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I'm having trouble understanding a specific step used to solve the integral below, where instead of replacing $g(x)$ by $u$, it is replaced by a function of $u$: $tan(u)$

$$\int sin(x)\sqrt{1+cos^2(x)}dx$$ The following substitution is used: $$cos(x) = tan(u)$$ $$\mathbf{-sin(x)dx = sec^2(u) du}$$ Resulting in the standard integral which can be solved : $$\int sec^2(U)\sqrt{1+tan^2(u)}du = \int sec^2(u) \sqrt{sec^2(u)} = \int sec^3(u) du $$

The substition in boldface is the one I am having trouble understanding. The left part is the same as usual, i.e. $\frac{du}{dx}dx = du$, yet it is set equal to the derivative of $tan(u)$. I understand this is due to replacing $cos(x)$ by $tan(u)$ instead of $u$, but I am having trouble understanding the exact steps. An explanation or name of the method would be greatly appreciated.

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Implicitly;

$$\frac{d}{du}(\cos x) = -\sin x\frac{dx}{du}$$ $$\frac{d}{du}(\tan u) = \sec^2 u$$ They equate so:

$$-\sin x \frac{dx}{du} = \sec^2 u$$

Then multiply by $du$ which is allowed.

$$-\sin x \ dx = \sec^2 u \ du$$

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    $\begingroup$ Thank you for your answer. I'm having trouble understanding the first line. You're taking the derivative of $cos x$ with respect to u (assuming this is a typo)? Which results in: $\frac{d}{dx}(cos x) \frac{dx}{du} = -sin x \frac{dx}{du}$? How does the $\frac{dx}{du}$ come in? $\endgroup$ – Ross Feb 1 '19 at 20:11
  • $\begingroup$ It's no typo, this is how you differentiate with respect to another variable, and it is because of the chain rule: $$\frac{dy}{du}=\frac{dy}{dx}\cdot\frac{dx}{du}$$. In this case in my first line we take $y=\cos(x)$ which leads to $\frac{dy}{dx}=-\sin x$, then we must multiply by $\frac{dx}{du}$ to make this valid. $\endgroup$ – Rhys Hughes Feb 1 '19 at 20:16
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    $\begingroup$ Ah the chain rule, my apologies. If I'm understanding this correctly, this is the reasoning: $$cos(x) = tan(u)$$ then $$\frac{d}{du}(cos(x)) = \frac{d}{du}(tan(u))$$ After which your next steps apply. Is this correct? $\endgroup$ – Ross Feb 1 '19 at 20:20
  • $\begingroup$ Precisely that. $\endgroup$ – Rhys Hughes Feb 1 '19 at 20:33

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