<p>I have a test tommorow on anti-derivatives and have no clue how do to do these problems:</p>
<p>integral/swirly line of ((11cos^2)*5x)dx</p>
<p>and</p>
<p>integral/swirly line of ((5tan^2)*8x)dx</p>
<p>THANK YOU!</p>
<p>Swirly line? LOL.</p>
<p>Um ... haven't done these in forever. Let's see...</p>
<p>Holy god never mind. Can't do integrals with sin, cos, etc. in them. Could've done so a year ago. But, you can find the final answer through searching for tools on the Internet that automatically integrate stuff (I know they exist because I used to use them all the time).</p>
<p>I don't get your notation. </p>
<p>Like, for the first one, do you mean:
integral(11cos^2(5x)) or
integral(11(cos(x))^2(5x))?
Sorry. I'm bad at internet notation. Is the cosine squared of x or of 5x?</p>
<p>yea me too...</p>
<p>it's so hard...to read....</p>
<p>for the first one its 11cosine squared of 5x. i put parantheses because i didnt want you to think iits cosine to the 2 times 5x power.</p>
<p>11<em>int((cos(5x))^2)dx = 11/5</em>int((cos(u))^2)du, with u=5x = (11/5)<em>int(1/2 + (1/2)</em>cos(2u))du, from the identity (cos(u))^2=1/2 + (1/2)<em>cos(2u) = (11/10)</em>int(1/2 + (1/2)<em>cos(v))dv, with v=2u = (11/10)</em>(v/2 + sin(v)/2) = (11/10)<em>(5x + sin(10x)/2) = (11/10)</em>(sin(5x)<em>cos(5x) + 5x), because sin(2</em>5x) = 2<em>sin(5x)</em>cos(5x)</p>
<p>5<em>int((tan(8x))^2)dx = (5/8)</em>int((tan(u))^2)du, with u=8x = (5/8)<em>int((sec(u)^2) - 1)du, since (tan(u))^2=(sec(u))^2 - 1 = (5/8)</em>(tan(u) - u) = (5/8)*(tan(8x) - 8x)</p>
<p>So tedious, but I don't have much to do this morning.</p>