Excuse me, I tried to solve $\int \frac{1}{4\cdot \sin \left(x\right)-5\cdot \cos \left(x\right)}dx$ but i can't get the right answer .. any help? Excuse me, I tried to solve this problem but i can't get the right answer .. any help? 
$\int \frac{1}{4\cdot \sin \left(x\right)-5\cdot \cos \left(x\right)}dx$

A function of the form $f(x)=a\sin x+b\cos x$ can always be put in the form
$$
f(x)=A\sin(x+\varphi)
$$
for suitable $A$ and $\varphi$. Indeed, take $A=\sqrt{a^2+b^2}$ and find $\varphi$ such that
$$
\cos\varphi=\frac{a}{A},\quad \sin\varphi=\frac{b}{A}
$$
which is always possible, because
$$\frac{a^2}{A^2}+\frac{b^2}{A^2}=1.$$
Then your integral becomes
$$
\frac{1}{A}\int\frac{1}{\sin(x+\varphi)}dx
$$
that can be reduced to computing
$$
\int\frac{1}{\sin u}du
$$
via the substitution $x+\varphi=u$. This is known:
\begin{align}
\int\frac{1}{\sin u}du
&=\int\frac{\cos^2(u/2)+\sin^2(u/2)}{2\sin(u/2)\cos(u/2)}du\\[2ex]
&=\frac{1}{2}\int\frac{\cos(u/2)}{\sin(u/2)}du+\frac{1}{2}\int\frac{\sin(u/2)}{\cos(u/2)}du\\[2ex]
&=\log|\sin(u/2)|-\log|\cos(u/2)|+C\\[2ex]
&=\log\left|\tan\frac{u}{2}\right|+C
\end{align}
The final expression for your integral is now just a matter of back substituting (if you really want to write it).
