Can You Crack the Trigonometric Integration Challenge?

[anemone]

Here is this week's POTW:

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Find $$\int_{}^{} \dfrac{c\cos x+d\sin x}{a\cos x+b\sin x}\,dx$$ given $a,\,b,\,c,\,d$ are constants such that $a^2+b^2=c^2+d^2=1$.

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[anemone]

Congratulations to the following members for their correct solution:):

1. greg1313
2. Ackbach
3. kaliprasad

Partial credit goes to lfdahl.

Solution from Ackbach:

Spoiler

We are asked to compute
$$\int\frac{c \cos(x)+d \sin(x)}{a \cos(x)+b \sin(x)} \, dx, \qquad a^2+b^2=c^2+d^2=1.$$
Because $a^2+b^2=c^2+d^2=1$, we may use a slight change of constants. That is, there must exist $\theta$ and $\varphi$ such that
\begin{align*}
a&=\sin(\varphi) \\
b&=\cos(\varphi) \\
c&=\sin(\theta) \\
d&=\cos(\theta).
\end{align*}
The integral becomes
\begin{align*}
\int\frac{c \cos(x)+d \sin(x)}{a \cos(x)+b \sin(x)} \, dx &= \int\frac{\sin(\theta) \cos(x)+\cos(\theta) \sin(x)}{\sin(\varphi) \cos(x)+\cos(\varphi) \sin(x)} \, dx \\
&=\int\frac{\sin(x+\theta)}{\sin(x+\varphi)} \, dx, \qquad y=x+\varphi \\
&=\int\frac{\sin(y+\theta-\varphi)}{\sin(y)} \, dy, \qquad \alpha=\theta-\varphi \\
&=\int\frac{\sin(y+\alpha)}{\sin(y)} \, dy \\
&=\int\frac{\sin(y)\cos(\alpha)+\cos(y) \sin(\alpha)}{\sin(y)} \, dy \\
&=\int [\cos(\alpha)+\sin(\alpha) \cot(y)] \, dy \\
&=y \cos(\alpha)+\sin(\alpha) \ln|\sin(y)|+C \\
&=(x+\varphi) \cos(\theta-\varphi)+\sin(\theta-\varphi)\ln|\sin(x+\varphi)|+C \\
&=(x+\arcsin(a)) (bd+ac)+(bc-ad) \ln|b \sin(x)+a \cos(x)|+C.
\end{align*}
We could absorb the $\arcsin(a) \, (bd+ac)$ into the constant to obtain the slightly simpler expression
$$\boxed{x(bd+ac)+(bc-ad) \ln|b \sin(x)+a \cos(x)|+C.}$$

Check via differentiation:
\begin{align*}
\d{}{x} \left[ x(bd+ac)+(bc-ad) \ln|b \sin(x)+a \cos(x)| \right] &=bd+ac+(bc-ad) \, \frac{-a \sin(x)+b \cos(x)}{a \cos(x)+b \sin(x)} \\
&=\frac{(bd+ac)(a \cos(x)+b\sin(x))+(bc-ad)(-a\sin(x)+b\cos(x))}{a\cos(x)+b\sin(x)} \\
&=\frac{(abd+a^2c+b^2c-abd)\cos(x)+(b^2d+abc-abc+a^2d)\sin(x)}{a\cos(x)+b\sin(x)} \\
&=\frac{c\cos(x)+d\sin(x)}{a\cos(x)+b\sin(x)},
\end{align*}
which is the original integrand, as required. So, as long as we can avoid the denominator being zero, which occurs when
\begin{align*}
a\cos(x)+b\sin(x)&=0 \\
a\cos(x)&=-b\sin(x) \\
-\frac{a}{b}&=\tan(x) \\
x&=-\arctan\left(\frac{a}{b}\right),
\end{align*}
then the antiderivative is given above.

Alternate solution from kaliprasad:

Spoiler

We have $\dfrac{d}{dx}(a\cos\,x+b\sin\,x)=-a\sin\,x+b\cos\,x\cdots(1)$

so let $c\cos\,x + d\sin\,x = A(a\cos\,x+b\sin\,x) + B(-a\sin\,x+b\cos\,x)$

hene comparing coefficients of $\cos\,x$ and $\sin\,x$ on both sides we see

$c= Aa + Bb \cdots(2)$

$d = Ab-aB\cdots(3)$

solving above 2 we get

$A=\dfrac{ca+db}{a^2+b^2} = ca+db$

and $B = \dfrac{bc-ad}{c^2+d^2} = bc-ad$

hence

$c \cos\,x + d\sin\,x= (ac+bd)(a\cos\,x+b\sin\,x) + (bc-ad)(-a\sin\,x+b\cos\,x)$

hence

$\dfrac{c \cos\,x + d\sin\,x}{\cos\,x+b\sin\,x} = (ac+bd) + (bc-ad) \dfrac{-a\sin\,x+b\cos\,x}{a\cos\,x+b\sin\,x}$

so integrating we get the result knowing (1)

= $(ac+bd) x + (bc-ad) ln |a\cos\,x+b\sin\,x| + C$