I really don't get the substitution rule

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SUMMARY

The discussion centers on the substitution rule in calculus, specifically regarding the integral of cos(3x). The correct integration process involves substituting u = 3x, leading to du = 3dx, which implies dx = (1/3)du. This substitution results in the integral ∫cos(3x)dx = (1/3)∫cos(u)du = (1/3)sin(u) + C = (1/3)sin(3x) + C. The confusion arises from the need to adjust for the derivative of the substitution, which introduces the factor of 1/3.

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tony873004
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I really don't get the substitution rule. This is supposed to be the easiest problem in the homework set: u=3x
\int {\cos \,3x\,\,dx\,\, = \,\,\int {\cos \,u\,\, = \,\,\sin \,u + C\,\, = \,\,\sin 3x + C} }

But the right answer is 1/3 sin(3x). Where did the 1/3 come from?
 
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Look at your second integral in here:
\int {\cos \,3x\,\,dx\,\, = \,\,\int {\cos \,u\,\,dx?\,\, = \,\,\sin \,u + C\,\, = \,\,\sin 3x + C} }

with respect to what variable? x?
or, did you mean to have a du in there?

If you meant for a du to be there, what makes you think that du can be substituted for dx? You wrote u=3x. if I take the derivative, with respect to x, I get du/dx = 3. Multiplying both sides by dx (thanks, Leibnitz for that notation), I get du = 3 dx. And from that, dx will be equal to 1/3 du
 
Actually, I think a "harder" integral will make this easier to see:
\int {\sec x \sec x \tan xdx
Let u = secx
du/dx = secxtanx
du = secxtanxdx

Now, when you substitute, u will take care of the first secx.
But, du will take care of the rest of it: secxtanxdx
 
I think I did it right, but I still have a question.

u = 3x,\,\,\,\,\frac{{du}}{{dx}} = \left( {3x} \right)^\prime = 3,\,\,\,\,dx = \frac{{du}}{\begin{array}{l}<br /> 3 \\ <br /> \\ <br /> \end{array}},\,\,\,\,\int {\cos \,3x\,\,dx\,\,} = \,\,\int {\cos \,u\,\,\frac{{du}}{3} = \frac{{\sin u}}{3}} = \frac{{\sin 3x}}{3}

So when I integrate, the du simply disappears?

I hate Leibnitz notation. I do not think it is intuitive. I wish in Calc 1 they would have spent a lecture on Leibnitz notation. Rather, they just started using it without describing it.
 
Last edited:
Yes, the du disappears with the integral sign (as would dx, dy, etc). Kind of like: 4 + 3 = 7, we don't leave the + there.
 
\int {\cos \,u\,\,\frac{{du}}{3}

At this point you should factor out the constant, 1/3, from the integral:

\frac{1}{3} \int {\cos \,u\,\,du
 
mattmns said:
...Kind of like: 4 + 3 = 7, we don't leave the + there.

:smile: Thanks to you, I will never forget that.

Thanks, everyone. :rolleyes: This was so confusing to me until I asked here. I just plowed through the next 10 problems with ease. But I'm stuck on the 11th, so I'm giving up for the night.
 
Man, that was an easy problem. Think forward of the answer. I guess light years will come up for your mind to catch up.
 
jermanie said:
Man, that was an easy problem.

It's only easy if you know the answer! When learning, any problem can be difficult if you can't see what to do!
 

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