How Do You Integrate (sec x)^4?

[teng125]

May i know how to integrate (sec x)^4 ??

the answer is tan x + 1/3 (tan x)^3

 

[Pyrrhus]

Try writing your integral as
$$\int (\tan^{2} x + 1) \sec^{2} x dx$$

 

[teng125]

okok...thanx

 

[fargoth]

try using $$u=tg(x)$$
$$du=\frac{dx}{cos^2(x)}$$
and $$\frac{1}{cos^2(x)}=1+tg^2(x)=1+u^2$$