The answer is supposedly $y^2 = 1 + \sqrt{x^2 - 16}$ I don't know where I went wrong cause I know for a fact that my substitution of $x = 4 \sec(\theta)$ is correct. I know for a fact that after substitution the integral becomes $\int \sec^2(\theta)= \tan(\theta)+C$. I am pretty sure that my substitution of $\tan(\theta)$ is correct. Am I missing something?

$2y \frac{dy}{dx} = \frac{x}{\sqrt{x^2 - 16}} \space \space \space y(5)=2$

$\int 2y \space dy = \int \frac{4\sec(\theta)}{\sqrt{(4\sec(\theta))^2-16}}\space d(\theta)$

$y^2 = \int \frac{4\sec(\theta)}{\sqrt{(\sec^2(\theta)-1)16}}\sec(\theta) \ tan(\theta)$

$y^2 = \frac{4\sec(\theta)}{4\tan(\theta)}\sec(\theta)\tan(\theta)$

$y^2 = \int sec^2(\theta)$

$y^2 = \tan(\theta) + c$

Using the reference triangle: Tangent is equal to $\frac{\sqrt{x^2-16}}{4}$

$y^2= \frac{\sqrt{x^2-16}}{4} + C$

$4 = \frac{3}{4}+ C$

$C = \frac{15}{4}$

$y^2= \frac{\sqrt{x^2-16}}{4} + \frac{15}{4}$


You have to use another $4$ .
$x = 4\sec\theta \implies dx = \color{red}4\sec\theta\tan\theta d\theta$

So, you'll have $y^2 = \sqrt{x^2-16} +c$

$2^2 = \sqrt{9} +c \implies c = 4-3= 1$


$$y^2 =\sqrt{x^2+16} +1$$


In integrating the RHS of in$$2y \frac{dy}{dx} = \frac{x}{\sqrt{x^2 - 16}} \space \space \space y(5)=2$$

it is advised to let $$u=x^2-16$$ and you have $$du=2xdx$$

Then you do not have to worry about trig substitution at all.


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.