# Part of another Trig Sub problem

• Oct 30th 2007, 08:50 PM
Got5onIt
Part of another Trig Sub problem
Solve for Y as a function of X:

$\displaystyle (x^2+1)^2 \frac{dy}{dx}=\sqrt{x^2+1}$

I isolate dy on one side, then take the integral of the right. After a couple of trig substitutions, I end up here but get stuck:

$\displaystyle y=\int \frac{1}{sec^3\theta} \\d\theta$

I think I'm on the right track. I went with a U substitution but it just didn't look right. Any help on how to finish it is greatly appreciated!
• Oct 30th 2007, 09:21 PM
Soroban
Hello, Got5onIt

Quote:

Solve: .$\displaystyle (x^2+1)^2 \frac{dy}{dx}\;=\;\sqrt{x^2+1}$

I ended up here: .$\displaystyle y\:=\:\int\frac{1}{\sec^3\!\theta}\,d\theta$

Did you forget the $\displaystyle {\color{blue}dx}$ ?

We have: .$\displaystyle dy \;=\;\frac{dx}{(x^2+1)^{\frac{3}{2}}}\quad\Rightar row\quad y \;=\;\int\frac{dx}{(x^2+1)^{\frac{3}{2}}}$

Let: $\displaystyle x \,=\,\tan\theta\quad\Rightarrow\quad dx \,=\,\sec^2\!\theta\,d\theta$

. . And: .$\displaystyle \sqrt{x^2+1} \:=\:\sqrt{\tan^2\!\theta+1} \:=\:\sqrt{\sec^2\!\theta} \:=\:\sec\theta$

Substitute: .$\displaystyle \int\frac{\sec^2\!\theta\,d\theta}{\sec^3\!\theta} \;=\;\int\frac{d\theta}{\sec\theta} \;=\;\int\cos\theta\,d\theta$

• Oct 30th 2007, 09:32 PM
Got5onIt
Quote:

Originally Posted by Soroban
Hello, Got5onIt

We have: .$\displaystyle dy \;=\;\frac{dx}{(x^2+1)^{\frac{3}{2}}}\quad\Rightar row\quad y \;=\;\int\frac{dx}{(x^2+1)^{\frac{3}{2}}}$

Let: $\displaystyle x \,=\,\tan\theta\quad\Rightarrow\quad dx \,=\,\sec^2\!\theta\,d\theta$

. . And: .$\displaystyle \sqrt{x^2+1} \:=\:\sqrt{\tan^2\!\theta+1} \:=\:\sqrt{\sec^2\!\theta} \:=\:\sec\theta$

Substitute: .$\displaystyle \int\frac{\sec^2\!\theta\,d\theta}{\sec^3\!\theta} \;=\;\int\frac{d\theta}{\sec\theta} \;=\;\int\cos\theta\,d\theta$

I must've forgotten a sec^2 somewhere. I will go back and recheck my work. Thanks so much!!!!!!!!!!!! :)