Integration of trigonometric function $intfrac{sin(2x)}{sin(x)-cos(x)}dx$
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$$intfrac{sin(2x)}{sin(x)-cos(x)}dx$$
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?
real-analysis integration trigonometry indefinite-integrals
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add a comment |
$begingroup$
$$intfrac{sin(2x)}{sin(x)-cos(x)}dx$$
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?
real-analysis integration trigonometry indefinite-integrals
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1
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As I see it, the half-angle tangent substitution, which you mentioned, is a very simple way here. Just try it - the numerator and denominator become very nice
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– Yuriy S
Aug 6 '16 at 11:10
add a comment |
$begingroup$
$$intfrac{sin(2x)}{sin(x)-cos(x)}dx$$
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?
real-analysis integration trigonometry indefinite-integrals
$endgroup$
$$intfrac{sin(2x)}{sin(x)-cos(x)}dx$$
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?
real-analysis integration trigonometry indefinite-integrals
real-analysis integration trigonometry indefinite-integrals
edited Dec 26 '18 at 23:20
DavidG
2,1121723
2,1121723
asked Aug 6 '16 at 11:07
Archis WelankarArchis Welankar
12k41642
12k41642
1
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As I see it, the half-angle tangent substitution, which you mentioned, is a very simple way here. Just try it - the numerator and denominator become very nice
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– Yuriy S
Aug 6 '16 at 11:10
add a comment |
1
$begingroup$
As I see it, the half-angle tangent substitution, which you mentioned, is a very simple way here. Just try it - the numerator and denominator become very nice
$endgroup$
– Yuriy S
Aug 6 '16 at 11:10
1
1
$begingroup$
As I see it, the half-angle tangent substitution, which you mentioned, is a very simple way here. Just try it - the numerator and denominator become very nice
$endgroup$
– Yuriy S
Aug 6 '16 at 11:10
$begingroup$
As I see it, the half-angle tangent substitution, which you mentioned, is a very simple way here. Just try it - the numerator and denominator become very nice
$endgroup$
– Yuriy S
Aug 6 '16 at 11:10
add a comment |
6 Answers
6
active
oldest
votes
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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 $u = x-frac{1}{4} pi$,
$$ int frac{1}{sqrt{2} sinleft(x-frac{1}{4} pi right) }dx = int frac{1}{sqrt{2} } csc(u)du $$
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There is a shortcut: write $sin (2x)=1-(sin (x) - cos (x))^2$. Is there a mistake in your sign? Shouln't it be $-sin (x)?$
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– Behnam Esmayli
Aug 6 '16 at 16:44
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@Behnam Nope. there is no sign mistake.
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– Zack Ni
Aug 7 '16 at 1:24
add a comment |
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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.
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add a comment |
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Let $u=x-fracpi4$, then
$$
begin{align}
intfrac{sin(2x)}{sin(x)-cos(x)},mathrm{d}x
&=intfrac{sinleft(2u+fracpi2right)}{sqrt2sin(u)},mathrm{d}u\
&=frac1{sqrt2}intfrac{cosleft(2uright)}{sin(u)},mathrm{d}u\
&=frac1{sqrt2}intfrac{1-2sin^2(u)}{sin(u)},mathrm{d}u\
&=frac1{sqrt2}intfrac{sin(u)}{1-cos^2(u)},mathrm{d}u-sqrt2intsin(u),mathrm{d}u\
&=sqrt2cos(u)-frac1{2sqrt2}intleft(frac1{1-cos(u)}+frac1{1+cos(u)}right),mathrm{d}cos(u)\
&=sqrt2cos(u)+frac1{2sqrt2}logleft(frac{1-cos(u)}{1+cos(u)}right)+C\
&=sqrt2cos(u)+frac1{sqrt2}log(tan(u/2))+C
end{align}
$$
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add a comment |
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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 $$I = frac{1}{sqrt{2}}ln left|tan left(frac{x-frac{pi}{4}}{2}right)right|+sqrt{2}cos left(x-frac{pi}{4}right)+mathcal{C}$$
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1
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Clever approach. How did you know that $sin x-cos x=sqrt{2}sin(x-frac{pi}{4})$. Did you use the sum identities and equate coefficients?
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– Ahmed S. Attaalla
Aug 6 '16 at 12:30
1
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There will a $sqrt(2)$ before $cos(x-pi/4)$ in last step.
