Using the substitution rule for definite integrals

[Hckyplayer8]

It took a long, long time for me to get semi proficient with the chain rule. I can tell the substitution rule will be likewise.

So far I have let u = delta/3 which means du = 1/3

Also the antiderivative of cos is sin.

Now what do I do?

 

[MarkFL]

If you let:

[MATH]u=\frac{\theta}{3}[/MATH]
Then:

[MATH]du=\frac{1}{3}\,d\theta\implies d\theta=3\,du[/MATH]
You will also need to rewrite your limits in terms of \(u\), so that your definite integral becomes:

[MATH]3\int_0^{\frac{\pi}{3}} \cos(u)\,du[/MATH]

 

[Harry_the_cat]

It's theta not delta BTW.

 

[pka]

View attachment 14748

Does anyone doubt that the anti-derivative here is some form of \(\displaystyle \sin\left(\frac{\theta}{3}\right)\)
Now \(\displaystyle D_{\theta}\left[\sin\left(\tfrac{\theta}{3}\right)\right]=\tfrac{1}{3}\cos\left(\tfrac{\theta}{3}\right)\)
\(\displaystyle \int_0^\pi {\cos \left( {\tfrac{\theta }{3}} \right)d\theta } = \left. {3\sin \left( {\tfrac{\theta }{3}} \right)} \right|_0^\pi=~?\)

 

[Steven G]

View attachment 14748

It took a long, long time for me to get semi proficient with the chain rule. I can tell the substitution rule will be likewise.

So far I have let u = delta/3 which means du = 1/3

Also the antiderivative of cos is sin.

Now what do I do?

Player, you must put the \(\displaystyle d\theta\) after the 1/3. The reason is that you have to account for each and every piece of the integral when you change over to u's. For example \(\displaystyle \cos (\frac{\theta}{3})\) becomes \(\displaystyle \cos(u)\) but what does \(\displaystyle d\theta\) become? What I am saying will become more obvious when the integrand is more complicated than just \(\displaystyle \cos\frac{\theta}{3}\). Don't forget to change the limits to u's or convert back to x's. You should try it both ways to see which you like better.

Saying Also the antiderivative of cos is sin is pure nonsense. Next time say that the antiderivative of cosx is sinx (you should even say +c)

 

[Hckyplayer8]

Thank you all for posting.

Based on all the feedback...

the lower limit of u is 0 and the upper limit is pi/3. cos (u) becomes sin (u) based off the common antiderivative rules. Plug in the bounds results in
sin (pi/3) - sin (0) = sin (pi/3) which simplifies to cubic root of three over two.

My only question left is why is the derivative of theta 1? Just cause it is a variable?

 

[pka]

My only question left is why is the derivative of theta 1? Just cause it is a variable?

If \(\displaystyle D_x(x)=1\) then \(\displaystyle D_{\theta}(\theta)=1\).