1
$\begingroup$

I'm trying to calculate $\int \frac{1+\cos2x}{\sin^2 2x}$ for the third time and I'm still getting the same result. As usual I tried substitution:

$u = 2x$

$$\frac{1}{2} \int \frac{1+\cos u}{\sin^2 u} = \frac{1}{2} *\int \csc^2 u + \int \frac{\cot u}{\sin u}$$

$$\frac{1}{2} *[-\cot u + \int \cos u ]$$

$$\frac{-\cot 2x}{2} + \frac{\sin 2x}{2}$$

The solution in my textbook is completly different, however. What's wrong with my method?

$\endgroup$
3
  • $\begingroup$ How did you get from $\int \frac {\cot u}{\sin u} \ du$ in one line to $\int \cos u \ du$ in the next. You should have $\int \csc u \cot u \ du$ instead $\endgroup$
    – Doug M
    Feb 1, 2018 at 18:06
  • $\begingroup$ @Doug Exactly same, i dont know what he did there $\endgroup$
    – King Tut
    Feb 1, 2018 at 18:12
  • $\begingroup$ @DougM Indeed, I messed up with the arithmetic $\endgroup$
    – Trey
    Feb 1, 2018 at 18:18

3 Answers 3

3
$\begingroup$

$${\displaystyle\int}\dfrac{\cos\left(2x\right)+1}{\sin^2\left(2x\right)}\,\mathrm{d}x ={\displaystyle\int}\dfrac{\csc^2\left(x\right)}{2}\,\mathrm{d}x =\class{steps-node}{\cssId{steps-node-1}{\dfrac{1}{2}}}{\displaystyle\int}\csc^2\left(x\right)\,\mathrm{d}x =-\dfrac{\cot\left(x\right)}{2} + C$$

$\endgroup$
1
  • $\begingroup$ Forgive my ignorance but what did you do with cos 2x? $\endgroup$
    – Trey
    Feb 1, 2018 at 18:22
2
$\begingroup$

$$\frac{1+\cos2x}{\sin^22x}=\frac{2\cos^2x}{4\sin^2x\cos^2x}=\frac{1}{2\sin^2x}$$ and we got $\cot$.

$\endgroup$
1
$\begingroup$

Again here is the problem $\int \frac{\cot (u)}{ \sin (u) } du= \int \csc( u) \cot( u) du = -\csc (u)$.

So final solution according to your method is :

$$-\frac{1}{2}\left(\cot(2x) + \csc(2x)\right) + c$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .