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Suppose we are seeking the primitive
$$int R(cos x, sin x) dx$$
where $R(u,v)=frac {P(u,v)}{ Q(u,v)}$ is a two-variable rational function. Show that

(a) if $R(−u,v)= R(u,v)$,then $R(u,v)$ has the form $R_1(u^2,v)$;

(b) if $R(−u, v) = −R(u, v)$, then $R(u, v) = u cdot R_2(u^2, v)$ and the substitution $t = sin x$ rationalizes the integral above;

(c) If $R(−u, −v) = R(u, v)$, then $R(u, v) = R_3( u , v^2)$, and the substitution $t =tan x$ rationalizes the integral above.

My attempt: Suppose
$$P(x)=sum_{i,jgeq0, i+jleq n}{a_{ij}x^iy^j}
=sum_{i,jgeq0, 2i+jleq n}{a_{2i,j}x^{2i}y^{j}}+sum_{i,jgeq0, 2i+jleq n}{a_{2i+1,j}x^{2i+1}y^{j}}=P_1(x^2,y)+xP_2(x^2,y)$$
Similarly,
$$Q(x)=Q_1(x^2,y)+xQ_2(x^2,y)$$
Then
$$P(x,y)=frac{P_1(x^2,y)+xP_2(x^2,y)}{Q_1(x^2,y)+xQ_2(x^2,y)}=frac{[P_1(x^2,y)+xP_2(x^2,y)][Q_1(x^2,y)-xQ_2(x^2,y)]}{[Q_1(x^2,y)]^2-[xQ_2(x^2,y)]^2}$$
$$=frac{P_3(x^2,y)+xP_4(x^2,y)}{Q_3(x^2,y)}$$
where
$$P_3(x^2,y)=P_1(x^2,y)Q_1(x^2,y)-x^2P_2(x,y^2)Q_2(x^2,y)$$
$$P_4(x^2,y)=-P_1(x^2,y)Q_2(x^2,y)+P_2(x^2,y)Q_1(x^2,y)$$
$$Q_3(x^2,y)=[Q_1(x^2,y)]^2+[xQ_2(x^2,y)]^2$$

For part (a), since $R(−u,v)= R(u,v)$, then
$$R(-u,v)=frac{P_3(u^2,v)-uP_4(u^2,v)}{Q_3(u^2,v)}$$
How could I eliminate $uP_4(u^2,v)$, or in the other way, this is alike an even function for single variable function, to show $R(u,v)$ in this form has no odd degree for $u$? If part (a) is done, then part (b) can be brought down using similar way as part (a).

Update:
Since $R(−u,v)= R(u,v)$, then
$$R(-u,v)=frac{P_3(u^2,v)-uP_4(u^2,v)}{Q_3(u^2,v)}=frac{P_3(u^2,v)+uP_4(u^2,v)}{Q_3(u^2,v)}=R(u,v)$$
implies $uP_4(u^2,v)=0$ for all $u$ and $v$. (Not necessary $u=0$). Then it is proven and follows part (a)? I am confused but this idea suddenly come
into my mind.

This topic was considered in a Russian book “Differential and Integral Calculus” by Grigorii Fichtenholz (v. II, 7-th edition, M.: Nauka, 1970). I copied the relevant pages: 74, 75, and 76.
According to Wikipedia
“book was translated, among others, into German, Chinese, and Persian however translation to English language has not been done still”. Below I translated the key moments relevant to your question.

(a) The conclusion $uP_4(u^2,v)=0$ looks OK. Indeed, even taking into account possible zeroes of the denominator $Q_3(u^2,v)$, we have a that a polynomial $S(u,v)= uP_4(u^2,v)Q_3(u^2,v)$ is zero for each $u,v$. Then $S(u,v)$ is the zero polynomial.
Indeed, assume to the contrary that $S(u,v)=s_0(u)+s_1(u)v+dots+s_m(u)v^m$ for some polynomials $s_0,dots, s_m$ such that $s_m(u)$ is not the zero polynomial. Since $s_m(u)$ has only finitely many roots, we can pick $u_0$ such that $s_m(u_0)ne 0$. Then $S(u_0,v)$ is
not the zero polynomial on $v$, so there exists $v_0$ such that $S(u_0,v_0)ne 0$, a contradiction.

