Why doesn't my method for integrating sinxcosx work?

[wahaj]

I know the way to do $\int sinxcos$ is by u-substitution but why doesn't the following work?
$$sin(2x) = 2sinxcosx \\ \frac{sin(2x)}{2}=sinxcosx \\ \int sinxcosx= \frac{1}{2} \int sin(2x) = -\frac{cos(2x)}{4}$$

 

[PeroK]

There's nothing wrong with that!

 

[Mark44]

I know the way to do $\int sinxcos$ is by u-substitution but why doesn't the following work?
$$sin(2x) = 2sinxcosx \\ \frac{sin(2x)}{2}=sinxcosx \\ \int sinxcosx= \frac{1}{2} \int sin(2x) = -\frac{cos(2x)}{4}$$

What makes you think it doesn't work?

You can always check the result from integrating by differentiating it. If you get the original integrand, then your integration was correct.

There are two things you are omitting: The dx in your integral and the constant of integration. You might be surprised to learn that the constant plays an important role in this problem.

 

[wahaj]

sorry about that. I have a habit of leaving those out when I am doing practice questions. Could the constant be $\frac{1}{4}$ by any chance?

 

[HallsofIvy]

In other words, you should have
$$\int sin(x)cos(x)dx= \frac{1}{2}\int sin(2x) dx= -\frac{cos(2x)}{4}+ C$$

What may be bothering you is that you could also do this integral with the substitution u= sin(x) so that du= cos(x) dx and that becomes
$$\int u du= \frac{u^2}{2}+ D= \frac{sin^2(x)}{2}+ D$$

Or use the substitution u= cos(x) so du= -sin(x)dx and the integral becomes
$$-\int u du= -\frac{u^2}{2}+ E= -\frac{cos^2(x)}{2}+ E$$

Those are, in fact, all the same. Use $cos^2(x)= 1- sin^2(x)$ to go from the third to the second and $cos(2x)= cos^2(x)- sin^2(x)= (1- sin^2(x))- sin^2(x)= 1- 2sin^2(x)$ and $cos(2x)= cos^2(x)-sin^2(x)= cos^2(x)- (1- cos^2(x))= 2cos^2(x)- 1$ to go from the first to the second and third form.

 

[wahaj]

I did try to solve the identity before posting here. I was leftover with that 1/4 term. I forgot that there was also a constant there. Thanks for the help, I think I can take it from here.