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Better way to solve integral
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I have the following integral
$$int_{0}^{pi / 2} mathrm{d} x ,, frac{cos^{3}(x/2) - cos^{4} (x)}{sin^{2} (x)}$$
My current solution is to use
$$v = tan left(frac{x}{4}right)$$
to obtain a rational function of $v$, and integrate this.
Is there a more practical / clever way of doing it?
integration
asked 2 days ago
Gordon
364
add a comment |
up vote
2
down vote
favorite
I have the following integral
$$int_{0}^{pi / 2} mathrm{d} x ,, frac{cos^{3}(x/2) - cos^{4} (x)}{sin^{2} (x)}$$
My current solution is to use
$$v = tan left(frac{x}{4}right)$$
to obtain a rational function of $v$, and integrate this.
Is there a more practical / clever way of doing it?
integration
asked 2 days ago
Gordon
364
Where did you see this integral?
– Frpzzd
2 days ago
Comes up in my work. I like the Weierstrass trick as it is a blanket method, but am looking for a better way to approach it
– Gordon
2 days ago
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I have the following integral
$$int_{0}^{pi / 2} mathrm{d} x ,, frac{cos^{3}(x/2) - cos^{4} (x)}{sin^{2} (x)}$$
My current solution is to use
$$v = tan left(frac{x}{4}right)$$
to obtain a rational function of $v$, and integrate this.
Is there a more practical / clever way of doing it?
integration
asked 2 days ago
Gordon
364
I have the following integral
$$int_{0}^{pi / 2} mathrm{d} x ,, frac{cos^{3}(x/2) - cos^{4} (x)}{sin^{2} (x)}$$
My current solution is to use
$$v = tan left(frac{x}{4}right)$$
to obtain a rational function of $v$, and integrate this.
Is there a more practical / clever way of doing it?
integration
integration
asked 2 days ago
Gordon
364
asked 2 days ago
Gordon
364
asked 2 days ago
Gordon
364
asked 2 days ago
Gordon
364
asked 2 days ago
Gordon
364
364
Where did you see this integral?
– Frpzzd
2 days ago
Comes up in my work. I like the Weierstrass trick as it is a blanket method, but am looking for a better way to approach it
– Gordon
2 days ago
add a comment |
Where did you see this integral?
– Frpzzd
2 days ago
Comes up in my work. I like the Weierstrass trick as it is a blanket method, but am looking for a better way to approach it
– Gordon
2 days ago
Where did you see this integral?
– Frpzzd
2 days ago
Where did you see this integral?
– Frpzzd
2 days ago
Comes up in my work. I like the Weierstrass trick as it is a blanket method, but am looking for a better way to approach it
– Gordon
2 days ago
Comes up in my work. I like the Weierstrass trick as it is a blanket method, but am looking for a better way to approach it
– Gordon
2 days ago
add a comment |
3 Answers
3
active
oldest
votes
up vote
0
down vote
While the substitution $v=tan(x/4)$ would work, I don't think it's a good method in this case, which just requires elementary antiderivatives.
First, split the fraction and notice that
$$
intfrac{cos^4x}{sin^2x},dx=intleft(frac{1}{sin^2x}-2+sin^2xright),dx
=-cot x-2x+frac{1}{2}(x-sin xcos x)
$$
This leaves the other piece:
$$
intfrac{cos^3(x/2)}{sin^2x},dx=[t=x/2]=2intfrac{cos^3t}{sin^22t},dt
=frac{1}{2}intfrac{cos t}{sin^2t},dt=-frac{1}{2sin t}
$$
Thus the integral is
$$
-frac{1}{2sin(x/2)}+cot x+2x-frac{1}{2}(x-sin xcos x)
$$
You can also note that
$$
cot x-frac{1}{2sin(x/2)}=frac{cos x}{2sin(x/2)cos(x/2)}-frac{1}{2sin(x/2)}=
frac{(cos(x/2)-1)(2cos(x/2)+1)}{2sin(x/2)}
$$
which has limit $0$ for $xto0$.
answered 2 days ago
egreg
173k1383198
add a comment |
up vote
0
down vote
$$frac{cos^3dfrac x2}{sin^2x}=frac{cosdfrac x2}{2sin^2dfrac x2}$$ is immediately integrable.
