How do you integrate #int x^2cos(3x)# by integration by parts method?

Question

William Abdalla · Accepted Answer

#1/27(9x^2-2)sin3x+(2x)/9cos3x+C#

Integration by parts formula

#I=intuv'dx=uv-intvu'dx#

#intx^2cos3xdx#

#u=x^2=>u'=2x#

#v'=cos3x=>v=1/3sin3x#

#:.I_1=x^2/3sin3x-int[(2x)/3sin3x]dx#

#:.I_1=x^2/3sin3x-2/3color(blue)(int[xsin3x]dx)#

we now have to use IBP on the integral highlighted in blue

#I_2=(int[xsin3x]dx)#

#u=x=>u'=1#

#v'=sin3x=>v=-1/3cos3x#

#I_2=(-x/3cos3x-int-1/3cos3xdx)#

#I_2=-x/3cos3x+1/9sin3x#

substituting back into #I_1#

#:.I_1=x^2/3sin3x-2/3(-x/3cos3x+1/9sin3x)+C#

#:.I_1=x^2/3sin3x+(2x)/9cos3x-2/27sin3x+C#

#I_1=(x^2/3-2/27)sin3x+(2x)/9cos3x+C#

#I_1=1/27(9x^2-2)sin3x+(2x)/9cos3x+C#

Ezekiel Alexander · Answer

undefined To integrate $ \int x^2 \cos(3x) $ using integration by parts:

Let $ u = x^2 $ and $ dv = \cos(3x) \, dx $.

Then, $ du = 2x \, dx $ and $ v = \frac{1}{3} \sin(3x) $.

Applying the integration by parts formula:

$$ \int u \, dv = uv - \int v \, du $$

We have:

$$ \int x^2 \cos(3x) \, dx = \frac{x^2}{3} \sin(3x) - \int \frac{1}{3} \sin(3x) \cdot 2x \, dx $$

Integrate the remaining integral, $ \int \frac{1}{3} \sin(3x) \cdot 2x \, dx $, by parts again if necessary.