How to compute this integral

[dbalinov]

I want to know how to resolve this integral:

[tex]\int\frac{dx}{cos^4x}[/tex]

Thank you in advance!

[dbalinov]

I solved it

[tex]\int\frac{dx}{\cos^4x}=\\=\int\frac{1}{\cos^2x}d\tan{x}=\\=\int\sec^2xd\tan{x}=\\=\tan{x}\sec^2x-\int\tan{x}d\sec^2x=\\=\tan{x}\sec^2x-\int\tan{x}.2\tan{x}\sec^2x dx=\\=\tan{x}(1+\tan^2x)-2\int\tan^2xsec^2xdx=\\=\tan{x}+\tan^3x-2\int\tan^2xd\tan{x}=\\=\tan{x} + \frac{3}{3}\tan^3x-\frac{2}{3}\tan^3x+C=\\=\tan{x}+\frac{1}{3}\tan^3x+C=\\=\frac{1}{3}(\tan^2x+1+2)\tan{x}+C=\\=\frac{1}{3}(\sec^2x+2)\tan{x}+C=\\=\frac{1}{3}(\frac{1}{\cos^2x}+2)\tan{x}+C[/tex]

[MM]

Continue posting here.

[garion]

[tex]\int\frac{dx}{\cos^4x}=\\=\int\frac{1}{\cos^2x}d\tan{x}=\\=\int\frac{\cos^2x+\sin^2x}{\cos^2x}d\tan{x}=\\=\int(1+\tan^2x)d\tan{x}=\\=\tan{x}+\frac{\tan^3{x}}{3}+C[/tex]