how do you integrate tan^3(2x)sec^5(2x)dx?

Answer 1

Now the trick is to get rid of all tangents, then we can turn powers of secants into powers of U. The derivative of sec x is sec x(tan x), which we need to make the U substitution. So before we get rid of the tangents, factor out a tan*sec. tan^3(2x)sec^5(2x)dx = tan^2 (2x)*sec^4(2x)*(sec 2x*tan 2x)dx.

Now tan^2 = sec^x - 1, so our equation reduces to (sec^2 (2x) - 1)*sec^4 (2x)*(sec 2x*tan 2x)dx = (sec^6 (2x) - sec^4(2x))(sec 2x*tan 2x)dx. The expression is now written as a power series of secants multiplied by the derivative of secants.

Now let us make the substitution. U = sec(2x). dU = 2sec(2x)*tan(2x)dx. dx = dU/2sec(2x)*tan(2x)dx. So the integral is int(1/2(U^6 + U^4)du). This integrates to U^7/14 + U^5/10 + c = sec^7 (2x)/14 - sec^5 (2x)/10 + c.

Answer 2

Integrate Tan 2x

Answer 3

Integrate (tan^3 2x)(sec^5 2x) with respect to x.

First do some algebraic manipulation to make the expression easier to integrate.

(tan^3 2x)(sec^5 2x) = (tan 2x)[(sec^2 2x) - 1](sec^5 2x)

= (tan 2x)(sec^7 2x) - (tan 2x)(sec^5 2x)
___________

Now we can integrate.

∫[(tan^3 2x)(sec^5 2x)] dx

= ∫[(tan 2x)(sec^7 2x) - (tan 2x)(sec^5 2x)] dx

= (1/2)(1/7)(sec^7 2x) - (1/2)(1/5)(sec^5 2x) + C

= (sec^7 2x)/14 - (sec^5 2x)/10 + C