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I did:

\begin{align} & \int \csc^2(x) \,dx \\ = {} &\int \left(\frac 1 {\sin(x)}\right)^2 \, dx\\ = {} & \int (\sin(x)^{-1})^2 \, dx\\ = {} & \int \sin(x)^{-2} \, dx\\ = {} & \frac{\sin(x)^{-1}}{-1} \cdot \int \sin(x) \\ = {} & -\frac{1}{\sin(x)}\cdot -\cos(x) \\ = {} & \cot(x) \end{align}

But the right answer is $-\cot(x)$. Where did it go wrong?

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    $\begingroup$ I question how you went from the fourth to fifth line. $\endgroup$ – John Dec 4 '17 at 20:10
  • $\begingroup$ @John I tried to apply a kind of chain rule. I suppose that's wrong... how would you solve this? $\endgroup$ – Mark Read Dec 4 '17 at 20:12
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    $\begingroup$ math.stackexchange.com/questions/239808/… $\endgroup$ – K Split X Dec 4 '17 at 20:13
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    $\begingroup$ @KSplitX Well $u$ sub is the chain rule for integration. $\endgroup$ – user223391 Dec 4 '17 at 20:15
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    $\begingroup$ @KSplitX $u$ sub is literally the chain rule in reverse $\endgroup$ – user223391 Dec 4 '17 at 20:20
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Observe that

$$\frac {1}{\sin^2 (x)}=-\frac {-\sin^2(x)-\cos^2 (x)}{\sin^2 (x)} $$

$$=-\frac {d}{dx}\frac {\cos (x)}{\sin (x)} $$

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  • $\begingroup$ how'd you get to step 2? $\endgroup$ – Mark Read Dec 4 '17 at 20:19
  • $\begingroup$ @MarkRead $(u/v)'=(u'v-uv')/v^2$. $\endgroup$ – hamam_Abdallah Dec 4 '17 at 20:21
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I was puzzled by this step: $$ \int \sin(x)^{-2} \, dx = \frac{\sin(x)^{-1}}{-1} \cdot \int \sin(x) \,dx \text{ ?} $$ Then I realized you were trying to apply a sort of chain rule.

If you have something like $\displaystyle \int (5x+3)^{12} \,dx$ you can say that it's equal to $\dfrac{(5x+3)^{13}}{13}\cdot \dfrac 1 5$ and that's an application of the chain rule for differentiation. Or a bit more elaborately: $$ \int(5x+4)^{12} \, dx = \int u^{12} \, \left(\frac{du} 5\right) = \frac{u^{13}}{13} \cdot \frac 1 5 + \text{constant} $$ and like all "$u$-substitutions" that is an application of the chain rule.

But consider this: $$\require{cancel} \xcancel{\frac d {dx} \left( \frac{\sin(x)^{-1}}{-1} \cdot \int\sin(x)\, dx \right) = \sin(x)^{-2} \cdot \sin(x)} $$ The problem is that you're multiplying two functions and then differentiating what you get, and for that you need the product rule. You can't just differentiate the two separately.

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  • $\begingroup$ Wow! I didn't you can do that. $\endgroup$ – user8277998 Dec 4 '17 at 21:36

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