Question

integrate tan^4(x)sec(x)dx

Answer 1

I have it to a point where it's going to get really long...I think I'm having trouble with the reduction formulas (I have it split into 3 integrals right now)

Answer 2

\[\int\limits_{?}^{?}\sec^5xdx -2\int\limits_{?}^{?}\sec^3xdx + \int\limits_{?}^{?}secxdx\]

nevermind the ?'s on the intergrals...not sure how to get rid of them

Answer 3

I would write tan^4 = (sec^2 - 1)^2

Then expand the integral out into a number of integrals involving sec^n

Then find each of these in turn. You may find it helpful to use (and derive if needs be) the standard reduction for formula for

\[ I_n = \int \sec^n x \ dx = stuff + coefficient.\int \sec^{n-2} x \ dx \]

Answer 4

@JamesJ I wonder why don't we let u = tanx, ---> du = tanx secx dx
and we break the original function to \[\int\limits_{}^{} \tan^3 tanxsecxdx\] and then replace as \[\int\limits_{}^{} u^3 du = 3u^2 +C \] and replace back to x to get \[3\tan^3x + C\].is it wrong?

Answer 5

yes, it's wrong. The derivative of tan x is sec^2 x

Answer 6

thanks i got it :)

Answer 7

Thanks guys, forgot I asked this. Went back and looked at it and used reduction formulas (2 different ones for the sec^5 integral and the sec^3 integral) and just used the integral on the last (sec(x))...took a good sheet of paper but after combining like terms after I got the answers from the formulas, I came out with the right answer :D