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$$\int \frac{\mathrm{d}t}{( t^2 + 9)^2} = \frac {1}{81} \int \frac{\mathrm{d}t}{\left( \frac{t^2}{9} + 1\right)^2}$$

$t = 3\tan\theta\;\implies \; dt = 3 \sec^2 \theta \, \mathrm{d}\theta$

$$\frac {1}{81} \int \frac{3\sec^2\theta \mathrm{ d}\theta}{ \sec^4\theta} = \frac {1}{27} \int \frac{ \mathrm{ d}\theta}{ \sec^2\theta} = \dfrac 1{27}\int \cos^2 \theta\mathrm{ d}\theta $$

$$ =\frac 1{27}\left( \frac{1}{2} \theta + 2(\cos\theta \sin\theta)\right) + C$$

$\arctan \frac{t}{3} = \theta \;\implies$

$$\frac{1}{27}\left(\frac{1}{2} \arctan \frac{t}{3} + 2 \left(\frac{\sqrt{9 - x^2}}{3} \frac{t}{3}\right)\right) + C$$

This is a mess, and it is also the wrong answer.

I have done it four times, where am I going wrong?

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3 Answers 3

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Try returning to your integral: $$\int \cos^2\theta = \frac{\theta}{2} + \frac{\sin\theta\cos\theta}{2} + C$$

We get a factor of $\frac 12,$ and not $2$, multiplying the second term in the sum.

Note, more importantly, that your substitutions for $\cos\theta\sin\theta$ in terms of $\theta = \arctan(t/3)$ are also incorrect.

We have that $\theta = \arctan(t/3) \implies \tan \theta = \dfrac t3.\;$ Corresponding to this is $\;\cos\theta = \dfrac{3}{\sqrt{t^2 + 9}}$ and $\;\sin\theta = \dfrac t{\sqrt{t^2 + 9}}.$

That gives us $$\frac{1}{ 27}\left( \frac 12\cdot \arctan \frac{t}{3} + \frac 12 \underbrace{\frac{t}{\sqrt{t^2 + 9}}}_{\sin \theta}\cdot \underbrace{\frac{3}{\sqrt{t^2 + 9}}}_{\cos\theta}\right) + C$$ $$ = \frac{1}{54}\left( \arctan \frac{t}{3} + \frac{3t}{t^2+9}\right) + C$$

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  • $\begingroup$ How was the sub for cossin wrong? incorrect? $\endgroup$
    – Sailor Jim
    Jul 4, 2013 at 20:24
  • $\begingroup$ We get $\cos \theta = \dfrac 3{\sqrt{t^2 + 9}}$ and $\sin\theta = \dfrac t{\sqrt{t^2 + 9}}$. Multiply: $ = \dfrac{3t}{t^3 + 9}$ $\endgroup$
    – amWhy
    Jul 4, 2013 at 20:29
  • $\begingroup$ Recall: $\theta = \arctan\left(\frac t3\right)$, which means $\tan \theta = \dfrac t3$. This makes opp = t, adj = 3, hypotenuse = $\sqrt{t^2 + 9}$. So $\cos \theta = \dfrac{\text{adj}}{\text{hypot}}$ is given as immediately above, and likewise $\sin\theta = \dfrac{\text{ opp}}{\text{hypot}}$ $\endgroup$
    – amWhy
    Jul 4, 2013 at 20:33
  • $\begingroup$ Sleep tight, @Babak! Wishing you the best of dreams! $\endgroup$
    – amWhy
    Jul 22, 2013 at 19:57
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The integral $$ \int \cos^2 \theta \, \mathrm d\theta=\frac{\theta}{2}+\frac{1}{2} \sin \theta \cos \theta+C. $$ (You had the coefficient wrong)

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    $\begingroup$ $+C$ of course. :) $\endgroup$ Jul 4, 2013 at 19:07
  • $\begingroup$ Yeah sorry ${}{}{}$ $\endgroup$
    – user1337
    Jul 4, 2013 at 19:07
  • $\begingroup$ I still get the wrong answer so where else did I mess up? $\endgroup$
    – Sailor Jim
    Jul 4, 2013 at 19:13
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    $\begingroup$ That's not entirely correct: the substitutions made by the OP, when back substituting, are also incorrect. The correct answer is much different than merely a change in coefficient: the correct result is: $$\frac{1}{54}\left( \arctan \frac{t}{3} + \frac{3t}{t^2+9}\right) + C$$ $\endgroup$
    – amWhy
    Jul 4, 2013 at 19:40
  • $\begingroup$ @Panda: Please delete your last comment: the OP's answer IS NOT FINE if merely changing the $2$ to $\frac 12$. The OP substituted the incorrect corresponding functions of $\cos \theta$ and $\sin \theta$ when expressing them in terms of $t$. Your comment gives false (wrong) assurance: see my answer and comments for explanation. $\endgroup$
    – amWhy
    Jul 5, 2013 at 15:15
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I would like to handle the integral by integration by parts. $$ \begin{aligned} \int \frac{d t}{\left(t^{2}+9\right)^{2}} &=-\int \frac{1}{2 t} d\left(\frac{1}{t^{2}+9}\right) \\ &=-\frac{1}{2 t\left(t^{2}+9\right)}-\frac{1}{2} \int \frac{1}{t^{2}\left(t^{2}+9\right)} d t \\ &=-\frac{1}{2 t\left(t^{2}+9\right)}-\frac{1}{18} \int\left(\frac{1}{t^{2}}-\frac{1}{t^{2}+9}\right) d t \\ &=-\frac{1}{2 t\left(t^{2}+9\right)}+\frac{1}{18 t}+\frac{1}{54} \arctan \left(\frac{t}{3}\right)+C \\ &=\frac{1}{54}\left[\frac{3 t}{t^{2}+9}+\arctan \left(\frac{t}{3}\right)\right]+C \end{aligned} $$

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