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– user220382
Aug 6 '16 at 13:05
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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 ! :-)
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add a comment |
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Assuming $$u=x-fracpi4$$
$$
begin{align}
intfrac{sin(2x)}{sin(x)-cos(x)},mathrm{d}x
&=intfrac{sinleft(2u+fracpi2right)}{sqrt2sin(u)},mathrm{d}u\
&=frac1{sqrt2}intfrac{cosleft(2uright)}{sin(u)},mathrm{d}u\
&=frac1{sqrt2}intfrac{1-2sin^2(u)}{sin(u)},mathrm{d}u\
&=frac1{sqrt2}intleft(csc u -2sin uright),mathrm{d}u\
&=frac1{sqrt2}int csc u text{du} -2sin u text{du} \
&=frac1{sqrt2}left(log left|tan frac{u}{2}right| +2cos uright) \
end{align}
$$
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There is a mistake in your final answer.Otherwise your method is nice!
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– user220382
Aug 6 '16 at 16:17
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@SanchayanDutta Fixed it
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– Aakash Kumar
Aug 6 '16 at 16:23
add a comment |
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6 Answers
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6 Answers
6
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$begingroup$
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 $u = x-frac{1}{4} pi$,
$$ int frac{1}{sqrt{2} sinleft(x-frac{1}{4} pi right) }dx = int frac{1}{sqrt{2} } csc(u)du $$
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There is a shortcut: write $sin (2x)=1-(sin (x) - cos (x))^2$. Is there a mistake in your sign? Shouln't it be $-sin (x)?$
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– Behnam Esmayli
Aug 6 '16 at 16:44
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@Behnam Nope. there is no sign mistake.
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– Zack Ni
Aug 7 '16 at 1:24
add a comment |
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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 $u = x-frac{1}{4} pi$,
$$ int frac{1}{sqrt{2} sinleft(x-frac{1}{4} pi right) }dx = int frac{1}{sqrt{2} } csc(u)du $$
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There is a shortcut: write $sin (2x)=1-(sin (x) - cos (x))^2$. Is there a mistake in your sign? Shouln't it be $-sin (x)?$
$endgroup$
– Behnam Esmayli
Aug 6 '16 at 16:44
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@Behnam Nope. there is no sign mistake.
$endgroup$
– Zack Ni
Aug 7 '16 at 1:24
add a comment |
$begingroup$
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 $u = x-frac{1}{4} pi$,
$$ int frac{1}{sqrt{2} sinleft(x-frac{1}{4} pi right) }dx = int frac{1}{sqrt{2} } csc(u)du $$
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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 $u = x-frac{1}{4} pi$,
$$ int frac{1}{sqrt{2} sinleft(x-frac{1}{4} pi right) }dx = int frac{1}{sqrt{2} } csc(u)du $$
edited Dec 27 '18 at 0:01
DavidG
2,1121723
2,1121723
answered Aug 6 '16 at 12:08
Zack NiZack Ni
3,466729
3,466729
$begingroup$
There is a shortcut: write $sin (2x)=1-(sin (x) - cos (x))^2$. Is there a mistake in your sign? Shouln't it be $-sin (x)?$
$endgroup$
– Behnam Esmayli
Aug 6 '16 at 16:44
$begingroup$
@Behnam Nope. there is no sign mistake.
$endgroup$
– Zack Ni
Aug 7 '16 at 1:24
add a comment |
$begingroup$
There is a shortcut: write $sin (2x)=1-(sin (x) - cos (x))^2$. Is there a mistake in your sign? Shouln't it be $-sin (x)?$
$endgroup$
– Behnam Esmayli
Aug 6 '16 at 16:44
$begingroup$
@Behnam Nope. there is no sign mistake.
$endgroup$
– Zack Ni
Aug 7 '16 at 1:24
$begingroup$
There is a shortcut: write $sin (2x)=1-(sin (x) - cos (x))^2$. Is there a mistake in your sign? Shouln't it be $-sin (x)?$
$endgroup$
– Behnam Esmayli
Aug 6 '16 at 16:44
$begingroup$
There is a shortcut: write $sin (2x)=1-(sin (x) - cos (x))^2$. Is there a mistake in your sign? Shouln't it be $-sin (x)?$
$endgroup$
– Behnam Esmayli
Aug 6 '16 at 16:44
$begingroup$
@Behnam Nope. there is no sign mistake.
$endgroup$
– Zack Ni
Aug 7 '16 at 1:24
$begingroup$
@Behnam Nope. there is no sign mistake.
$endgroup$
– Zack Ni
Aug 7 '16 at 1:24
add a comment |
$begingroup$
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.
