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Didn't get good notes on how to solve this problem so I wasn't able to solve it. The only thing I have to go off of is that a $\frac{1}{\cos^2(x)}$ was multiplied times it to give $\dfrac{\ln(\tan(x))}{\sin^2(x)/\cos^2(x)}$ but wasn't sure how this works. From that a substitution of $u = \tan(x)$ was used and from there its solvable. I was just unsure how we get the $\frac{1}{\cos^2(x)}$ seemingly for free? The final given answer was -1/tan(x)(ln(tan(x)-1/tan(x))+C. I'm not really looking for an answer, just an explanation of how my teacher did the problem

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  • $\begingroup$ $du = \sec^2(x) dx = \frac{1}{\cos^2(x)} dx$ $\endgroup$ Oct 22, 2019 at 2:56
  • $\begingroup$ Gonzalo Benavides I was wondering how we add du for seemingly free even though its not in the original integral? $\endgroup$ Oct 22, 2019 at 3:07

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Try integration by parts. Let $u=\log(\tan x)$ and $dv=\csc^2 x\, dx$. Then $du=\csc x\sec x\, dx$ and $v=-\cot x$. Thus, your integral turns into $$\begin{align} \int\frac{\log(\tan x)}{\sin^2 x}dx &= \int udv = uv-\int vdu \\ &= -\cot x\log(\tan x)+\int (-\csc^2 x)\, dx \\ &= -\cot x\, (\log(\tan x)+1)+c. \end{align}$$

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Hint:

$$\ln(\tan x)=-\ln(\cot x)$$

Now set $\cot x=y,-\csc^2x\ dx=dy$

Use https://www.khanacademy.org/math/old-integral-calculus/integration-techniques/integration-by-parts-ic/v/integral-of-ln-x?modal=1

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You have to multiply both the numerator and denominator of the fraction by $\frac{1}{cos^2 x}$ to make sure it doesn't change. That gives $\displaystyle\int \dfrac{\ln(\tan x)/\cos^2 x}{\sin^2 x/\cos^2 x}dx$. Let $u=\tan x$. Then $\dfrac{du}{dx}=\sec^2 xdx$ so substitution gives $\displaystyle\int \dfrac{\ln(\tan x)/\cos^2 x}{\sin^2 x/\cos^2 x}dx\\ =\displaystyle\int \dfrac{\ln(u)}{u^2}du\\ =-\frac{\ln u}{u}-\frac{1}{u}+C=-\frac{\ln \tan x}{\tan x}-\frac{1}{\tan x}\\ =-\cot x\cdot(\ln(\tan x)+1)+C$.

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