# Thread: Part of another Trig Sub problem

1. ## Part of another Trig Sub problem

Solve for Y as a function of X:

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

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:

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

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!

2. Hello, Got5onIt

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

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

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

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

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

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

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

3. Originally Posted by Soroban
Hello, Got5onIt

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

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

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

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

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