# Alternative Integral solution to $\sec^2(x) \sin(x)$

I understand that for solving the integral

$\int\sec^3(x)\sin(x) dx$ can be solved using $\tan(x)\sec^2(x) dx$

but why cant I solve this integral using U substitution?

$$\frac{1}{\cos^3(x)\sin(x)}$$

let $$u =\cos(x)\quad du= -\sin(x) dx \\ \quad \\ -du =\sin(x) dx \\ \int\frac{1}{u^3}\cdot(-du) \\ ∫-u^{-3} du \\ -u^{-2}/-2 = \\ \frac{\cos^{-2}(x)}{2}$$

• I'm finding it hard to understand your question. Please format it using MathJax: math.meta.stackexchange.com/questions/5020/… – mrnovice Apr 7 '17 at 22:45
• You can. But don 't forget your "+C". $\frac 12 \tan^2 x + c = \frac 12 \sec^2 x + c = \frac 12 \cos^{-2} + c$ – Doug M Apr 7 '17 at 22:46
• You can.. you answered it correctly.. – Rab Apr 7 '17 at 22:46
• And you've garbled at least two minus signs that cancelled out. – Chappers Apr 7 '17 at 22:47
• For basic information about writing math on MSE see here, here, here and here. – user409521 Apr 7 '17 at 22:59

This is actually a correct way to go about it and your answer is fine (except for the missing $+C$ constant, but we'll let it slide).