3
$\begingroup$

Trying to evaluate this using trig substitution:

$$\int \frac {1}{49x^2 + 25}\mathrm{d}x $$

Here's how I'm going about it, using $x = 5/7(\tan\theta)$

$$\int \frac {1}{49\left(x^2 + \frac{25}{49}\right)}\mathrm{d}x $$ $$=\int \frac {1}{49\left(\frac{25}{49}\tan^2\theta + \frac{25}{49}\right)} \mathrm{d}\theta$$ $$=\int \frac {1}{(25\tan^2\theta + 25)} $$ $$=\int \frac {1}{25(\tan^2\theta + 1)}\mathrm{d}\theta $$ $$=\int \frac {1}{25\sec^2\theta}\mathrm{d}\theta $$ $$=\int \frac {\cos^2\theta}{25}\mathrm{d}\theta $$ $$=\frac{1}{50}(\theta + \sin\theta + \cos\theta) $$ To generalize for $x$, $\theta = \arctan(7x/5)$ $$\frac{1}{50}\left(\arctan\left(\frac{7x}{5}\right) + \sin\left(\arctan\left(\frac{7x}{5}\right)\right) + \cos\left(\arctan\left(\frac{7x}{5}\right)\right)\right) $$ $$\frac{1}{50} \left(\frac{7x}{5\left(\frac{49x^2}{25}+1\right)} + \arctan\left(\frac{7x}{5}\right)\right)$$ But taking the derivative of this gets me: $$ \frac{35}{(49x^2 +25)^2}$$ Where is my mistake?

$\endgroup$
2
  • $\begingroup$ Where is your differential element $dx$? $\endgroup$ Feb 20, 2021 at 17:11
  • $\begingroup$ Why was this removed from the hot network questions? $\endgroup$ Aug 3, 2021 at 21:42

3 Answers 3

3
$\begingroup$

Welcome to MSE. The issue is that you left off $dx$: $$\int \frac1{49x^2 + 25} dx = \int \frac1{49(5/7\tan(\theta))^2 + 25} d(5/7 \tan(\theta) = \frac57\int \frac1{25\tan^2(\theta) + 25} \sec^2(\theta)d\theta = \cdots $$

$\endgroup$
1
  • 1
    $\begingroup$ Well damn, I had completely forgotten about that. Thank you $\endgroup$
    – herpderp
    Feb 20, 2021 at 17:20
1
$\begingroup$

You could also evaluate the integral in this way:

$$\int \frac {1}{49x^2 + 25}dx=\frac1{49}\int\frac1{x^2+(\tfrac57)^2}dx$$ By knowing that $\int\frac{1}{x^2+a^2}dx=\frac1a\tan^{-1}(\frac xa)$ we have:

$$\frac1{49}\int\frac1{x^2+(\tfrac57)^2}dx=\frac1{49}\times\frac75\tan^{-1}\left(\frac{7x}{5}\right)+C=\frac1{35}\tan^{-1}\left(\frac{7x}5\right)+C$$

$\endgroup$
1
$\begingroup$

We can calculate a more general integral of the form $$I =\int\frac{1}{a^2x^2+b^2}\,\mathrm{d}x.$$ In your example $a=7$ and $b=5$. First of all do the non-trigonometrical substitution $u=ax/b$. That will give you $$\int\frac{b}{a(b^2u^2+b^2)}\,\mathrm{d}x=\frac{1}{ab}\int\frac{1}{u^2+1}\,\mathrm{d}u.$$ You should be familiar with this integral. It's equal to $\arctan u$. Substituting back yields $$I = \dfrac{\arctan(ax/b)}{ab}+C.$$

$\endgroup$

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.