<p>can someone explain in depth how to find the integral of this:</p>
<p>t*sin(t^2)</p>
<p>set t^2 as your "u" and diff. it. Then pull the 1/2 out of the integral and take the integral of sin(u)</p>
<p>more in depth...sry</p>
<p>tsin(t^2)
u = t^2
du = 2tdt
du/2 = tdt
(1/2)integral of sin(u)du = 1/2(-cos(u)) = -1/2cos(t^2)</p>
<p>can someone check my math</p>
<p>thats right...your fourth step should be du/2t = dt i think...looks good</p>
<p>ok thx esco</p>
<p>can't be du/2t = dt, you need the tdt in there to substitute the du in.</p>
<p>np obeserabbit</p>
<p>Or you can save some time and take the integral of sin(t^2) first. So you will get -cos(t^2) times 2t, but you dont want the 2t so take that reciprocal and multiply it by 1/(2t). Finally, the t's will cancel out from the original and you will be left with -1/2 cost(t^2).</p>
<p>or that :D</p>
<p>i always prefer "u" substitution...idk why</p>