Trig function integration problem (substitution)

Question

I got this practice problem:

$$\int^\ 9sin(x) cos^{-3}(x)dx$$

so what I did is substitute u for cos(x), and du=-sin(x)dx

and I then got

$$\int^\ -9u^{-3}du$$

which is equal to

$$ \frac{9}2u^{-2}+C$$

I then substituted u=cosx back in and got

$$ \frac{9}2cos(x)^{-2}+C$$

But it is not right.

I don't see where I went wrong so if anyone can help me that would be great

Answer 1

There is nothing wrong with your answer.

$$\frac{9\cos^{-2}x}{2}=\frac{9}{2\cos^{2}x}=\frac{9}{2}\cdot\frac{1}{\cos^2x}=\frac{9\sec^2x}{2}$$ which is likely the answer in the book.

Answer 2

Hint: $$\int^\ 9sin(x) cos^{-3}(x)dx=\int^\ 9\tan(x) \sec^{2}(x)dx$$ let $u=\tan(x)$ $du=\sec^2(x)dx$ $$\int^\ 9udu=9u^2/2+C=\frac{9}{2}\tan^2(x)+C=\frac{9}{2}(\sec^2x-1)+C$$ $$=\frac{9}{2}\sec^2 x+(C-\frac{9}{2})=\frac{9}{2}\sec^2 x+K=\frac{9}{2}\cos^{-2} x+K$$ so your answer is correct