At 1:18 when you cancel out the sin(x) you must declare that sin(x) cannot be zero (that is x is not 0 degrees or 180 degrees, etc). There is nothing in the problem that declares that this is OK to do so technically this is algebraically incorrect
what I first thought was integration by parts, we write ∫sin^3(x) as ∫sin^2(x)sinx then using the integration by parts we write it as: sin^2(x)(-cos(x) )+ 2∫sinxcos^2(x)=> sin^2(x)(-cos(x)) + 2 ∫sin(x)(1-sin^2(x))... then we get ∫sin^3(x) on both sides which we can take as 'I' then solve the rest of the integral (which is easy) to get ∫sin^3(x), i dont know if this will be easier or not but this is what I think might be quickest for me.
My first try for integrals is kinda guessing part of the integral and try to fix differences. Though that fails a good amount of times, it's a pretty fast approach and in case that fails you can try the more systematical, but slower approaches, without having lost much time: Guess that cos^3(x) is part of the integral, so derive it to see the differences. d/dx (cos^3(x)) = 3 cos^2(x) (-sin(x)) = 3 (1 - sin^2(x)) (-sin(x)) = 3 sin^3(x) - 3 sin(x) Here the differences are easily fixable. integral sin^3(x) dx = integral 1/3 (3 sin^3(x) - 3 sin(x)) + sin(x) dx = 1/3 cos^3(x) - cos(x) + c
At 1:18 when you cancel out the sin(x) you must declare that sin(x) cannot be zero (that is x is not 0 degrees or 180 degrees, etc). There is nothing in the problem that declares that this is OK to do so technically this is algebraically incorrect
what I first thought was integration by parts, we write ∫sin^3(x) as ∫sin^2(x)sinx then using the integration by parts we write it as: sin^2(x)(-cos(x) )+ 2∫sinxcos^2(x)=> sin^2(x)(-cos(x)) + 2 ∫sin(x)(1-sin^2(x))... then we get ∫sin^3(x) on both sides which we can take as 'I' then solve the rest of the integral (which is easy) to get ∫sin^3(x), i dont know if this will be easier or not but this is what I think might be quickest for me.
@@scutyardwilliamgate nice I think this should work nicely as well! Neat!!
My first try for integrals is kinda guessing part of the integral and try to fix differences. Though that fails a good amount of times, it's a pretty fast approach and in case that fails you can try the more systematical, but slower approaches, without having lost much time:
Guess that cos^3(x) is part of the integral, so derive it to see the differences.
d/dx (cos^3(x)) = 3 cos^2(x) (-sin(x)) = 3 (1 - sin^2(x)) (-sin(x)) = 3 sin^3(x) - 3 sin(x)
Here the differences are easily fixable.
integral sin^3(x) dx = integral 1/3 (3 sin^3(x) - 3 sin(x)) + sin(x) dx = 1/3 cos^3(x) - cos(x) + c
Nice! Is there a name for this technique? I've definitely used something similar to integrate things like xe^x
@@JPiMaths I am not aware that that technique has a name, though i doubt it has one, given that you highly rely on intuitition on the first step.