Question

Evaluate the indefinite integral.

1/cos^5(x) dx

Answer 1

\[∫\frac{1}{\cos^5x}dx=∫(sec^5 x dx)\]
Let w = sec x, so sec^5 x = w^5/2
dw/dx = sec x tanx
dw = sec x tan x dx
\[dw = 2w \sqrt{w^2 - 1}dx\]
Do integration by substitution from here

Answer 2

oh okay thanks for the help

Answer 3

is dw the same with du?!

Answer 4

Yeah

Answer 5

ok thanks

Answer 6

No, this isn't a helpful substitution, since you will nave to reintroduce trig functions to get any further with it.

Instead try this: Apply integration by parts, stripping out a sec^2 term and integrating it to tan x, to find show:

\[\int\limits \sec^n x \ dx = {1 \over n-1} \sec^{n-2} x \tan x + {n-2 \over n-1} \int\limits \sec^{n-2} x \ dx\]


Then apply it to your integrand, sec^5 x, twice.