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

Question

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).

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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.

Answer 1

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

Answer 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 $\blacksquare$