# Primitive of $\int R(\cos x, \sin x) dx$

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 $$P(x)=\sum_{i,j\geq0, i+j\leq n}{a_{ij}x^iy^j} =\sum_{i,j\geq0, 2i+j\leq n}{a_{2i,j}x^{2i}y^{j}}+\sum_{i,j\geq0, 2i+j\leq n}{a_{2i+1,j}x^{2i+1}y^{j}}=P_1(x^2,y)+xP_2(x^2,y)$$ 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

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

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

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

$$R(u,v)=R\left(\frac uvv,v\right)=R^*\left(\frac uv,v\right).$$

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

$$R^*\left(\frac uv,-v\right)= R^*\left(\frac uv,v\right),$$

then, as we know,

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

Thus

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

that is

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

Here a substitution $$t=\tan x$$ $$\left(-\frac\pi{2} reaches the goal, because $$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 $$\sum_{i=0\\i\text{ is even}}^{m}\sum_{j=0}^{n}a_{ij}u^iv^j+\sum_{i=0\\i\text{ is odd}}^{m}\sum_{j=0}^{n}a_{ij}u^iv^j=\sum_{i=0\\i\text{ is even}}^{m}\sum_{j=0}^{n}a_{ij}(-u)^iv^j+\sum_{i=0\\i\text{ is odd}}^{m}\sum_{j=0}^{n}a_{ij}(-u)^iv^j$$from which we obtain $$\sum_{i=0\\i\text{ is odd}}^{m}\sum_{j=0}^{n}a_{ij}u^iv^j=0$$for any $$u,v$$. Therefore we can write $$S(u,v){=\sum_{i=0}^{m}\sum_{j=0}^{n}a_{ij}u^iv^j\\=\sum_{i=0\\i\text{ is even}}^{m}\sum_{j=0}^{n}a_{ij}u^iv^j\\=\sum_{i'=0}^{m'}\sum_{j=0}^{n}a_{i'j}u^{2i'}v^j\\=S_1(u^2,v)}$$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 $$\blacksquare$$