How do you find the integral of #f(x)=x^2sinx# using integration by parts?

Question

Emily Ammerman · Accepted Answer

The formula for intergration by parts is the following: #int g(x) h'(x) dx = g(x) h(x) - int g'(x) h(x) dx# This means, that with your product #x^2 * sin(x)# you must determine which factor is #g(x)# (which you will differentiate later) and which factor is #h'(x)# (which you will integrate later). Here, the choice is easy since #sin# or #cos# don't get more complicated when being integrated, so it's generally an excellent choice to pick this term as the term-to-be-inegrated. So, if I were you, I would totally choose 
# g(x) = x^2# and #h'(x) = sin(x)#. Based on this choice, let's compute #g'(x)# and #h(x)#: #g(x) = x^2 => g'(x) = 2x#
#h'(x) = sin(x) => h(x) = - cos(x)# So, let's apply the rule: #int x^2 sin(x) = x^2 * (-cos(x)) - int x* (- cos(x)) dx#
# = - x^2 cos(x) + int x  cos(x) dx# Believe it or not, this now easier. 
However, we still need to apply integrations by parts once again. Here, again, we would like to differentiate #x# and to integrate #cos(x)#, so it's  #g(x) = x => g'(x) = 1# and 
#h'(x) = cos(x) => h(x) = sin(x)#. So, 
#int x cos(x) dx = x sin(x) - int 1 * sin(x) dx#
# = x sin(x) - (-cos(x))#
# = x sin(x) + cos(x)# In total, we get:
#int x^2 sin(x)#
# = - x^2 cos(x) + int x  cos(x) dx#
# = - x^2 cos(x) + x sin(x) + cos(x)#. Hope that this helped!

Cameron Alexander · Answer

#intx^2sin(x)dx=-x^2cos(x)+2xsin(x)+2cos(x)+c#,

where #c# is the constant of integration.

#intx^2sin(x)dx=-intx^2frac{d}{dx}(cos(x))dx# #=-[x^2cos(x)-intfrac{d}{dx}(x^2)cos(x)dx]# #=-x^2cos(x)+int(2x)cos(x)dx# #=-x^2cos(x)+2intxfrac{d}{dx}(sin(x))dx# #=-x^2cos(x)+2[xsin(x)-intfrac{d}{dx}(x)sin(x)dx]# #=-x^2cos(x)+2xsin(x)-2intsin(x)dx# #=-x^2cos(x)+2xsin(x)+2cos(x)+c#, where #c# is the constant of integration.

Axel Alvarez · Answer

undefined To find the integral of $ f(x) = x^2 \sin(x) $ using integration by parts, we use the formula:

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

where $ u $ and $ dv $ are chosen such that $ du $ and $ v $ can be easily computed.

Let's choose:

$$ u = x^2 $$  
$$ dv = \sin(x) \, dx $$

Then, compute $ du $ and $ v $:

$$ du = 2x \, dx $$  
$$ v = -\cos(x) $$

Now, apply the integration by parts formula:

$$ \int x^2 \sin(x) \, dx = -x^2 \cos(x) - \int -2x \cos(x) \, dx $$

This leaves us with another integral that can be computed. Integrating $ -2x \cos(x) $ by parts again:

$$ \int -2x \cos(x) \, dx = -2x \sin(x) - \int -2 \sin(x) \, dx $$

Now, integrate $ -2 \sin(x) $ to get:

$$ \int -2 \sin(x) \, dx = 2\cos(x) $$

Substitute this back into the previous expression:

$$ \int -2x \cos(x) \, dx = -2x \sin(x) - 2\cos(x) $$

Now, substitute this result back into the original equation:

$$ \int x^2 \sin(x) \, dx = -x^2 \cos(x) + 2x \sin(x) + 2\cos(x) + C $$

Where $ C $ is the constant of integration.