And in
$$frac{cos^4x}{sin^2x}=frac1{sin^2x}-2+sin^2x,$$ the first two terms are also immediate, and
$$sin^2x=frac{1-cos2x}2.$$
answered 2 days ago
Yves Daoust
121k668216
add a comment |
up vote
0
down vote
$$I=intfrac{cos^3(frac x2)-cos^4x}{sin^2x}mathrm{d}x$$
$$I=intfrac{cos^3(frac x2)}{sin^2x}mathrm{d}x-intfrac{cos^4x}{sin^2x}mathrm{d}x$$
$$I=big(I_1-I_2big)bigg|_0^{pi/2}$$
$$I_1=intfrac{cos^3(frac x2)}{sin^2x}mathrm{d}x$$
$u=x/2$:
$$I_1=2intfrac{cos^3u}{sin^22u}mathrm{d}u$$
$$I_1=2intfrac{cos^3u}{4sin^2ucos^2u}mathrm{d}u$$
$$I_1=frac12intfrac{cos u}{sin^2u}mathrm{d}u$$
$t=sin u$:
$$I_1=frac12intfrac{mathrm{d}t}{t^2}$$
$$I_1=-frac1{2t}$$
$$I_1=-frac1{2sin(x/2)}$$
$$I_2=intfrac{cos^4x}{sin^2x}mathrm{d}x$$
$$I_2=intfrac{cos^2x(1-sin^2x)}{sin^2x}mathrm{d}x$$
$$I_2=intcot^2x mathrm{d}x-intcos^2x mathrm{d}x$$
$$I_2=-bigg(cot x+frac12cos x,sin x+frac32xbigg)$$
$$I=bigg(-frac12csc(x/2)+cot x+frac12cos xsin x+frac32xbigg)bigg|_0^{pi/2}$$
$$I=frac{3pi-2sqrt{2}}4$$
answered 2 days ago
clathratus
1,839219
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
While the substitution $v=tan(x/4)$ would work, I don't think it's a good method in this case, which just requires elementary antiderivatives.
First, split the fraction and notice that
$$
intfrac{cos^4x}{sin^2x},dx=intleft(frac{1}{sin^2x}-2+sin^2xright),dx
=-cot x-2x+frac{1}{2}(x-sin xcos x)
$$
This leaves the other piece:
$$
intfrac{cos^3(x/2)}{sin^2x},dx=[t=x/2]=2intfrac{cos^3t}{sin^22t},dt
=frac{1}{2}intfrac{cos t}{sin^2t},dt=-frac{1}{2sin t}
$$
Thus the integral is
$$
-frac{1}{2sin(x/2)}+cot x+2x-frac{1}{2}(x-sin xcos x)
$$
You can also note that
$$
cot x-frac{1}{2sin(x/2)}=frac{cos x}{2sin(x/2)cos(x/2)}-frac{1}{2sin(x/2)}=
frac{(cos(x/2)-1)(2cos(x/2)+1)}{2sin(x/2)}
$$
which has limit $0$ for $xto0$.
answered 2 days ago
egreg
173k1383198
add a comment |
up vote
0
down vote
While the substitution $v=tan(x/4)$ would work, I don't think it's a good method in this case, which just requires elementary antiderivatives.
