Integral of 1500/(100+(t-5)^2)
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The derivative you calculated is incorrect.
Please post your work about calculating the derivative.
Thanks, Subhotosh. I found chain rule was missing:
Since \(\displaystyle \frac{d}{dx}arctan(x)=\frac{1}{1+x^2}\),
\(\displaystyle \frac{d}{dt}150\cdot arctan(\frac{t-5}{10})=150\cdot \frac{1}{1+\frac{(t-5)^2}{10^2}}\cdot \frac{1}{10}=15\cdot \frac{1}{\frac{10^2+(t-5)^2}{10^2}}=\frac{1500}{{10^2+(t-5)^2}}\)
Taking the integral is more difficult, though. Please show how you do it.
Since \(\displaystyle \frac{d}{dx}arctan(x)=\frac{1}{1+x^2}\),
\(\displaystyle \frac{d}{dt}150\cdot arctan(\frac{t-5}{10})=150\cdot \frac{1}{1+\frac{(t-5)^2}{10^2}}\cdot \frac{1}{10}=15\cdot \frac{1}{\frac{10^2+(t-5)^2}{10^2}}=\frac{1500}{{10^2+(t-5)^2}}\)
Taking the integral is more difficult, though. Please show how you do it.
Dr.Peterson
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Have you learned the method of substitution (or u-substitution)? Let `u = (t-5)/10` and see what happens. You need to have learned the integral of `1/(a^2 + u^2)`.
This could also be done with trigonometric substitution, which is commonly taught later. Let `t-5 = 10 tan(u)`.
Steven G
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Fine so here is a quick lesson. [math]\int \dfrac{1}{a^2+u^2}du= \dfrac{1}{a}arctan\dfrac{u}{a} + c.[/math]Now you should (in my opinion) NEVER believe anything that anyone tells in math, especially your teacher! I am not saying that your teacher is wrong! You should verify everything in math that you are told so that you understand it. Hence, please go ahead and take the derivative of my answer to verify that you get back [math]\dfrac{1}{a^2+u^2}.[/math]
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\(\displaystyle \int \frac{du}{a^2 + u^2}\)
substitute:
u = a*tan(x)
du = a*sec2(x) dx ........... then
\(\displaystyle \int \frac{a * sec^2(x)}{a^2 + a^2 * tan^2(x)}dx\)
continue........
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1 + tan2(x) = ??
Steven G
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You really should supply the then part.
Since you were told to make this substitution it means that things should become simpler. At this point I would thing identities (as Sir Khan pointed out)
\(\displaystyle 1+tan^2(x)=sec^2(x)\)
Then:
\(\displaystyle ∫\frac{sec^2(x)}{a(1+tan^2(x))}dx=∫\frac{1}{a} dx=\frac{x}{a}\) + C
But it cannot be generalized that \(\displaystyle ∫\frac{du}{a^2+u^2}=\frac{x}{a}\)
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Back substitute now....
u = a*tan(x) → u/a = tan(x) → x = tan-1(u/a)
This is grade-school algebra......
Thank you very much, I got the proof now.
Hi Jomo, yes, I just did and realized the proof. Thank you for the quick lesson, clear and to the point, I did verify it.
By the way, I found this about your interesting quote: https://quoteinvestigator.com/2015/02/15/hidden-cat/
This thread did shed light upon my path of learning calculus.
Best regards, gentlemen.
skeeter
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Integral of 1500/(100+(t-5)^2)
How do you take the integral of $\frac{1500}{100+(t-5)^2}$? It is supposed to be $150\cdot arctan(\frac{t-5}{10})$ but its derivative I find is $\frac{15000}{100+(t-5)^2}$. What am I missing?
mathforums.com
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You mean these MULTIPLE threads. You got the detailed answer at long before I wasted time working with you ........
Integral of 1500/(100+(t-5)^2)
How do you take the integral of $\frac{1500}{100+(t-5)^2}$? It is supposed to be $150\cdot arctan(\frac{t-5}{10})$ but its derivative I find is $\frac{15000}{100+(t-5)^2}$. What am I missing?
mathforums.com
You mean these MULTIPLE threads. You got the detailed answer at long before I wasted time working with you ........
Integral of 1500/(100+(t-5)^2)
How do you take the integral of $\frac{1500}{100+(t-5)^2}$? It is supposed to be $150\cdot arctan(\frac{t-5}{10})$ but its derivative I find is $\frac{15000}{100+(t-5)^2}$. What am I missing?
Not really, the proof wasn't provided in the other thread. Also, they ran parallel, so maybe "long before" is a little exaggerated. Thanks for your time, I really appreciate it.
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Oh ... the part about multiplying by 100/100 - to get the final expression??
Well that is really far away from the u = a*tan(x) substitution you presented to begin the proof with. Thanks again.
Steven G
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The point is tutors feel the they are being used when students post their questions on multiple site. The reason is that it is quite possible that there are two tutors giving the same advise when instead of both helping you one of the tutors could have been helping something else. Sounds a bit selfish, doesn't it? If this didn't happen this time then it was by pure luck. The fact that you think it is not a big deal is enough for me not to help you ever again and to ask a moderator to ban you from the forum.
When you do your research you certainly look to different sources, which provide different perspectives of the same topic. Furthermore, the threads can be viewed by other students who will learn from them. The fact that you are giving so much attention making a big deal of it is wasting way more time than the question itself. So, thank you and have a nice Sunday.