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June 3rd, 2017, 09:43 AM   #1
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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.
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June 3rd, 2017, 04:49 PM   #2
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anyone? I know you guys practice a lot
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June 3rd, 2017, 05:34 PM   #3
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June 3rd, 2017, 07:05 PM   #4
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Not an easy problem, perhaps.

$\displaystyle \int_{0}^{\pi} \dfrac{2\sin x+3\cos x-3}{13\cos x-5} \,dx$
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June 3rd, 2017, 09:55 PM   #5
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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.
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June 6th, 2017, 08:08 PM   #6
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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.

For more information about the substitution used, see here.

Last edited by greg1313; June 6th, 2017 at 08:12 PM.
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June 7th, 2017, 01:33 PM   #7
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Quote:
Originally Posted by greg1313 View Post
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.
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June 7th, 2017, 07:05 PM   #8
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Quote:
Originally Posted by greg1313 View Post
$$\sin x=2\sin\frac x2\cos\frac x2\Rightarrow\sin^2x=4\sin^2\frac x2\cos^2\frac x2=4\frac{1-\cos^2\frac x2}{1+t^2}=\frac{4t^2}{(1+t^2)^2}\implies\sin x=\frac{2t}{1+t^2}$$
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*}
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June 8th, 2017, 09:30 AM   #9
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Quote:
Originally Posted by v8archie View Post
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 $2\pi$?
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