Question

(a) Evaluate:
Integrate(x² sin ² x )dx​

Answer 1

maths question

The answer is

=16x3−x24sin2x

−4cos2x

−18sin2x+C

Explanation:

Reminder

cos2x=1−2sin2x

sin

2x=12(1−cos2x)

Integration by parts

∫uv'

=uv−

∫u.v

The integral is

I=∫x2sin2xdx

=12∫x2(1-cos2x)dx

=12

∫x2−12

∫x2cos2xdx

=16x3−12∫x2cos2xdx

Perform the second integral by parts

u=x2

, ⇒

u'=2x

v'=cos2x

⇒

v=12sin2x

So,

∫x2cos2xdx=x22sin2x−∫xsin2xdx

Calculate the third intehral by parts

u=x

⇒

,

u'=1

v'=sin2x

,

⇒

, v=−12cos2x

So,

∫xsin2xdx=−x2cos2x+12

∫cos2xdx=−x2cos2x+14sin2x

Putting it all together

I=16x3−x24sin2x−x4cos2x−18sin2x+C

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