Xrfgtjtk

$$intfrac{sin(2x)}{sin(x)-cos(x)}dx$$

My attempt: Firstly, $sin(2x)=2sin(x)cos(x)$.

After that, eliminate the $cos(x)$ seen in both the numerator and denominator to get

$$2intfrac{sin(x)}{tan(x)-1} dx.$$

From here onwards, should I convert $sin(x)$, $tan(x)$ to half-angles and use $tan(x/2)=t$?

But this would be a time consuming method. Any suggestions?

Hint:
$$intfrac{sin(2x)}{sin(x)-cos(x)}dx = intfrac{2sin(x)cos(x) }{sin(x)-cos(x)}dx = intfrac{sin(x)cos(x)+cos(x)sin(x) }{sin(x)-cos(x)}dx\
=intfrac{sin(x)cos(x)-sin^2(x)+cos(x)sin(x)-cos^2(x)+1 }{sin(x)-cos(x)}dx \=int -sin(x)dx+int cos(x)dx + int frac{1}{sin(x) -cos(x)}dx \= cos(x) + sin(x) +int frac{1}{sin(x) -cos(x)}dx $$

For $int frac{1}{sin(x) -cos(x)}dx$:

Notice that $$int frac{1}{sin(x) -cos(x)}dx = int frac{1}{sqrt{2} sin(x-frac{1}{4} pi ) }dx$$

Let $u = x-frac{1}{4} pi$,

$$ int frac{1}{sqrt{2} sinleft(x-frac{1}{4} pi right) }dx = int frac{1}{sqrt{2} } csc(u)du $$

HINT:

$$intfrac{sin(2x)}{sin(x)-cos(x)}spacetext{d}x=$$

Use $sin(2x)=2sin(x)cos(x)$:

$$2intfrac{sin(x)cos(x)}{sin(x)-cos(x)}spacetext{d}x=$$

Sustitute $u=tanleft(frac{x}{2}right)$ and $text{d}u=frac{xsec^2left(frac{x}{2}right)}{2}spacetext{d}x$:

$$-8intfrac{u(u^2-1)}{(u^2+1)^2(u^2+2u-1)}spacetext{d}u$$

Now, use partial fractions.

Let $u=x-fracpi4$, then
$$
begin{align}
intfrac{sin(2x)}{sin(x)-cos(x)},mathrm{d}x
&=intfrac{sinleft(2u+fracpi2right)}{sqrt2sin(u)},mathrm{d}u\
&=frac1{sqrt2}intfrac{cosleft(2uright)}{sin(u)},mathrm{d}u\
&=frac1{sqrt2}intfrac{1-2sin^2(u)}{sin(u)},mathrm{d}u\
&=frac1{sqrt2}intfrac{sin(u)}{1-cos^2(u)},mathrm{d}u-sqrt2intsin(u),mathrm{d}u\
&=sqrt2cos(u)-frac1{2sqrt2}intleft(frac1{1-cos(u)}+frac1{1+cos(u)}right),mathrm{d}cos(u)\
&=sqrt2cos(u)+frac1{2sqrt2}logleft(frac{1-cos(u)}{1+cos(u)}right)+C\
&=sqrt2cos(u)+frac1{sqrt2}log(tan(u/2))+C
end{align}
$$

Let $$I = intfrac{sin 2x}{sin x-cos x}dx = frac{1}{sqrt{2}}intfrac{sin 2x}{sin left(x-frac{pi}{4}right)}dx$$

Now Put $displaystyle x- frac{pi}{4} = t;,$ Then $dx = dt$

So $$I = frac{1}{sqrt{2}}intfrac{cos 2t}{sin t}dt = frac{1}{sqrt{2}}int frac{1-2sin^2 t}{sin t}dt$$

So $$I = frac{1}{sqrt{2}}int csc t dt-sqrt{2}int sin t dt$$

So $$I = frac{1}{sqrt{2}}ln left|tan frac{t}{2}right|+sqrt{2}cos t+mathcal{C}$$

So $$I = frac{1}{sqrt{2}}ln left|tan left(frac{x-frac{pi}{4}}{2}right)right|+sqrt{2}cos left(x-frac{pi}{4}right)+mathcal{C}$$

I like trigonometric substitutions so I will try to implement that !

Use $sin(x)-cos(x)=z$.Find numerator in terms of $z^2$.

Then $(cos(x)+sin(x))dx=dz$

$sin(x)-cos(x)=z$
so,$z^2=1-sin(2x)$

Also $(sin(x)+cos(x))^2+ (sin(x)-cos(x))^2=2$

So the integral boils down to

$$frac{1-z^2}{z(sqrt{2-z^2})}dz$$

Substitute $z=sqrt{2}sin(y)$
The integral becomes $$int frac{cos(2y)}{sqrt{2}sin(y)}dy$$
$$int frac{csc(y)-2sin(y)}{sqrt{2}}dy$$
which equals $$frac{1}{sqrt{2}}ln|tan(y/2)|+sqrt{2}cos(y)+C$$

On resubstituting original variables we get $$frac{1}{sqrt{2}}ln left|tan left(frac{x-frac{pi}{4}}{2}right)right|+sqrt{2}cos left(x-frac{pi}{4}right)+mathcal{C}$$

Hurray ! :-)

Assuming $$u=x-fracpi4$$
$$
begin{align}
intfrac{sin(2x)}{sin(x)-cos(x)},mathrm{d}x
&=intfrac{sinleft(2u+fracpi2right)}{sqrt2sin(u)},mathrm{d}u\
&=frac1{sqrt2}intfrac{cosleft(2uright)}{sin(u)},mathrm{d}u\
&=frac1{sqrt2}intfrac{1-2sin^2(u)}{sin(u)},mathrm{d}u\
&=frac1{sqrt2}intleft(csc u -2sin uright),mathrm{d}u\
&=frac1{sqrt2}int csc u text{du} -2sin u text{du} \
&=frac1{sqrt2}left(log left|tan frac{u}{2}right| +2cos uright) \
end{align}
$$