1.33.1 that the answer you have to mutiply
I hope this helps you
perimeter of square =4x
4x=equilateral triangle perimeter
4x=3.1
x=3/4
Answer:
B
Step-by-step explanation:
-1/4 cos 2x + C
<span>or </span>1/2 sin^2 x + C
or -1/2 cos^2 x + C
well, <span><span>sinx</span><span>cosx</span>=<span><span><span>sin2</span>x</span>2</span></span> so you are looking at<span><span>12</span>∫ <span>sin2</span>x <span>dx</span>=<span>(<span>12</span>)</span><span>[<span>(<span>12</span>)</span><span>(−<span>cos2</span>x)</span>+C]</span>=−<span>14</span><span>cos2</span>x+C'</span>
or maybe easier you can notice the pattern that <span><span>(<span><span>sinn</span>x</span>)</span>'=n<span><span>sin<span>n−1</span></span>x</span><span>cosx</span></span> and pattern match. here <span>n−1=1</span> so n = 2 so we trial <span><span>(<span><span>sin2</span>x</span>)</span>'</span> which gives us <span>2<span>sinx</span><span>cosx</span></span> so we now that the anti deriv is <span><span>12</span><span><span>sin2</span>x</span>+C</span>
the other pattern also works ie<span><span>(<span><span>cosn</span>x</span>)</span>'=n<span><span>cos<span>n−1</span></span>x</span><span>(−<span>sinx</span>)</span>=−n<span><span>cos<span>n−1</span></span>x</span><span>sinx</span></span>
so trial solution <span><span>(−<span><span>cos2</span>x</span>)</span>'=−2<span>cosx</span><span>(−<span>sinx</span>)</span>=2<span>cosx</span><span>sinx</span></span> so the anti deriv is <span>−<span>12</span><span><span>cos2</span>x</span>+<span>C</span></span>
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