by Eulerian last updated on 20/Jan/22

$$$$$$\mathbf{Solution}:$$$$1.\int \frac{\sin \mathrm{x}}{{\sin }^2\mathrm{x}}\mathrm{dx}=\int \frac{\sin \mathrm{x}}{1-{\cos }^2\mathrm{x}}=-{\tanh }^{-1}\left(\cos \mathrm{x}\right)+\mathrm{C}$$$$$$$$2.\int \frac{\cos \mathrm{x}}{{\cos }^2\mathrm{x}}\mathrm{dx}=\int \frac{\cos \mathrm{x}}{1-{\sin }^2\mathrm{x}}\mathrm{dx}={\tanh }^{-1}\left(\sin \mathrm{x}\right)+\mathrm{C}$$$$$$

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