Hello could anyone integrate the following with respect to x
(1-2sin2x)/(-cos2x) dx
Could you please show the working as well.
Thank you.
The first thing I would do is multiply numerator and denominator by cos(2x). That gives me a "cosine" in the numerator that I can use with the "dx" and $cos^2$ in the numerator that can be changed to $1- sin^2(2x)$:
$-\int \frac{1- 2sin(2x)}{cos(2x)}dx= -\int\frac{1- 2sin(2x)}{cos^2(2x)}(cos(2x)dx)= -\int\frac{1- 2sin(2x)}{1- sin^2(2x)}(cos(2x) dx)$
Now let $u= sin(2x)$ so that $du= 2 cos(2x) dx$ and the integral becomes
$-\frac{1}{2}\int \frac{1- 2u}{1- u^2}du$.
$$ \begin{aligned} \int \frac{1-2\sin {(2x)}}{-\cos {(2x)}} \, \mathrm dx &= - \int \sec{(2x)} - 2\tan{(2x)} \, \mathrm dx \\ &= -\bigg( \tfrac12 \ln{\big|\sec{(2x)} + \tan{(2x)}\big|} + \ln{\big|\cos{(2x)}\big|} \bigg) +c \\ &= - \bigg( \tfrac12 \ln{\big|\sec{(2x)} + \tan{(2x)}\big|} + \tfrac12 \ln{\big|\cos^2{(2x)}\big|} \bigg) +c \\[8pt] &= - \tfrac12 \ln{\big|\cos{(2x)} + \sin{(2x)}\cos{(2x)}\big|} +c \\[4pt] &= - \tfrac12 \ln{\bigg|\cos{(2x)}\big(1 + \sin{(2x)}\big)\bigg|} +c \end{aligned} $$
The integrals of the two terms are ones that you should remember as standard.
(For some reason, this wouldn't render under [tex][/tex] tags).