5
$\begingroup$

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

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

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

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

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

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

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

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

$\endgroup$
1
4
$\begingroup$

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)

$\endgroup$
6
  • 2
    $\begingroup$ $+C$ of course. :) $\endgroup$ – Cameron Williams Jul 4 '13 at 19:07
  • $\begingroup$ Yeah sorry ${}{}{}$ $\endgroup$ – user1337 Jul 4 '13 at 19:07
  • $\begingroup$ I still get the wrong answer so where else did I mess up? $\endgroup$ – Sailor Jim Jul 4 '13 at 19:13
  • 2
    $\begingroup$ 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 '13 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 '13 at 15:15
5
$\begingroup$

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

$\endgroup$
4
  • $\begingroup$ How was the sub for cossin wrong? incorrect? $\endgroup$ – Sailor Jim Jul 4 '13 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 '13 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 '13 at 20:33
  • $\begingroup$ Sleep tight, @Babak! Wishing you the best of dreams! $\endgroup$ – amWhy Jul 22 '13 at 19:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.