integrate sec

 

from Mathematica:

 

ok thanks guys

 

Knowing the answer is not nearly as important as understanding the technique required to arrive at the answer. This is a common problem to practice integration by parts. If you write:

sec^3(x) =
sec(x) * sec^2(x) =
sec(x) * (1 + tan^2(x)) =
sec(x) + sec(x)tan^2(x) =
sec(x) + (sec(x)tan(x))tan(x)

the antiderivative of sec(x) is ln[sec(x) + tan(x)], and the antiderivative of (sec(x)tan(x))tan(x) is found using integration by parts:

f'(x) = sec(x)tan(x)
g(x) = tan(x)

f(x) = sec(x)
g'(x) = sec^2(x)

so the antiderivative is sec(x)tan(x) - Int[sec^3(x)]. Putting it all together, you get:

Int[sec^3(x)] = ln[sec(x) + tan(x)] + sec(x)tan(x) - Int[sec^3(x)].

Add Int[sec^3(x)] to both sides to get:

2*Int[sec^3(x)] = ln[sec(x) + tan(x)] + sec(x)tan(x)

Finally, divide by 2 to arrive at:

Int[sec^3(x)] = 1/2 {ln[sec(x) + tan(x)] + sec(x)tan(x)}

This is a bit simpler than pll's; Mathematica gives the correct answer, but not always the most elegant answer.

Remember to add the constant of integration.

 

wtf.... damn... this is what I'm going to have to go through... F me....