Hi I tried these three questions, on trig substitution and I didn't get the answer for all three..I'm sorry I cannot show my whole working, latex is hard for me and I'm struggling trying my best to type the more 'important' parts in it. Hopefully it turns out fine..

1.) Integrate $\displaystyle \frac{1}{x^2\sqrt{16-x^2}}dx$

I had let x be $\displaystyle 4\sin\theta$, dx = $\displaystyle 4\cos\theta$ d(delta sign)

Subbing it in and simplifying, I got $\displaystyle \frac{1}{16\sin\theta}$ left to integrate..After integrating I got $\displaystyle \frac{-cot\theta}{16}+c $ and since x = $\displaystyle 4\sin\theta$, $\displaystyle \sin\theta = \frac{x}{4} $..

Substituting, I nearly got the answer,

My answer is $\displaystyle \frac{\-sqrt{16-x^2}}{4x} + c $ but it's wrong..

2.) Integrate $\displaystyle \frac{1}{x\(x^2+16)}dx$

I let x be $\displaystyle 4\tan\theta$, dx = $\displaystyle 4\sec^2\theta$ d(delta sign)

Subbing in and simplifying, I got to integrating $\displaystyle \frac{tan\theta}{16}$, used the substitution method to tackle this integration of $\displaystyle tan\theta$, got $\displaystyle \frac{1}{16}\ln|cos\theta|}+c$ but then I tried to put x in terms of delta, it didn't work out for I get really weird looking answers...!

3.) Integrate x^2(sqrt{4-x^2}), I had let x be $\displaystyle 4\sin\theta$, dx = $\displaystyle 2\cos\theta$ d(delta sign)

Subbing and simplifying, I was left with integrating $\displaystyle \16\cos\^2\theta\-\16\cos\^4\theta}dx$. Using the double angle expression, I simplified, integrated and got $\displaystyle 4\sin\2\theta\-\2\sin\4\theta + c$... continued with it but I'm getting nowhere!

Thank you so much for your time!

P.s. the trigo functions with values like 2 and 4 with that little ^ above them, sorry for my bad skills, it actually means that it is eg. (cos(x))^2 etc!