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add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
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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.
answered Aug 6 '16 at 11:13
JanJan
21.8k31240
21.8k31240
add a comment |
add a comment |
$begingroup$
Let $u=x-fracpi4$, then
$$
begin{align}
intfrac{sin(2x)}{sin(x)-cos(x)},mathrm{d}x
&=intfrac{sinleft(2u+fracpi2right)}{sqrt2sin(u)},mathrm{d}u\
&=frac1{sqrt2}intfrac{cosleft(2uright)}{sin(u)},mathrm{d}u\
&=frac1{sqrt2}intfrac{1-2sin^2(u)}{sin(u)},mathrm{d}u\
&=frac1{sqrt2}intfrac{sin(u)}{1-cos^2(u)},mathrm{d}u-sqrt2intsin(u),mathrm{d}u\
&=sqrt2cos(u)-frac1{2sqrt2}intleft(frac1{1-cos(u)}+frac1{1+cos(u)}right),mathrm{d}cos(u)\
&=sqrt2cos(u)+frac1{2sqrt2}logleft(frac{1-cos(u)}{1+cos(u)}right)+C\
&=sqrt2cos(u)+frac1{sqrt2}log(tan(u/2))+C
end{align}
$$
$endgroup$
add a comment |
$begingroup$
Let $u=x-fracpi4$, then
$$
begin{align}
intfrac{sin(2x)}{sin(x)-cos(x)},mathrm{d}x
&=intfrac{sinleft(2u+fracpi2right)}{sqrt2sin(u)},mathrm{d}u\
&=frac1{sqrt2}intfrac{cosleft(2uright)}{sin(u)},mathrm{d}u\
&=frac1{sqrt2}intfrac{1-2sin^2(u)}{sin(u)},mathrm{d}u\
&=frac1{sqrt2}intfrac{sin(u)}{1-cos^2(u)},mathrm{d}u-sqrt2intsin(u),mathrm{d}u\
&=sqrt2cos(u)-frac1{2sqrt2}intleft(frac1{1-cos(u)}+frac1{1+cos(u)}right),mathrm{d}cos(u)\
&=sqrt2cos(u)+frac1{2sqrt2}logleft(frac{1-cos(u)}{1+cos(u)}right)+C\
&=sqrt2cos(u)+frac1{sqrt2}log(tan(u/2))+C
end{align}
$$
$endgroup$
add a comment |
$begingroup$
Let $u=x-fracpi4$, then
$$
begin{align}
intfrac{sin(2x)}{sin(x)-cos(x)},mathrm{d}x
&=intfrac{sinleft(2u+fracpi2right)}{sqrt2sin(u)},mathrm{d}u\
&=frac1{sqrt2}intfrac{cosleft(2uright)}{sin(u)},mathrm{d}u\
&=frac1{sqrt2}intfrac{1-2sin^2(u)}{sin(u)},mathrm{d}u\
&=frac1{sqrt2}intfrac{sin(u)}{1-cos^2(u)},mathrm{d}u-sqrt2intsin(u),mathrm{d}u\
&=sqrt2cos(u)-frac1{2sqrt2}intleft(frac1{1-cos(u)}+frac1{1+cos(u)}right),mathrm{d}cos(u)\
&=sqrt2cos(u)+frac1{2sqrt2}logleft(frac{1-cos(u)}{1+cos(u)}right)+C\
&=sqrt2cos(u)+frac1{sqrt2}log(tan(u/2))+C
end{align}
$$
$endgroup$
Let $u=x-fracpi4$, then
$$
begin{align}
intfrac{sin(2x)}{sin(x)-cos(x)},mathrm{d}x
&=intfrac{sinleft(2u+fracpi2right)}{sqrt2sin(u)},mathrm{d}u\
&=frac1{sqrt2}intfrac{cosleft(2uright)}{sin(u)},mathrm{d}u\
&=frac1{sqrt2}intfrac{1-2sin^2(u)}{sin(u)},mathrm{d}u\
&=frac1{sqrt2}intfrac{sin(u)}{1-cos^2(u)},mathrm{d}u-sqrt2intsin(u),mathrm{d}u\
&=sqrt2cos(u)-frac1{2sqrt2}intleft(frac1{1-cos(u)}+frac1{1+cos(u)}right),mathrm{d}cos(u)\
&=sqrt2cos(u)+frac1{2sqrt2}logleft(frac{1-cos(u)}{1+cos(u)}right)+C\
&=sqrt2cos(u)+frac1{sqrt2}log(tan(u/2))+C
end{align}
$$
edited Aug 6 '16 at 13:37
answered Aug 6 '16 at 13:30
robjohn♦robjohn
268k27308633
268k27308633
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$begingroup$
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 $$I = frac{1}{sqrt{2}}ln left|tan left(frac{x-frac{pi}{4}}{2}right)right|+sqrt{2}cos left(x-frac{pi}{4}right)+mathcal{C}$$
$endgroup$
1
$begingroup$
Clever approach. How did you know that $sin x-cos x=sqrt{2}sin(x-frac{pi}{4})$. Did you use the sum identities and equate coefficients?
