I'm asked to evaluate $$\int_{\pi/4}^{\pi/2} \frac{3+5\cot(x)}{5-3\cot(x)}$$.

If this were just an integral for $\displaystyle\frac{5\cot(x)}{-3\cot(x)}$, I would be able to solve it just fine, but I'm a bit stuck on this one. I can't use u-substitution (or if I can, I don't see how) because there's no relation between the numerator and denominator. I also can't split the fraction in two because those constants being added to the cotangent terms keep getting in the way.

Could someone please offer a hint as to how I should get started with this problem?


Notice that

\begin{align} \int \frac{5 \cot(x) + 3 }{5-3 \cot(x) } ~\text{d}x &= \int \frac{5 \cot(x) + 3 }{5-3 \cot(x) } \cdot \frac{\tan(x) \frac{1}{\cos(x)^2}}{\tan(x) \frac{1}{\cos(x)^2}}~\text{d}x \\ &=\int \frac{(5+3 \tan(x)) \frac{1}{\cos(x)^2}}{(5\tan(x)-3)(\tan(x)^2+1)}~\text{d}x \\ &=\int \frac{5+3y}{(5y+3)(y^2+1)} ~\text{d}y \end{align}

where we used that $\tan(x)^2+1=\frac{1}{\cos(x)^2}$ and now use partial fraction decomposition


to evaluate this integral.

  • $\begingroup$ Could you please clarify where the $tan^{2}x + 1$ in the denominator came from? Shouldn't it also be a $\frac{1}{cos^{2}(x)}$? $\endgroup$ – AleksandrH Dec 7 '16 at 20:28
  • $\begingroup$ @AleksandrH Yes, but this is the same since $$\tan(x)^2+1=\frac{\sin(x)^2}{\cos(x)^2}+1=\frac{\sin(x)^2+\cos(x)^2}{\cos(x)^2}=\frac{1}{\cos(x)^2}$$ $\endgroup$ – Fritz Dec 7 '16 at 20:28
  • $\begingroup$ Ah...man, this is quite a confusing and difficult problem. I would have never thought of doing this. $\endgroup$ – AleksandrH Dec 7 '16 at 20:33
  • $\begingroup$ @AleksandrH Yeah, the essential part is to prepare all this so we can substitute $y=\tan(x)$ and notice that $$\text{d} y = \frac{1}{\cos(x)^2} ~\text{d}x.$$ By this we get this easier integral. $\endgroup$ – Fritz Dec 7 '16 at 20:35

As way of enrichment, notice you are trying to calculate $$\int \frac{3\sin (x)+5\cos(x)}{5\sin (x)-3\cos (x)}dx=\int \ln (5\sin (x)-3\cos (x))'dx$$ You can reach this by writing out $\cot(x)$ as $\frac {\cos(x)}{\sin(x)}$, simplfying fractions and canceling the remaining $\sin(x)$


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