3
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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

$$\frac{5+3y}{(5y+3)(y^2+1)}=\frac{5}{5y-3}-\frac{y}{y^2+1}$$

to evaluate this integral.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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 $\sin(x)$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.