$endgroup$
– Ahmed S. Attaalla
Aug 6 '16 at 12:30
1
$begingroup$
There will a $sqrt(2)$ before $cos(x-pi/4)$ in last step.
$endgroup$
– user220382
Aug 6 '16 at 13:05
add a comment |
$begingroup$
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 $$I = frac{1}{sqrt{2}}ln left|tan left(frac{x-frac{pi}{4}}{2}right)right|+sqrt{2}cos left(x-frac{pi}{4}right)+mathcal{C}$$
$endgroup$
1
$begingroup$
Clever approach. How did you know that $sin x-cos x=sqrt{2}sin(x-frac{pi}{4})$. Did you use the sum identities and equate coefficients?
$endgroup$
– Ahmed S. Attaalla
Aug 6 '16 at 12:30
1
$begingroup$
There will a $sqrt(2)$ before $cos(x-pi/4)$ in last step.
$endgroup$
– user220382
Aug 6 '16 at 13:05
add a comment |
$begingroup$
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 $$I = frac{1}{sqrt{2}}ln left|tan left(frac{x-frac{pi}{4}}{2}right)right|+sqrt{2}cos left(x-frac{pi}{4}right)+mathcal{C}$$
$endgroup$
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 $$I = frac{1}{sqrt{2}}ln left|tan left(frac{x-frac{pi}{4}}{2}right)right|+sqrt{2}cos left(x-frac{pi}{4}right)+mathcal{C}$$
edited Aug 6 '16 at 13:46
answered Aug 6 '16 at 12:09
juantheronjuantheron
34.3k1147142
34.3k1147142
1
$begingroup$
Clever approach. How did you know that $sin x-cos x=sqrt{2}sin(x-frac{pi}{4})$. Did you use the sum identities and equate coefficients?
$endgroup$
– Ahmed S. Attaalla
Aug 6 '16 at 12:30
1
$begingroup$
There will a $sqrt(2)$ before $cos(x-pi/4)$ in last step.
$endgroup$
– user220382
Aug 6 '16 at 13:05
add a comment |
1
$begingroup$
Clever approach. How did you know that $sin x-cos x=sqrt{2}sin(x-frac{pi}{4})$. Did you use the sum identities and equate coefficients?
$endgroup$
– Ahmed S. Attaalla
Aug 6 '16 at 12:30
1
$begingroup$
There will a $sqrt(2)$ before $cos(x-pi/4)$ in last step.
$endgroup$
– user220382
Aug 6 '16 at 13:05
1
1
$begingroup$
Clever approach. How did you know that $sin x-cos x=sqrt{2}sin(x-frac{pi}{4})$. Did you use the sum identities and equate coefficients?
$endgroup$
– Ahmed S. Attaalla
Aug 6 '16 at 12:30
$begingroup$
Clever approach. How did you know that $sin x-cos x=sqrt{2}sin(x-frac{pi}{4})$. Did you use the sum identities and equate coefficients?
$endgroup$
– Ahmed S. Attaalla
Aug 6 '16 at 12:30
1
1
$begingroup$
There will a $sqrt(2)$ before $cos(x-pi/4)$ in last step.
$endgroup$
– user220382
Aug 6 '16 at 13:05
$begingroup$
There will a $sqrt(2)$ before $cos(x-pi/4)$ in last step.
$endgroup$
– user220382
Aug 6 '16 at 13:05
add a comment |
$begingroup$
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 ! :-)
$endgroup$
add a comment |
$begingroup$
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 ! :-)
$endgroup$
add a comment |
$begingroup$
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 ! :-)
$endgroup$
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 ! :-)
edited Aug 6 '16 at 13:06
answered Aug 6 '16 at 12:09
user220382
add a comment |
add a comment |
$begingroup$
Assuming $$u=x-fracpi4$$
$$
begin{align}
intfrac{sin(2x)}{sin(x)-cos(x)},mathrm{d}x
&=intfrac{sinleft(2u+fracpi2right)}{sqrt2sin(u)},mathrm{d}u\
&=frac1{sqrt2}intfrac{cosleft(2uright)}{sin(u)},mathrm{d}u\
&=frac1{sqrt2}intfrac{1-2sin^2(u)}{sin(u)},mathrm{d}u\
&=frac1{sqrt2}intleft(csc u -2sin uright),mathrm{d}u\
&=frac1{sqrt2}int csc u text{du} -2sin u text{du} \
&=frac1{sqrt2}left(log left|tan frac{u}{2}right| +2cos uright) \
end{align}
$$
$endgroup$
$begingroup$
There is a mistake in your final answer.Otherwise your method is nice!
$endgroup$
– user220382
Aug 6 '16 at 16:17
$begingroup$
@SanchayanDutta Fixed it
$endgroup$
– Aakash Kumar
Aug 6 '16 at 16:23
add a comment |
$begingroup$
Assuming $$u=x-fracpi4$$
$$
begin{align}
intfrac{sin(2x)}{sin(x)-cos(x)},mathrm{d}x
&=intfrac{sinleft(2u+fracpi2right)}{sqrt2sin(u)},mathrm{d}u\
&=frac1{sqrt2}intfrac{cosleft(2uright)}{sin(u)},mathrm{d}u\
&=frac1{sqrt2}intfrac{1-2sin^2(u)}{sin(u)},mathrm{d}u\
&=frac1{sqrt2}intleft(csc u -2sin uright),mathrm{d}u\
&=frac1{sqrt2}int csc u text{du} -2sin u text{du} \
&=frac1{sqrt2}left(log left|tan frac{u}{2}right| +2cos uright) \
end{align}
$$
$endgroup$
$begingroup$
There is a mistake in your final answer.Otherwise your method is nice!
$endgroup$
– user220382
Aug 6 '16 at 16:17
$begingroup$
@SanchayanDutta Fixed it
$endgroup$
– Aakash Kumar
Aug 6 '16 at 16:23
add a comment |
$begingroup$
Assuming $$u=x-fracpi4$$
$$
begin{align}
intfrac{sin(2x)}{sin(x)-cos(x)},mathrm{d}x
&=intfrac{sinleft(2u+fracpi2right)}{sqrt2sin(u)},mathrm{d}u\
&=frac1{sqrt2}intfrac{cosleft(2uright)}{sin(u)},mathrm{d}u\
&=frac1{sqrt2}intfrac{1-2sin^2(u)}{sin(u)},mathrm{d}u\
&=frac1{sqrt2}intleft(csc u -2sin uright),mathrm{d}u\
&=frac1{sqrt2}int csc u text{du} -2sin u text{du} \
&=frac1{sqrt2}left(log left|tan frac{u}{2}right| +2cos uright) \
end{align}
$$
$endgroup$
Assuming $$u=x-fracpi4$$
$$
begin{align}
intfrac{sin(2x)}{sin(x)-cos(x)},mathrm{d}x
&=intfrac{sinleft(2u+fracpi2right)}{sqrt2sin(u)},mathrm{d}u\
&=frac1{sqrt2}intfrac{cosleft(2uright)}{sin(u)},mathrm{d}u\
&=frac1{sqrt2}intfrac{1-2sin^2(u)}{sin(u)},mathrm{d}u\
&=frac1{sqrt2}intleft(csc u -2sin uright),mathrm{d}u\
&=frac1{sqrt2}int csc u text{du} -2sin u text{du} \
&=frac1{sqrt2}left(log left|tan frac{u}{2}right| +2cos uright) \
end{align}
$$
edited Aug 6 '16 at 16:18
answered Aug 6 '16 at 15:43
Aakash KumarAakash Kumar
2,5211519
2,5211519
$begingroup$
There is a mistake in your final answer.Otherwise your method is nice!
$endgroup$
– user220382
Aug 6 '16 at 16:17
$begingroup$
@SanchayanDutta Fixed it
$endgroup$
– Aakash Kumar
Aug 6 '16 at 16:23
add a comment |
$begingroup$
There is a mistake in your final answer.Otherwise your method is nice!
$endgroup$
– user220382
Aug 6 '16 at 16:17
$begingroup$
@SanchayanDutta Fixed it
$endgroup$
– Aakash Kumar
Aug 6 '16 at 16:23
$begingroup$
There is a mistake in your final answer.Otherwise your method is nice!
$endgroup$
– user220382
Aug 6 '16 at 16:17
$begingroup$
There is a mistake in your final answer.Otherwise your method is nice!
$endgroup$
– user220382
Aug 6 '16 at 16:17
$begingroup$
@SanchayanDutta Fixed it
$endgroup$
– Aakash Kumar
Aug 6 '16 at 16:23
$begingroup$
@SanchayanDutta Fixed it
$endgroup$
– Aakash Kumar
Aug 6 '16 at 16:23
add a comment |
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As I see it, the half-angle tangent substitution, which you mentioned, is a very simple way here. Just try it - the numerator and denominator become very nice
$endgroup$
– Yuriy S
Aug 6 '16 at 11:10