Is this an acceptable way to integrate?

Question

I am supposed to find: $$ \int \sec(1-x)\tan(1-x) dx $$

I then set $ u = \sec(1-x) $ $$ du = -\tan(1-x)\sec(1-x)\ dx $$ therefore

$$ \frac{-du}{\sec(1-x)} = \tan(1-x)\ dx$$ Which when applied gives me this: $$\int \sec(1-x) \frac{-du}{\sec(1-x)} = -\int 1\ du \\ = -\sec(1-x) + C $$

which is correct, but i feel like this is like cheating for some reason, is this not acceptable?

Answer 1

You've done just fine.

Alternately,

since $\dfrac{d}{dx}\left(\sec \theta\right) = \sec \theta \tan \theta$, it makes sense that $\int \sec\theta \tan\theta\,d\theta = \sec\theta + C$.

Now can you more readily recognize that your integrand is equal $\;-\dfrac{d}{dx}\big(\sec(1-x)\big),\:$ and what that means for $\;\int \sec(1-x)\tan(1-x)\,dx\;?$