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

Question

17932 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-19T15:00:17

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

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.

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