Find the indefinate integral of sin^4mx(dx)?

Question

with out the MX i get 1/4 ((x-sin2x)/(2)) + (sin4x)/(x). what happends when you add in the mx????

Answer 1

∫sin^4(mx) dx
= (1/m) ∫sin^4(mx) d(mx)
= (1/m)∫[(1/2)(1-cos(2mx)]^2 d(mx)
= [1/(4m)]∫1 - 2cos(2mx) + (1/2)(1+cos(4mx)) d(mx)
= [1/(4m)][mx - sin(2mx) + (1/2)(mx + (1/4)sin(4mx)] + c

Answer 2

Find the indefinate integral of sin^4(mx) (dx).

First let's put the expression into a form easier to integrate.

sin^4(mx) = {[1 - cos(2mx)]/2}²
= (1/4)[1 - 2cos(2mx) + cos²(2mx)]
= 1/4 - (1/2)cos(2mx) + (1/4)cos²(2mx)
= 1/4 - (1/2)cos(2mx) + (1/4){[1 + cos(4mx)]/2}
= 1/4 - (1/2)cos(2mx) + 1/8 + (1/8)cos(4mx)
= 3/8 - (1/2)cos(2mx) + (1/8)cos(4mx)

Now we can integrate.

∫sin^4(mx) dx = ∫[3/8 - (1/2)cos(2mx) + (1/8)cos(4mx)] dx

= ⅜ x - ½ [1/(2m]sin(2mx) + ⅛ [1/(4m)]sin(4mx) + C

= ⅜ x - 1/(4m) sin(2mx) + 1/(32m) sin(4mx) + C

Answer 3

? ?tan(7x) sec²(7x) dx = observe that the integrand carries the two the function tan(7x) (whether decrease than root) and something (i.e. sec²(7x)) you may unquestionably replace into the spinoff of tan(7x); consequently permit: tan(7x) = u differentiate bth factors: d[tan(7x)] = du ? 7sec²(7x) dx = du ? sec²(7x) dx = (a million/7)du consequently, substituting, you get: ? ?tan(7x) sec²(7x) dx = ? ?u (a million/7)du = pull out the consistent: (a million/7) ? ?u du = (a million/7) ? u^(a million/2) du = (a million/7) {u^[(a million/2) +a million]} /[(a million/2) +a million] + C = (a million/7) [u^[(3/2)] /(3/2) + C = (a million/7)(2/3) ?u³ + C = (2/21) u?u + C then, substituting lower back u = tan(7x), you get: ? ?tan(7x) sec²(7x) dx = (2/21) tan(7x)?tan(7x) + C i'm hoping it facilitates... Bye!