For integrals of the kinds involving fractional trigonometric integrands
$$∫\frac{a\sin x+b\cos x+c}{d\sin x+e\cos x+f}dx$$
$$∫\frac{a\sin x+b\cos x}{c\sin x+d\cos x}dx$$
$$∫\frac{dx}{a\sin x+\cos x}$$
what are the relations between the numerator in the denominator, and what is the general pattern to solve these type of questions?
I did not treat this question like a homework question, since it is not tagged homework.
What are the relations between the numerator in [and] the denominator
Both are linear functions of $\sin x,\cos x$
what is the general pattern to solve these type of questions?
The universal standard substitution to evaluate an integral of a rational fraction in $\sin x,\cos x$, i.e. a rational fraction of the form
$$R(\sin x,\cos x)=\frac{P(\sin x,\cos x)}{Q(\sin x,\cos x)},$$
where $P,Q$ are polynomials in $\sin x,\cos x$ is a trigonometric substitution known as the Weierstrass substitution
$$\tan \frac{x}{2}=t,\qquad x=2\arctan t.\tag{*}$$
Differentiating both sides of $(^*)$ w.r.t. $t$, we get
$$\frac{dx}{dt}=\frac{2}{1+t^2},\qquad dx=\frac{2}{1+t^2}dt.$$
Since we know$^1$ from trigonometry that $$\cos x =\frac{1-\tan ^{2}\frac{x }{2}}{1+\tan ^{2}\frac{ x}{2}}=\frac{1-t^2}{1+t^2},\qquad \sin x =\frac{2\tan \frac{x }{2}}{1+\tan ^{2} \frac{x }{2}}=\frac{2t}{1+t^2},$$
we see that in general, the integrand becomes a rational fraction in $t$, whose standard integration technique is the partial fractions decomposition:
$$\int R(\sin x,\cos x)\, dx=\int R\Big(\frac{2t}{1+t^2},\frac{1-t^2}{1+t^2} \Big)\frac{2}{1+t^2}\, dt.$$
For instance this substitution in the 2nd. integral yields $$ \begin{eqnarray*} \int \frac{a\sin x+b\cos x}{c\sin x+d\cos x}dx &=&\int \dfrac{a\dfrac{2t}{ 1+t^{2}}+b\dfrac{1-t^{2}}{1+t^{2}}}{c\dfrac{2t}{1+t^{2}}+d\dfrac{1-t^{2}}{ 1+t^{2}}}\dfrac{2}{1+t^{2}}dt \\ &=&2\int \frac{bt^{2}-2at-b}{\left( dt^{2}-2ct-d\right) \left( 1+t^{2}\right) }dt. \end{eqnarray*} $$
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$^1$ A possible proof is the following one, which uses the double-angle formulas and the identity $\cos ^{2}\frac{x}{2}+\sin ^{2}\frac{x}{2}=1$:
$$ \begin{eqnarray*} \cos x &=&\cos ^{2}\frac{x}{2}-\sin ^{2}\frac{x}{2}=\frac{\frac{\cos ^{2} \frac{x}{2}-\sin ^{2}\frac{x}{2}}{\cos ^{2}\frac{x}{2}}}{\frac{\cos ^{2} \frac{x}{2}+\sin ^{2}\frac{x}{2}}{\cos ^{2}\frac{x}{2}}}=\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2}\frac{x}{2}}, \\ && \\ \sin x &=&2\sin \frac{x}{2}\cos \frac{x}{2}=\frac{\frac{2\sin \frac{x}{2} \cos \frac{x}{2}}{\cos ^{2}\frac{x}{2}}}{\frac{\cos ^{2}\frac{x}{2}+\sin ^{2}% \frac{x}{2}}{\cos ^{2}\frac{x}{2}}}=\frac{2\tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}. \end{eqnarray*} $$
You can use the following way to calculate all of them once. Let $$ A=\int \frac{\sin xdx}{d\sin x+e\cos x+f}, B=\int \frac{\cos xdx}{d\sin x+e\cos x+f}, C=\int \frac{dx}{d\sin x+e\cos x+f}. $$ It is easy to check $$ dA+eB+fC=1, dB-eB=\ln|d\sin x+e\cos x+f|+Const. $$ So $$ A=\frac{d(1-fC)}{d^2+e^2}-\frac{e\ln|d\sin x+e\cos x|}{d^2+e^2}+Const, B=\frac{e(1-fC)}{d^2+e^2}+\frac{f\ln|d\sin x+e\cos x|}{d^2+e^2}+Const.$$ Thus we only need to get $C$. Let $$\tan \frac{x}{2}=t,x=2\arctan t. $$ Then $$\sin x=\frac{2t}{1+t^2},\cos x=\frac{1-t^2}{1+t^2},dx=\frac{2}{1+t^2}dt.$$ Then we have $$ C=\int \frac{dx}{d\sin x+e\cos x+f}=2\int\frac{dt}{(f-e)t^2+2dt+(f+e)}$$ which is not hard to get.
Decompose the integrand as follows
\begin{align} I=&\int \frac{a\sin x+b\cos x+c}{d\sin x+e\cos x+f}\ dx\\ =&\ \int P+ Q\frac{d\cos x-e\sin x}{d\sin x+e\cos x+f}+ \frac{c-Pf}{d\sin x+e\cos x+f}\ dx \end{align} with $P= \frac{ad+be}{d^2+e^2}$ and $Q= \frac{bd-ae}{d^2+e^2}$. While the first and second integrals are easily evaluated, the third one trifurcates according to $\Delta=f^2-d^2-e^2$ \begin{align} J_\Delta =&\int \frac{1}{d\sin x+e\cos x+f}\ dx\\ =& \ \begin{cases}\frac1{\sqrt{\Delta}}\tan^{-1}\frac{d^2+e^2+f(d\sin x+e\cos x)}{\sqrt{\Delta}\ (d\cos x-e\sin x)} &\Delta>0 \\ \frac{d\sin x+e\cos x - \sqrt{d^2+e^2}}{\sqrt{d^2+e^2}(d\cos x-e\sin x)}\>\>\>&\Delta=0\\ \frac{-1}{\sqrt{-\Delta}}\coth^{-1}\frac{d^2+e^2+f(d\sin x+e\cos x)}{\sqrt{-\Delta}\ (d\cos x-e\sin x)} &\Delta<0 \end{cases} \end{align} Thus, the general integral $I$ evaluates to \begin{align} I=\ Px+Q\ln |d\sin x+e\cos x+f |+(c-Pf)J_\Delta \end{align} In the special case of $c=f=0$, it reduces to $$\int \frac{a\sin x+b\cos x}{d\sin x+e\cos x}\ dx=Px + Q\ln |d\sin x+e\cos x| $$
HINT: Write $$a\sin x+b\cos x+c=A \frac{(d\sin x+e cos x+f)}{dx}+B(d\sin x+e cos x+f)+C$$ where $A,B,C$ are arbitrary constant whose values can be determined by comparing the coefficients $\sin x,\cos x$ and constants
Now, for $$\int \frac{dx}{d\sin x+e\cos x+f}$$ use Weierstrass substitution.
