<p>I am having trouble with these integrals:</p>
<p>sinxcosx</p>
<p>(cos2x)^2. i thought it would be 1/6(cos2x)^3 but it wasnt one of the choices.</p>
<p>2xe^(-2x^2)</p>
<p>Any help would be greatly appreciated.</p>
<p>integral of sinxcosx can be solved by u-substitution.</p>
<p>u=sinx, du=cosx dx, </p>
<p>so you can do it as: integral u du, which is u^2/2,
which is (sinx)^2/2.</p>
<p>same for the other one. </p>
<p>u=-2x^2, du=-4x dx. Since the integral has to stay the same this would be:</p>
<p>-1/2 integral sign e^u du, which would be -1/2e^u, which would be (-1/2)e^(-2x^2)</p>
<p>Hope you understood. Read u-substitution section in your book. it's not that hard.</p>
<p>hey thanks...yeah, I thought of that later on. but for sinxcosx, the choices were -(sinx)^2/2 or (cosx)^2/2. I got -(cosx)^2/2 and the answer you got. </p>
<p>What about (cos2x)^2? My answer also wasnt in the choices. </p>
<p>The third one..yeah I completely forgot about u substitution and used the product rule, which didnt work. Thanks.</p>
<p>(cos2x)^2dx = 2(-2sin2x) = -4sin2x</p>
<p>You took the derivative...I was asking about the integral. </p>
<p>wouldnt it be (1/6)(sin2x)^3</p>
<p>actually no..thats wrong. So how do you do it?</p>
<p>It looks like you are trying to use u-sub and integrating cosu simultaneously to get the anti-derivative, but there is no such method. The anti-derivative of (cosu)^2du is 1/2<em>u + 1/4</em>sin2u + C, so with u=2x, you have 1/2<em>x + 1/8</em>sin4x + C.</p>
<p>duality is right, but the actual reasoning is quite simple...</p>
<p>trig identity:
(cosu)^2 = 1/2(1+cos2u), where u = 2x in this case.</p>
<p>which gives you the integral:</p>
<p>1/2(1 + cos(4x))dx</p>
<p>doing the integral yields the result...</p>
<p>x/2 + sin(4x)/8 + C</p>