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 October 17th, 2017, 10:32 AM #1 Member   Joined: Nov 2016 From: Ireland Posts: 84 Thanks: 3 Definite integrals Hi guys I'm working on these integrals. See this one second from the bottom, with the cos 2x dx? My lecturer has given the answer of $\displaystyle [1/2 \sin 2x]$ I'm thinking this should be $\displaystyle [2/2 \sin 2x]$ Last edited by skipjack; October 17th, 2017 at 01:00 PM. October 17th, 2017, 10:36 AM #2 Senior Member   Joined: Sep 2015 From: USA Posts: 2,531 Thanks: 1390 Your lecturer is correct. Thanks from Kevineamon October 17th, 2017, 11:05 AM #3 Member   Joined: Nov 2016 From: Ireland Posts: 84 Thanks: 3 Lol thx Rom - Ok say it was the 2x on it's own = $\displaystyle 2x^1$ Integral $\displaystyle 2x^1$ = $\displaystyle (2x^2)/2$ Right? So what am I missing here? how does the trigonometric function change that? October 17th, 2017, 11:36 AM   #4
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 Originally Posted by Kevineamon Lol thx Rom - Ok say it was the 2x on it's own = $\displaystyle 2x^1$ Integral $\displaystyle 2x^1$ = $\displaystyle (2x^2)/2$ Right? So what am I missing here? how does the trigonometric function change that?
oy...

you can't apply polynomial derivative rules to trig functions.

Just take the derivative of $\sin(2x)$ vs. $\dfrac 1 2 \sin(2x)$ and see which one gets you $\cos(2x)$ October 17th, 2017, 12:32 PM #5 Member   Joined: Nov 2016 From: Ireland Posts: 84 Thanks: 3 Hmmm I'm a bit of a n00bee with these Rom According to my calculations this is the derivative of: sin(2x)  October 17th, 2017, 12:55 PM #6 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,675 Thanks: 2655 Math Focus: Mainly analysis and algebra You know that $\sin(2x)$ is the sine of $2x$? It's not $\sin{}$ times $2x$. $\sin{}$ on its own is meaningless. It needs an argument. You'd use the chain rule for that derivative. Thanks from topsquark Last edited by v8archie; October 17th, 2017 at 12:58 PM. October 17th, 2017, 01:34 PM #7 Member   Joined: Nov 2016 From: Ireland Posts: 84 Thanks: 3 Hmm k - I wonder if someone could give me the answers to Roms questions, in a short step format. I seem to have unlearned everything I thought I learned. Sorry guys October 17th, 2017, 01:53 PM #8 Senior Member   Joined: Sep 2015 From: USA Posts: 2,531 Thanks: 1390 $\dfrac {d}{dx}\left( \sin(2x)\right) = \cos(2x)\cdot (2) = 2 \cos(2x)$ $\dfrac {d}{dx}\left( \dfrac 1 2 \sin(2x)\right) = \dfrac 1 2 \cos(2x) \cdot 2 = \cos(2x)$ soooo $\displaystyle \int \cos(2x)~dx = \dfrac 1 2 \sin(2x)$ October 17th, 2017, 03:40 PM #9 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 If your lecture actually said that the "answer" to this problem is $\displaystyle (1/2)\sin(2x)$ that is incorrect! This is, as you say in the title of this thread, a definite integral so the "answer" is a number, not a function of x! To integrate $\displaystyle \int_{\pi/4}^{\pi/2} \cos(2x) dx$, let $\displaystyle u= 2x$. Then $\displaystyle du= 2dx$ so $\displaystyle dx= \frac{1}{2}du$. When $\displaystyle x= \pi/4$, $\displaystyle u= 2(\pi/4)= \pi/2$ and when $\displaystyle x= \pi/2$ $\displaystyle u= 2(\pi/2)= \pi$. The integral becomes $\displaystyle \frac{1}{2}\int_{\pi/2}^{\pi} \cos(u)du= \left[(1/2) \sin(u)\right]_{\pi/2}^{\pi}= \frac{1}{2}(\sin(\pi)- \sin(\pi/2))= \frac{1}{2}(0- 1)= -\frac{1}{2}$. If you really, as v8archie suggested, were trying to do this as "sin" times "2x", that is very troubling. You should have immediately recognized that "sin" without any "x" is not a function at all! It is just three meaningless letters! Last edited by skipjack; October 18th, 2017 at 11:29 AM. Tags definite, integrals Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Agata78 Calculus 6 January 19th, 2013 02:05 PM Agata78 Calculus 18 January 18th, 2013 12:39 PM jakeward123 Calculus 10 February 28th, 2011 12:18 PM Aurica Calculus 2 May 10th, 2009 05:05 PM Agata78 Abstract Algebra 0 December 31st, 1969 04:00 PM

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