What are the indefinite integrals of cos x and sin x over cos x plus sin x?

[MarkFL]

Hello MHB Community,

anemone couldn't be with us this week, but she conscientiously made provisions for me to post this week's POTW for Secondary School/High School Students in her stead.

So, with no further ado...

The indefinite integrals $I_1$ and $I_2$ are defined by:

$\displaystyle I_1=\int \dfrac{\cos x}{\cos x+\sin x} dx$

and

$\displaystyle I_2=\int \dfrac{\sin x}{\cos x+\sin x} dx$

Determine $I_1$, $I_2$ and hence:

$\displaystyle \int \dfrac{\sin x}{a\cos x+b\sin x} dx$

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[anemone]

Hi MHB,

Thanks to MarkFL for his help in posting last week's POTW.! :)

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

1. laura123
2. kaliprasad

Solution from laura123:

Spoiler

$\begin{aligned}\displaystyle I_1&=\int \dfrac{\cos x}{\cos x+\sin x} dx=\int \dfrac{2\cos x}{2(\cos x+\sin x)} dx=\\
&=\int \dfrac{2\cos x+\sin x-\sin x}{2(\cos x+\sin x)}=\dfrac{1}{2}\int \dfrac{\cos x+\cos x+\sin x-\sin x}{\cos x+\sin x} dx=\\
&=\dfrac{1}{2}\int\dfrac{\cos x+\sin x}{\cos x+\sin x}dx+\dfrac{1}{2}\int\dfrac{\cos x-\sin x}{\cos x+\sin x}dx=\\
&=\dfrac{1}{2}x+\dfrac{1}{2}\log|\cos x+\sin x|+k.
\end{aligned}$

$\begin{aligned}\displaystyle I_2&=\int \dfrac{\sin x}{\cos x+\sin x} dx=\int \dfrac{\sin x+\cos x-\cos x}{\cos x+\sin x} dx=\\
&=\int \left(1-\dfrac{\cos x}{\cos x+\sin x}\right)dx=x-I_1=\\
&=\dfrac{1}{2}x-\dfrac{1}{2}\log|\cos x+\sin x|+k.
\end{aligned}$

$\begin{aligned}\displaystyle \int \dfrac{\sin x}{a\cos x+b\sin x} dx&=\dfrac{1}{a^2+b^2}\int \dfrac{(a^2+b^2)\sin x}{a\cos x+b\sin x} dx=\\
&=\dfrac{1}{a^2+b^2}\int \dfrac{a^2\sin x+b^2\sin x+ab\cos x-ab\cos x}{a\cos x+b\sin x} dx=\\
&=\dfrac{1}{a^2+b^2}\int \dfrac{a^2\sin x-ab\cos x}{a\cos x+b\sin x} dx+\dfrac{1}{a^2+b^2}\int\dfrac{b^2\sin x+ab\cos x}{a\cos x+b\sin x}dx=\\
&=\dfrac{1}{a^2+b^2}\int \dfrac{a(a\sin x-b\cos x)}{a\cos x+b\sin x} dx+\dfrac{1}{a^2+b^2}\int\dfrac{b(b\sin x+a\cos x)}{a\cos x+b\sin x}dx=\\
&=-\dfrac{a}{a^2+b^2}\log|a\cos x+b\sin x|+\dfrac{b}{a^2+b^2}x+k.
\end{aligned}$

Solution from kaliprasad:

Spoiler

$∫ \dfrac{\sin\ x}{\sin \ x + \ cos\ x} dx $

= $∫\dfrac{1}{2} \cdot \dfrac{2\sin\ x}{\sin \ x + \ cos\ x} dx $

= $∫\dfrac{1}{2} \cdot \dfrac{(\sin\ x+\cos \ x) +(\sin\ x-\cos \ x) }{\sin \ x + \ cos\ x} dx $

= $∫\dfrac{1}{2} (1+ \dfrac{\sin\ x-\cos \ x }{\sin \ x + \ cos\ x} dx $

= $∫\dfrac{1}{2}dx + ∫\dfrac{1}{2} \dfrac{\sin\ x-\cos \ x }{\sin \ x + \ cos\ x} dx $

= $\dfrac{1}{2}x - \dfrac{1}{2} ln \mid (\sin \ x + \ cos\ x)\mid +c_1$

as $\dfrac{\cos\ x}{\sin \ x + \ cos\ x} $

= $1- \dfrac{\sin\ x}{\sin \ x + \ cos\ x} $

so $∫ \dfrac{\cos \ x}{\sin \ x + \ cos\ x} dx $

= $\dfrac{1}{2}x + \dfrac{1}{2} ln \mid (\sin \ x + \ cos\ x)\mid +c_2$

now we need to inetgrate

$\dfrac{\sin\ x}{a \cos \ x + b\ sin\ x} $

as derivate of $a \cos \ x + b\sin \ x $ is $- a \sin \ x + b\cos \ x $

we can put $\sin\ x= m(a \cos \ x + b\ sin\ x) + n (- a \sin \ x + b\cos \ x) $



comparing coefficients we get 1 = ma + nb and 0=mb-na



solving these 2 we get



$m=\dfrac{b}{b^2+a^2}$ and $n= \dfrac{- a}{b^2+a^2}$



so $\sin\ x=\dfrac{b}{b^2+a^2}(a \cos \ x + b\ sin\ x) - \dfrac{a}{b^2+a^2} (- a \sin \ x + b\cos \ x) $



so $\dfrac{\sin\ x}{a \ cos \ x + b\sin \ x }= \dfrac{b}{b^2+a^2} - \dfrac{a}{b^2+a^2} \dfrac{- a \sin \ x + b\cos \ x} {a \cos \ x + b\sin \ x } $



so integral of above = $ \dfrac{b}{b^2+a^2} x - \dfrac{a}{b^2+a^2} ln \mid a \cos \ x + b\sin \ x \mid +C$

where $C$ is the constant of integration.