Question

question

Answer 1

\[\int\limits \frac{\cos}{1-cosx}dx\]\[\int\limits \frac{\cos^2\frac{x}{2}-\sin^2\frac{x}{2}}{2\sin^2\frac{x}{x}}dx\]\[\frac{1}{2}[\int\limits (\cot^2\frac{x}{2}-1)dx]\]\[\frac{1}{2}[\int\limits (cosec^2\frac{x}{2}-2)dx]\]\[\frac{1}{2}\int\limits cosec^2\frac{x}{2}dx-\int\limits dx\]\[-\cot \frac{x}{2}-x+C\]
OK?

Answer 2

@mathmath333

Answer 3

\(\Huge \begin{align} \color{black}{
\checkmark \hspace{.33em}\\~\\
}\end{align}\)

Answer 4

I was confused since the answer that was given was...
\[-cosecx-cotx-x+C\] @mathmath333

Answer 5

Ahh, nailed it \[\frac{-1}{sinx}+\frac{-cosx}{\sin}=\frac{-1}{sinx}(1+cosx)=\frac{-2\cos^2\frac{x}{2}}{2\sin \frac{x}{2}\cos \frac{x}{2}}\]\[=-\cot \frac{x}{2}\]

Answer 6

\(\large \begin{align} \color{black}{
-\cot x-x \hspace{.33em}\\~\\
=-\dfrac{\cos x}{\sin x} -x \hspace{.33em}\\~\\
=-\dfrac{2\cos^2 x}{2\sin x\cos x} -x \hspace{.33em}\\~\\
=-\dfrac{2(\frac{1+\cos x}{2})}{sinx} -x \hspace{.33em}\\~\\
=-\dfrac{1+\cos x}{sinx} -x \hspace{.33em}\\~\\
=-cosec ~ x-\cot x -x \hspace{.33em}\\~\\
}\end{align}\)