The derivative you calculated is incorrect.How do you take the integral of \(\displaystyle \frac{1500}{100+(t-5)^2}\)?
It is supposed to be \(\displaystyle 150\cdot arctan(\frac{t-5}{10})\) but its derivative I find is \(\displaystyle \frac{15000}{100+(t-5)^2}\).
What am I missing?
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)`.Taking the integral is more difficult, though. Please show how you do it.
\(\displaystyle \int \frac{du}{a^2 + u^2}\)Thank you, Dr. Peterson.
I did learn the method of substitution but not the integral of \(\displaystyle \frac{1}{a^2+u^2}\).
\(\displaystyle \int \frac{du}{a^2 + u^2}\)
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........
1 + tan2(x) = ??\(\displaystyle ∫\frac{sec^2(x)}{a(1+tan^2(x))}dx\)
Then...?
You really should supply the then part.\(\displaystyle ∫\frac{sec^2(x)}{a(1+tan^2(x))}dx\)
Then...?
1 + tan2(x) = ??
\(\displaystyle 1+tan^2(x)=sec^2(x)\)
Then:
\(\displaystyle ∫\frac{sec^2(x)}{a(1+tan^2(x))}dx=∫\frac{1}{a} dx=\frac{x}{a}\)
But it cannot be generalized that \(\displaystyle ∫\frac{du}{a^2+u^2}=\frac{x}{a}\)
Back substitute now....substitute:
u = a*tan(x)
Back substitute now....
u = a*tan(x) → u/a = tan(x) → x = tan-1(u/a)
This is grade-school algebra......
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)
You mean these MULTIPLE threads. You got the detailed answer at long before I wasted time working with you ........Thank you very much, I got the proof now.
This thread did shed light upon my path of learning calculus.
Best regards, gentlemen.
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
Oh ... the part about multiplying by 100/100 - to get the final expression??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.
Oh ... the part about multiplying by 100/100 - to get the final expression??
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.