# Thread: Basic questions about integration by substitution

1. ## Basic questions about integration by substitution

1. How does it possible for sin(x)cos(x) dx to have two different solutions, namely (1/2)*(sin2x)+c and -(1/2)*(cos2x)+c, so that would bring to (rediculous) conclusion that sin2x+cos2x=0?
2. Integration by substitution makes use of differentials (not simply derivatives), thus is it purely anti-derivative?
Thank you in advance

2. ## Re: Basic questions about integration by substitution

Originally Posted by jojo7777777
1. How does it possible for sin(x)cos(x) dx to have two different solutions, namely (1/2)*(sin2x)+c and -(1/2)*(cos2x)+c, so that would bring to (rediculous) conclusion that sin2x+cos2x=0?
You are assuming that the 2 "c"s are the same. Yes, $\displaystyle \int sin(x)cos(x) dx= (1/2)sin^2(x)+ c$, for some constant, c. Yes, $\displaystyle \int sin(x)cos(x)dx= -(1/2)cos^2(x)+ C$ where "c" and "C" are generally NOT the same. Now that reduces to $\displaystyle sin^2(x)+ cos^2(x)=c- C$ which tells us that $\displaystyle sin^2(x)+ cos^2(x)$ is a constant- NOT "ridiculous".

2. Integration by substitution makes use of differentials (not simply derivatives), thus is it purely anti-derivative?
Thank you in advance
I don't know what you mean by"purely anti-derivative". It is, of course, an anti-derivateve. What does the "purely" indicate? And integration by substitution makes use of both differentials and derivatives which have a "one-to-one correpondence" so it really doesn't matter.

3. ## Re: Basic questions about integration by substitution

Thank you very much...!!!

4. ## Re: Basic questions about integration by substitution

Moreover 1/2 ( sin^2X) + C = 1/2 ( 1 - cos^2x ) + C = 1/2 -1/2 ( cos^2x) + C = -1/2 (cos^2x) + C +1/2 = -1/2 ( cos^2x) + c where c = C + 1/2
Thus both the solutions are the same.