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Alex Ar [27]
3 years ago
8

Find the antiderivative of cos x + sin x?

Mathematics
1 answer:
IrinaK [193]3 years ago
6 0

-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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\tt{ \: H \: E \: Y \: \:  T \: H \: E \: R\: E \: } !

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