How do I use U-Trig substitution to evaluate the indefinite integral of dx/(9+x^2) and (sq root(16-x^2) /x)dx?

Question

kainui · Answer

So it's basically playing off the idea that $$\tan^2\theta +1=\sec^2\theta $$ You can easily multiply both sides of this equation by 9 to get:
$$9\tan^2\theta +9=9\sec^2\theta $$ and we can see if we make $$x=3\tan \theta $$ then that means that $$x^2=9\tan^2\theta$$

That way we can turn it into 3sec(theta) in the denominator, do you see? Don't forget that you must turn dx into dtheta in order to do the integral in terms of theta, then translate it back to terms of x just like you did with u substitution.

anonymous · Answer

I did that and got $$1/3*\int\limits_{}^{}3\cos ^{2}(\tan ^{-1} x/3)dx$$, how do I simplify