First, split the fraction and notice that
$$
intfrac{cos^4x}{sin^2x},dx=intleft(frac{1}{sin^2x}-2+sin^2xright),dx
=-cot x-2x+frac{1}{2}(x-sin xcos x)
$$
This leaves the other piece:
$$
intfrac{cos^3(x/2)}{sin^2x},dx=[t=x/2]=2intfrac{cos^3t}{sin^22t},dt
=frac{1}{2}intfrac{cos t}{sin^2t},dt=-frac{1}{2sin t}
$$
Thus the integral is
$$
-frac{1}{2sin(x/2)}+cot x+2x-frac{1}{2}(x-sin xcos x)
$$
You can also note that
$$
cot x-frac{1}{2sin(x/2)}=frac{cos x}{2sin(x/2)cos(x/2)}-frac{1}{2sin(x/2)}=
frac{(cos(x/2)-1)(2cos(x/2)+1)}{2sin(x/2)}
$$
which has limit $0$ for $xto0$.
answered 2 days ago
egreg
173k1383198
add a comment |
up vote
0
down vote
up vote
0
down vote
While the substitution $v=tan(x/4)$ would work, I don't think it's a good method in this case, which just requires elementary antiderivatives.
First, split the fraction and notice that
$$
intfrac{cos^4x}{sin^2x},dx=intleft(frac{1}{sin^2x}-2+sin^2xright),dx
=-cot x-2x+frac{1}{2}(x-sin xcos x)
$$
This leaves the other piece:
$$
intfrac{cos^3(x/2)}{sin^2x},dx=[t=x/2]=2intfrac{cos^3t}{sin^22t},dt
=frac{1}{2}intfrac{cos t}{sin^2t},dt=-frac{1}{2sin t}
$$
Thus the integral is
$$
-frac{1}{2sin(x/2)}+cot x+2x-frac{1}{2}(x-sin xcos x)
$$
You can also note that
$$
cot x-frac{1}{2sin(x/2)}=frac{cos x}{2sin(x/2)cos(x/2)}-frac{1}{2sin(x/2)}=
frac{(cos(x/2)-1)(2cos(x/2)+1)}{2sin(x/2)}
$$
which has limit $0$ for $xto0$.
answered 2 days ago
egreg
173k1383198
While the substitution $v=tan(x/4)$ would work, I don't think it's a good method in this case, which just requires elementary antiderivatives.
First, split the fraction and notice that
$$
intfrac{cos^4x}{sin^2x},dx=intleft(frac{1}{sin^2x}-2+sin^2xright),dx
=-cot x-2x+frac{1}{2}(x-sin xcos x)
$$
This leaves the other piece:
$$
intfrac{cos^3(x/2)}{sin^2x},dx=[t=x/2]=2intfrac{cos^3t}{sin^22t},dt
=frac{1}{2}intfrac{cos t}{sin^2t},dt=-frac{1}{2sin t}
$$
Thus the integral is
$$
-frac{1}{2sin(x/2)}+cot x+2x-frac{1}{2}(x-sin xcos x)
$$
You can also note that
$$
cot x-frac{1}{2sin(x/2)}=frac{cos x}{2sin(x/2)cos(x/2)}-frac{1}{2sin(x/2)}=
frac{(cos(x/2)-1)(2cos(x/2)+1)}{2sin(x/2)}
$$
which has limit $0$ for $xto0$.
answered 2 days ago
egreg
173k1383198
answered 2 days ago
egreg
173k1383198
answered 2 days ago
egreg
173k1383198
answered 2 days ago
egreg
173k1383198
173k1383198
add a comment |
add a comment |
up vote
0
down vote
$$frac{cos^3dfrac x2}{sin^2x}=frac{cosdfrac x2}{2sin^2dfrac x2}$$ is immediately integrable.
And in
$$frac{cos^4x}{sin^2x}=frac1{sin^2x}-2+sin^2x,$$ the first two terms are also immediate, and
$$sin^2x=frac{1-cos2x}2.$$
answered 2 days ago
Yves Daoust
121k668216
add a comment |
up vote
0
down vote
$$frac{cos^3dfrac x2}{sin^2x}=frac{cosdfrac x2}{2sin^2dfrac x2}$$ is immediately integrable.
And in
$$frac{cos^4x}{sin^2x}=frac1{sin^2x}-2+sin^2x,$$ the first two terms are also immediate, and
$$sin^2x=frac{1-cos2x}2.$$
answered 2 days ago
Yves Daoust
121k668216
add a comment |
up vote
0
down vote
up vote
0
down vote
$$frac{cos^3dfrac x2}{sin^2x}=frac{cosdfrac x2}{2sin^2dfrac x2}$$ is immediately integrable.
