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 June 3rd, 2017, 09:43 AM #1 Member   Joined: Jan 2017 From: California Posts: 80 Thanks: 8 Random calculus practice problems Hi guys, Is there a software or website where I can practice random problems? I don't like using the book problems because when I do them I know exactly in which section I am and that there's only one way to do it. For example, all the problems related to integration by parts would be in the section for integration part. So I already know that. I want something that will allow me to solve the problem without already knowing the section. Let me know please. Last edited by skipjack; June 3rd, 2017 at 11:02 AM. June 3rd, 2017, 04:49 PM #2 Member   Joined: Jan 2017 From: California Posts: 80 Thanks: 8 anyone? I know you guys practice a lot  June 3rd, 2017, 05:34 PM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,614 Thanks: 2603 Math Focus: Mainly analysis and algebra Thanks from dthiaw June 3rd, 2017, 07:05 PM #4 Global Moderator   Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,923 Thanks: 1122 Math Focus: Elementary mathematics and beyond Not an easy problem, perhaps. $\displaystyle \int_{0}^{\pi} \dfrac{2\sin x+3\cos x-3}{13\cos x-5} \,dx$ June 3rd, 2017, 09:55 PM   #5
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 Originally Posted by greg1313 Not an easy problem, perhaps. $\displaystyle \int_{0}^{\pi} \dfrac{2\sin x+3\cos x-3}{13\cos x-5} \,dx$
I just can't see it sir. Last edited by skipjack; June 6th, 2017 at 10:58 PM. June 6th, 2017, 08:08 PM   #6
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 Originally Posted by greg1313 Not an easy problem, perhaps. $\displaystyle \int_{0}^{\pi} \dfrac{2\sin x+3\cos x-3}{13\cos x-5} \,dx$
First, make the sub $t=\tan\frac x2$ so that $1+t^2=\sec^2\frac x2$

Then

$$\cos^2\frac x2=\frac{1+\cos x}{2}\Rightarrow\frac{1}{1+t^2}=\frac{1+\cos x}{2}\implies\cos x=\frac{1-t^2}{1+t^2}$$

$$\sin x=2\sin\frac x2\cos\frac x2\Rightarrow\sin^2x=4\sin^2\frac x2\cos^2\frac x2=4\frac{1-\cos^2\frac x2}{1+t^2}=\frac{4t^2}{(1+t^2)^2}\implies\sin x=\frac{2t}{1+t^2}$$

and

$$dt=\frac{\sec^2\frac x2}{2}\,dx\implies\frac{2dt}{1+t^2}=dx$$

Now, using the above, make the necessary substitutions to express the integral in terms of $t$, simplify,
apply partial fractions and adjust the bounds of integration. The resulting improper integral
may be integrated directly, then the bounds may be applied.

Last edited by greg1313; June 6th, 2017 at 08:12 PM. June 7th, 2017, 01:33 PM   #7
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 Originally Posted by greg1313 First, make the sub $t=\tan\frac x2$ so that $1+t^2=\sec^2\frac x2$ Then $$\cos^2\frac x2=\frac{1+\cos x}{2}\Rightarrow\frac{1}{1+t^2}=\frac{1+\cos x}{2}\implies\cos x=\frac{1-t^2}{1+t^2}$$ $$\sin x=2\sin\frac x2\cos\frac x2\Rightarrow\sin^2x=4\sin^2\frac x2\cos^2\frac x2=4\frac{1-\cos^2\frac x2}{1+t^2}=\frac{4t^2}{(1+t^2)^2}\implies\sin x=\frac{2t}{1+t^2}$$ and $$dt=\frac{\sec^2\frac x2}{2}\,dx\implies\frac{2dt}{1+t^2}=dx$$ Now, using the above, make the necessary substitutions to express the integral in terms of $t$, simplify, apply partial fractions and adjust the bounds of integration. The resulting improper integral may be integrated directly, then the bounds may be applied. For more information about the substitution used, see here.
I would have never seen this in a million years. am trying to see where does this fit in in my calculus lifesaver textbook.  June 7th, 2017, 07:05 PM   #8
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Quote:
 Originally Posted by greg1313 $$\sin x=2\sin\frac x2\cos\frac x2\Rightarrow\sin^2x=4\sin^2\frac x2\cos^2\frac x2=4\frac{1-\cos^2\frac x2}{1+t^2}=\frac{4t^2}{(1+t^2)^2}\implies\sin x=\frac{2t}{1+t^2}$$
Squaring and rooting is not a water-tight derivation.

\begin{align*}
\sin x &= 2\sin \tfrac x2 \cos \tfrac x2 &&= 2\frac{\sin \tfrac x2}{\cos \tfrac x2} \cos^2 \tfrac x2 \\
&= \frac{2\tan \tfrac x2}{\sec^2 \tfrac x2} &&= \frac{2\tan \tfrac x2}{1 + \tan^2 \tfrac x2} \\
&= \frac{2t}{1 + t^2}
\end{align*} June 8th, 2017, 09:30 AM   #9
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Quote:
 Originally Posted by v8archie Squaring and rooting is not a water-tight derivation.
But is it worth noting that $\sin x$ is negative wherever $t$ is negative and positive wherever $t$ is positive on the interval $(-\pi,\,\pi)$ and that both have a period of $2\pi$? Tags calculus, practice, problems, random 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 alymrod Probability and Statistics 2 March 30th, 2016 07:23 PM messaoud2010 Probability and Statistics 1 July 18th, 2014 11:26 AM matthematical Real Analysis 0 October 2nd, 2011 11:42 PM NeuroFuzzy Probability and Statistics 1 March 21st, 2011 06:47 AM roncarlston Calculus 1 February 4th, 2011 03:28 PM

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