Estude! Com Prof. Caju

[tex3]\int \sec x \ \mathrm{d}x =\int \frac{1}{\cos x} \ \mathrm{d}x = \int \frac{\cos x}{\cos^2 x} \ \mathrm{d}x = \int \frac{\cos x}{1-\sen^2 x} \ \mathrm{d}x [/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\int \sec x \ \mathrm{d}x =\int \frac{1}{\cos x} \ \mathrm{d}x = \int \frac{\cos x}{\cos^2 x} \ \mathrm{d}x = \int \frac{\cos x}{1-\sen^2 x} \ \mathrm{d}x [/tex3]

Faça [tex3]\sen x = u \implies \mathrm{d}u = \cos x\ \mathrm{d}x[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\sen x = u \implies \mathrm{d}u = \cos x\ \mathrm{d}x[/tex3]

[tex3]\int \frac{\cos x}{1-\sen^2 x} \ \mathrm{d}x = \int \frac{1}{1-u^2}\mathrm{d}u[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\int \frac{\cos x}{1-\sen^2 x} \ \mathrm{d}x = \int \frac{1}{1-u^2}\mathrm{d}u[/tex3]

[tex3]\frac{1}{1-u^2} = \frac{1}{2(u+1)} - \frac{1}{2(u-1)} [/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\frac{1}{1-u^2} = \frac{1}{2(u+1)} - \frac{1}{2(u-1)} [/tex3]

[tex3]\int \frac{1}{1-u^2}\mathrm{d}u = \int \(\frac{1}{2(u+1)} - \frac{1}{2(u-1)}\)\mathrm{d}u[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\int \frac{1}{1-u^2}\mathrm{d}u = \int \(\frac{1}{2(u+1)} - \frac{1}{2(u-1)}\)\mathrm{d}u[/tex3]

Só continuar, o restante é trivial