Originally Posted by

**Elvis** I'm trying to find the integral $\displaystyle \int\sec^2{\theta}\tan{\theta}\; {d\theta}$. Here is what I did:

Let $\displaystyle u = \tan{\theta}$. Then $\displaystyle \dfrac{du}{d\theta} = \sec^2\theta \Rightarrow {\color{blue}d{\theta} = \dfrac{du}{\sec^2\theta}} $ so that $\displaystyle \int\sec^2{\theta}\tan{\theta}\;{d\theta} = \int\dfrac{\sec^2{\theta}u}{sec^2\theta}\;{du} = \int{u}\;{du} = \frac{u^2}{2}+C$ $\displaystyle \Rightarrow \int\sec^2{\theta}\tan{\theta}\;{d\theta} = \frac{\tan^2\theta}{2}+C$, which is right.

My first question is, is there something wrong with the highlighted step? I have seen people raising questions about $\displaystyle \dfrac{dy}{dx}$ being treated as if they were numbers.

My second question is to do with when I tried to substitute $\displaystyle u$ for $\displaystyle \sec^2\theta $ and got different results than the above:

Let $\displaystyle u = \sec^2\theta$. Then $\displaystyle \dfrac{du}{d\theta} = 2\sec^2\theta\tan\theta \Rightarrow {\color{blue}d{\theta} = \dfrac{du}{2\sec^2\theta\tan\theta}} $ so that $\displaystyle \int\sec^2{\theta}\tan{\theta}\;{d\theta} = \int\dfrac{u\tan\theta}{2u\tan\theta}\;{du} = \frac{1}{2}\int\frac{u}{u}\;{du} = \frac{u}{2}+C$ $\displaystyle \; \Rightarrow \int\sec^2{\theta}\tan{\theta}\;{d\theta} = \frac{\sec^2\theta}{2}+C$.

Getting two different results for the same integral is something I can't grasp at all.