And in
$$frac{cos^4x}{sin^2x}=frac1{sin^2x}-2+sin^2x,$$ the first two terms are also immediate, and
$$sin^2x=frac{1-cos2x}2.$$
answered 2 days ago
Yves Daoust
121k668216
$$frac{cos^3dfrac x2}{sin^2x}=frac{cosdfrac x2}{2sin^2dfrac x2}$$ is immediately integrable.
And in
$$frac{cos^4x}{sin^2x}=frac1{sin^2x}-2+sin^2x,$$ the first two terms are also immediate, and
$$sin^2x=frac{1-cos2x}2.$$
answered 2 days ago
Yves Daoust
121k668216
answered 2 days ago
Yves Daoust
121k668216
answered 2 days ago
Yves Daoust
121k668216
answered 2 days ago
Yves Daoust
121k668216
121k668216
add a comment |
add a comment |
up vote
0
down vote
$$I=intfrac{cos^3(frac x2)-cos^4x}{sin^2x}mathrm{d}x$$
$$I=intfrac{cos^3(frac x2)}{sin^2x}mathrm{d}x-intfrac{cos^4x}{sin^2x}mathrm{d}x$$
$$I=big(I_1-I_2big)bigg|_0^{pi/2}$$
$$I_1=intfrac{cos^3(frac x2)}{sin^2x}mathrm{d}x$$
$u=x/2$:
$$I_1=2intfrac{cos^3u}{sin^22u}mathrm{d}u$$
$$I_1=2intfrac{cos^3u}{4sin^2ucos^2u}mathrm{d}u$$
$$I_1=frac12intfrac{cos u}{sin^2u}mathrm{d}u$$
$t=sin u$:
$$I_1=frac12intfrac{mathrm{d}t}{t^2}$$
$$I_1=-frac1{2t}$$
$$I_1=-frac1{2sin(x/2)}$$
$$I_2=intfrac{cos^4x}{sin^2x}mathrm{d}x$$
$$I_2=intfrac{cos^2x(1-sin^2x)}{sin^2x}mathrm{d}x$$
$$I_2=intcot^2x mathrm{d}x-intcos^2x mathrm{d}x$$
$$I_2=-bigg(cot x+frac12cos x,sin x+frac32xbigg)$$
$$I=bigg(-frac12csc(x/2)+cot x+frac12cos xsin x+frac32xbigg)bigg|_0^{pi/2}$$
$$I=frac{3pi-2sqrt{2}}4$$
answered 2 days ago
clathratus
1,839219
add a comment |
up vote
0
down vote
$$I=intfrac{cos^3(frac x2)-cos^4x}{sin^2x}mathrm{d}x$$
$$I=intfrac{cos^3(frac x2)}{sin^2x}mathrm{d}x-intfrac{cos^4x}{sin^2x}mathrm{d}x$$
$$I=big(I_1-I_2big)bigg|_0^{pi/2}$$
$$I_1=intfrac{cos^3(frac x2)}{sin^2x}mathrm{d}x$$
$u=x/2$:
$$I_1=2intfrac{cos^3u}{sin^22u}mathrm{d}u$$
$$I_1=2intfrac{cos^3u}{4sin^2ucos^2u}mathrm{d}u$$
$$I_1=frac12intfrac{cos u}{sin^2u}mathrm{d}u$$
$t=sin u$:
$$I_1=frac12intfrac{mathrm{d}t}{t^2}$$
$$I_1=-frac1{2t}$$
$$I_1=-frac1{2sin(x/2)}$$
$$I_2=intfrac{cos^4x}{sin^2x}mathrm{d}x$$
$$I_2=intfrac{cos^2x(1-sin^2x)}{sin^2x}mathrm{d}x$$
$$I_2=intcot^2x mathrm{d}x-intcos^2x mathrm{d}x$$
$$I_2=-bigg(cot x+frac12cos x,sin x+frac32xbigg)$$
$$I=bigg(-frac12csc(x/2)+cot x+frac12cos xsin x+frac32xbigg)bigg|_0^{pi/2}$$
