Question

pls integral of xsin^22xdx

Answer 1

You mean:
\[\int x.sin^2(2x).dx\]^This?

Answer 2

yes i mean this

Answer 3

\[\int\limits x \sin ^2(2 x) \, dx\]
\[\int\limits x(sin(2x))^2 \, dx\]
Use \(sin(2x)=2sin(x)cos(x)\),
\[\int\limits x(2sinxcosx)^2 \, dx\]
\[\frac{1}{2}\int\limits x (1-\cos (4 x)) \, dx\]
Expand
\[\frac{1}{2}\int\limits (x-x \cos (4 x)) \, dx\]
Split
\[\frac{1}{2}\int\limits x \, dx-\frac{1}{2}\int\limits x \cos (4 x) \, dx\]
Integration by parts for xcos(4x)
u=x dv=cos(4x)
du=dx v=(1/4)sin(4x)
\[-\frac{1}{8} x \sin (4 x)+\frac{1}{2}\int\limits x \, dx+\frac{1}{8}\int\limits \sin (4 x) \, dx\]

Answer 4

Then for sin(4x)
Let t=4x
dt=4dx
\[\frac{1}{32}\int\limits \sin (t) \, dt-\frac{1}{8} x \sin (4 x)+\frac{1}{2}\int\limits x \, dx\]
Integrate

Answer 5

how did you come up with
1/2 \[\int\limits_{}^{}x(1-\cos4x)dx\]

Answer 6

\[\int\limits\limits x(2sinxcosx)^2 \, dx\]
To
\[\frac{1}{2}\int\limits\limits x (1-\cos (4 x)) \, dx\]?

Answer 7

yes

Answer 8

That's actually a trigonometry identity,
\[\sin ^2(2 x)=\frac{1}{2} (1-\cos (4 x))\]

Answer 9

ok