<li>[sec^2(piroot(x)]/(rootx) dx</li>
<li>8sin(2x)cos^4(2x) dx</li>
<li>12cos^3(w) dw</li>
</ol>
<p>thanks for any help</p>
<ol>
<li> This is just general power rule. Sec^2 (piroot(x))/(rootx)<br>
The derivative of pirootx = pi/(2rootx) so, we have this in the integral, except we have to multiply by pi/2. Then it’s easy to solve.</li>
</ol>
<p>edit: I’m self-studying calculus BC… does anyone know if I need to know hyperbolic functions for the exam? Thanks!</p>
<p>Hyperbolics aren’t on the exam; my BC class skipped them.</p>
<ol>
<li>is straightforward u substitution. Let u = cos(2x) </li>
<li>is integration by parts by looking at it, but it might be something else because AB doesn’t cover that.</li>
</ol>
<ol>
<li>u-substitution - u=πsqrt(x). So you get (2/π)tan(πsqrtx)</li>
<li>u-substitution - u=cos(2x). So you get -.8(cos2x)^5</li>
<li>don’t believe you can do this with AB stuff - it’s integration by parts</li>
</ol>
<ol>
<li> Rewrite [cos(w)]^2 as 1 - [sin(w)]^2. Then:</li>
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<p>integral([cos(w)]^3 dw)
= integral(cos(w)[1 - (sin(w))^2] dw)
= integral([cos(w) - cos(w)[sin(w)]^2] dw).</p>
<p>The first term of the integral is straightforward, and the second term of the integral can be solved using u-substitution with u = sin w.</p>
<p>That being said questions like #3 that require trig identities to rewrite them are not on the AP Exam, even if they later turn into u-substitution questions.</p>
<p>^Right that works too. Trig stuff like that is still BC material though.</p>
<p>Oh please, #3? Parts. Get serious.</p>
<p>Ignore the 12 since it is a constant.</p>
<p>cos^3
(1 - sin^2)cos
u = sin
du = cos </p>
<p>Pretty straightforward.</p>
<p>Then again, I think Mu Alpha Theta has spoiled me.</p>
<p>thanks all</p>