How do you integrate #int ( 6x^3+3x+sqrt7 ) sin( 5x ) dx#?

1 Answer
Oct 23, 2015

#I = (-(6x^3)/5-(39x)/125-sqrt7/5)cos5x+ ((18x^2)/25-21/625)sin5x + C#

Explanation:

Let's solve (using Integration By Parts):

#I_n = int x^nsinaxdx#

#u=x^n => du = nx^(n-1)dx#

#dv = sinaxdx => v = int sinaxdx = -1/acosax#

#I_n = -x^n/acosax + n/a int x^(n-1)cosaxdx#

#u=x^(n-1) => du=(n-1)x^(n-2)dx#

#dv= cosaxdx => v = int cosaxdx = 1/asinax#

#I_n = -x^n/acosax + n/a [x^(n-1)/asinax-(n-1)/a int x^(n-2)sinaxdx]#

#I_n = -x^n/acosax + (nx^(n-1))/a^2 sinax-(n(n-1))/a^2 I_(n-2)#

This is recurrent formula, we need to find #I_1# and #I_2# in general case. For our task, we need #I_1#:

#I_1 = int xsinaxdx#

#u=x => du=dx#

#dv = sinaxdx => v = int sinaxdx = -1/acosaxdx#

#I_1 = -x/acosax + 1/aint cosaxdx = -x/acosax + 1/a^2sinax#

#a=5 => I_1 = -x/5cos5x+1/25sin5x#

#n = 3 => I_3 = -x^3/5cos5x+ (3x^2)/25sin5x-6/25I_1#

#I_3 = -x^3/5cos5x+ (3x^2)/25sin5x-6/25(-x/5cos5x+1/25sin5x)#

#I_3 = -x^3/5cos5x+ (3x^2)/25sin5x + (6x)/125cos5x - 6/625sin5x#

#I = int (6x^3+3x+sqrt7)sin5xdx#

#I = 6intx^3sin5xdx + 3int xsin5xdx +sqrt7 int sin5xdx#

#I = 6I_3+3I_1-sqrt7/5cos5x+C#

#I = 6( -x^3/5cos5x+ (3x^2)/25sin5x + (6x)/125cos5x - 6/625sin5x)+3(-x/5cos5x+1/25sin5x)-sqrt7/5cos5x+C#

#I = (-(6x^3)/5+(36x)/125-(3x)/5-sqrt7/5)cos5x+ ((18x^2)/25-36/625+3/25)sin5x + C#

#I = (-(6x^3)/5-(39x)/125-sqrt7/5)cos5x+ ((18x^2)/25-21/625)sin5x + C#