$$I=frac{3pi-2sqrt{2}}4$$
answered 2 days ago
clathratus
1,839219
add a comment |
up vote
0
down vote
up vote
0
down vote
$$I=intfrac{cos^3(frac x2)-cos^4x}{sin^2x}mathrm{d}x$$
$$I=intfrac{cos^3(frac x2)}{sin^2x}mathrm{d}x-intfrac{cos^4x}{sin^2x}mathrm{d}x$$
$$I=big(I_1-I_2big)bigg|_0^{pi/2}$$
$$I_1=intfrac{cos^3(frac x2)}{sin^2x}mathrm{d}x$$
$u=x/2$:
$$I_1=2intfrac{cos^3u}{sin^22u}mathrm{d}u$$
$$I_1=2intfrac{cos^3u}{4sin^2ucos^2u}mathrm{d}u$$
$$I_1=frac12intfrac{cos u}{sin^2u}mathrm{d}u$$
$t=sin u$:
$$I_1=frac12intfrac{mathrm{d}t}{t^2}$$
$$I_1=-frac1{2t}$$
$$I_1=-frac1{2sin(x/2)}$$
$$I_2=intfrac{cos^4x}{sin^2x}mathrm{d}x$$
$$I_2=intfrac{cos^2x(1-sin^2x)}{sin^2x}mathrm{d}x$$
$$I_2=intcot^2x mathrm{d}x-intcos^2x mathrm{d}x$$
$$I_2=-bigg(cot x+frac12cos x,sin x+frac32xbigg)$$
$$I=bigg(-frac12csc(x/2)+cot x+frac12cos xsin x+frac32xbigg)bigg|_0^{pi/2}$$
$$I=frac{3pi-2sqrt{2}}4$$
answered 2 days ago
clathratus
1,839219
$$I=intfrac{cos^3(frac x2)-cos^4x}{sin^2x}mathrm{d}x$$
$$I=intfrac{cos^3(frac x2)}{sin^2x}mathrm{d}x-intfrac{cos^4x}{sin^2x}mathrm{d}x$$
$$I=big(I_1-I_2big)bigg|_0^{pi/2}$$
$$I_1=intfrac{cos^3(frac x2)}{sin^2x}mathrm{d}x$$
$u=x/2$:
$$I_1=2intfrac{cos^3u}{sin^22u}mathrm{d}u$$
$$I_1=2intfrac{cos^3u}{4sin^2ucos^2u}mathrm{d}u$$
$$I_1=frac12intfrac{cos u}{sin^2u}mathrm{d}u$$
$t=sin u$:
$$I_1=frac12intfrac{mathrm{d}t}{t^2}$$
$$I_1=-frac1{2t}$$
$$I_1=-frac1{2sin(x/2)}$$
$$I_2=intfrac{cos^4x}{sin^2x}mathrm{d}x$$
$$I_2=intfrac{cos^2x(1-sin^2x)}{sin^2x}mathrm{d}x$$
$$I_2=intcot^2x mathrm{d}x-intcos^2x mathrm{d}x$$
$$I_2=-bigg(cot x+frac12cos x,sin x+frac32xbigg)$$
$$I=bigg(-frac12csc(x/2)+cot x+frac12cos xsin x+frac32xbigg)bigg|_0^{pi/2}$$
$$I=frac{3pi-2sqrt{2}}4$$
answered 2 days ago
clathratus
1,839219
answered 2 days ago
clathratus
1,839219
answered 2 days ago
clathratus
1,839219
answered 2 days ago
clathratus
1,839219
1,839219
add a comment |
add a comment |
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Where did you see this integral?
– Frpzzd
2 days ago
Comes up in my work. I like the Weierstrass trick as it is a blanket method, but am looking for a better way to approach it
– Gordon
2 days ago