How do you integrate #int x^2sin^2x#?

Question

Ezekiel Alexander · Accepted Answer

#(x^3)/6+(x*cos2x)/4-(x^2-2)*(sin2x)/4+C#

#int x^2*(sinx)^2*dx#

#=1/2*int x^2*(1-cos2x)*dx#

#=int (x^2*dx)/2#-#1/2*int x^2*cos2x*dx#

After using tabular integration for second integral.

#int x^2*cos2x*dx#

=#x^2*(sin2x)/2-2x*(cos2x)/(-4)+2*(sin2x)/(-8)#

=#(2x^2-4)*(sin2x)/4+(x*cos2x)/2#

Thus,

#int x^2*(sinx)^2*dx#

=#(x^3)/6+C#-#1/2*[(2x^2-4)*(sin2x)/4+(x*cos2x)/2]#

=#(x^3)/6+(x*cos2x)/4-(x^2-2)*(sin2x)/4+C#

Isabella Ackerman · Answer

undefined To integrate $ \int x^2 \sin^2(x) $, you can use integration by parts twice.

Let $ u = x^2 $ and $ dv = \sin^2(x) dx $.
Then, $ du = 2x \, dx $ and $ v = \frac{1}{2}(x - \sin(2x)) $.

Using the integration by parts formula $ \int u \, dv = uv - \int v \, du $, you get:

$$
\int x^2 \sin^2(x) \, dx = \frac{1}{2}x^2\sin(2x) - \int x \left(x - \frac{1}{2}\sin(2x)\right) \, dx
$$

Integrating the remaining integral by parts again:

Let $ u = x $ and $ dv = \left(x - \frac{1}{2}\sin(2x)\right) dx $.
Then, $ du = dx $ and $ v = \frac{1}{2}\left(x^2 - \frac{1}{4}\cos(2x)\right) $.

Applying integration by parts again, you get:

$$
\int x^2 \sin^2(x) \, dx = \frac{1}{2}x^2\sin(2x) - \left(\frac{1}{2}x^2\sin(2x) - \frac{1}{4}\int(2x\cos(2x)) \, dx\right)
$$

Simplify the expression:

$$
\int x^2 \sin^2(x) \, dx = \frac{1}{4}x^2\sin(2x) + \frac{1}{8}\cos(2x) - \frac{1}{4}x\sin(2x) + C
$$

Where $ C $ is the constant of integration.