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$$∫\sin(x)\cos(x)dx$$ My question is: why are there 2 answers for this integral?

  1. we take $\sin x$ and $dx$ $\sin x$ is $\cos x$, then by power rule, integral is $\sin x^2 / 2 + C$
  2. if we multiply it by $2/2$, now it is $2\sin x \cos x = \sin 2x$ , which becomes $-\cos2x / 2 + C$

Could someone Explain Please.

thanks

Sebastiano
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RobbaN
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    Hint: use the double angle formula to show that these are, in fact, the same (don't forget the free additive constant). – lulu Jan 01 '24 at 17:59
  • @lulu isn't sinx^2 = 1-cos2x / 2 ? where is the 1 and 1/2 ? – RobbaN Jan 01 '24 at 18:02
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    The additive $1$ is irrelevant...it gets absorbed into the free additive constant. The multiplicative factor matters, but you made a computational error and it should be $-\frac {\cos(2x)}4$. – lulu Jan 01 '24 at 18:04
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    You didn't show your calculation but my guess would be that, when you multiplied by $1=\frac 22$, you forgot the $2$ in the denominator. – lulu Jan 01 '24 at 18:06
  • Each answer has infinitely many functions: $C$ is an arbitrary constant, and not keeping that in mind gives the illusion that you have two different answers when in fact both answers are identical: the $C$ in one answer does not match the $C$ in the other answer. You should graph $\sin^2(x)/2$ and $-\cos(2x)/4$ to see they are vertical shifts of each other. – KCd Jan 01 '24 at 19:57

3 Answers3

3

This is just a question on indefinite integration, and if your two answer hase a difference of a constant, they are both possible answers.
But i found that your two answers are not in that case, taht means you at least made a mistake in one of them.
And i got you have made mistake in the second method.
$\begin{aligned}&=\int sin(x)cos(x)dx\\&=\int \frac{2sin(x)cos(x)}{2}dx\\&=\int\frac{sin(2x)}{2}dx\\&=\int\frac{sin(2x)}{4}2dx\\&=\int\frac{sin(2x)}{4}d2x\\&=-\frac{cos(2x)}{4}+c\end{aligned}$
And i think your first answer is right.
So i will check if they really have a difference as a constant.
$\begin{aligned}differnce&=\frac{sin^2(x)}{2}-[-\frac{cos(2x)}{4}]\\&=\frac{sin^2(x)}{2}+\frac{cos(2x)}{4}\\&=\frac{sin^2(x)}{2}+\frac{1-2sin^2(x)}{4}\\&=\frac{1}{4}\end{aligned}$
I hope it will help.

Sando Jarvis
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welcome to MSE ;For the first one $y_1=\int =\frac{\sin^2 x}{2}+c_1$ for the second one $y_2=\int = \frac {\cos (2x)}{4}+c_2$ now if you take it right , there must be $$(y_1-y_2)'=0$$ id it true ?
$$(y_1-y_2)'=(\frac{\sin^2 x}{2}+c_1-(\frac {-\cos (2x)}{4}+c_2))'=\\ (\frac{2\sin x \cos x}{2}+0-(\frac{+2\sin2x }{4}+0))\\=(\sin x \cos x -\frac{4\sin x \cos x}{4})\\=0$$

Khosrotash
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Hint

$2\cos{x}\sin{x}=\sin{2x}\Leftrightarrow \cos{x}\sin{x}=\frac{\sin{2x}}{2}$

Hence

$\int\cos{x}\sin{x}\ dx=\int\frac{\sin{2x}}{2}\ dx $

1123581321
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