Question

How do you evaluate the integral of `int t sin 2t dt`?

Answer 1

`int t*sin 2t*dt=-1/2t*cos 2t+1/4*sin 2t+C`

Explanation:

From the given integral `int t* sin 2t *dt`, the solution is by
Integration by Parts

`int u*dv=uv-int v*du`

Let `u=t`
Let `dv=sin 2t *dt`
Let `v=-1/2*cos 2t`
Let `du=dt`

`int u*dv=uv-int v*du`
`int t*sin 2t*dt=t*(-1/2cos 2t)-int -1/2cos 2t*dt`

`int t*sin 2t*dt=t*(-1/2cos 2t)+1/2int cos 2t*dt`

`int t*sin 2t*dt=-1/2t*cos 2t+1/4int cos 2t*2dt`

`int t*sin 2t*dt=-1/2t*cos 2t+1/4*sin 2t+C`

God bless....I hope the explanation is useful.