(b) Fichtenholz says that the first part follows from (a) applid to the function $frac {R(u,v)}u$. The second part is I at at p. 75 which states

$$R(sin x,cos x)dx=R_0(sin^2 x,cos x)sin x dx=-R_0(1-cos^2 x,cos x) dcos x.$$

(c) I translated III from pp. 75-76:

Substtuting $u$ by $frac uvv$, we have

$$R(u,v)=Rleft(frac uvv,vright)=R^*left(frac uv,vright).$$

By the property of the function $R$ with the respect to change of signs of $u$ and $v$ (which does not change the fraction $frac uv$),

$$R^*left(frac uv,-vright)= R^*left(frac uv,vright),$$

then, as we know,

$$R^*left(frac uv,vright)=R_1^*left(frac uv,v^2right).$$

Thus

$$R(sin x,cos x)=R^*_1(tan x,cos^2 x)= R^*_1left(tan x,frac 1{1+tan^2 x}right),$$

that is

$$R(sin x,cos x)=tilde R(tan x).$$

Here a substitution $t=tan x$ $left(-fracpi{2}<x<fracpi{2} right)$ reaches the goal, because $$R(sin x,cos x)dx=tilde R(t)frac{dt}{1+t^2}$$

For part a), note that if $$R(u,v)=R(-u,v)$$then $$P(-u,v)Q(u,v)=P(u,v)Q(-u,v)$$Define $S(u,v)=P(-u,v)Q(u,v)$. Then $S(u,v)$ is a polynomial function of $u,v$ since it is the product of two other polynomials and therefore can be expressed as $$S(u,v)=sum_{i=0}^{m}sum_{j=0}^{n}a_{ij}u^iv^j$$from $S(u,v)=S(-u,v)$ we obtain$$sum_{i=0}^{m}sum_{j=0}^{n}a_{ij}u^iv^j=sum_{i=0}^{m}sum_{j=0}^{n}a_{ij}(-u)^iv^j$$which yields to $$sum_{i=0\itext{ is even}}^{m}sum_{j=0}^{n}a_{ij}u^iv^j+sum_{i=0\itext{ is odd}}^{m}sum_{j=0}^{n}a_{ij}u^iv^j=sum_{i=0\itext{ is even}}^{m}sum_{j=0}^{n}a_{ij}(-u)^iv^j+sum_{i=0\itext{ is odd}}^{m}sum_{j=0}^{n}a_{ij}(-u)^iv^j$$from which we obtain $$sum_{i=0\itext{ is odd}}^{m}sum_{j=0}^{n}a_{ij}u^iv^j=0$$for any $u,v$. Therefore we can write $$S(u,v){=sum_{i=0}^{m}sum_{j=0}^{n}a_{ij}u^iv^j\=sum_{i=0\itext{ is even}}^{m}sum_{j=0}^{n}a_{ij}u^iv^j\=sum_{i'=0}^{m'}sum_{j=0}^{n}a_{i'j}u^{2i'}v^j\=S_1(u^2,v)}$$This means that $S(u,v)$ is a polynomial of $u^2$ and $v$ so is $P(-u,v)Q(u,v)$. Then both $P(-u,v)$ and $Q(u,v)$ must be polynomials of $u^2$ and $v$ (otherwise at least one term as $u^{2k+1}v^l$ would be appeared in $S(u,v)$) and by dividing $P(u,v)$ on $Q(u,v)$ we conclude that $$R(u,v)=R_1(u^2,v)$$The other parts can be proved easily.