Mixing
Primitive of a function with $sin frac{1}{x}$
$begingroup$
I have the next integral:
$$intbiggl({frac{sin frac{1}{x}}{x^2sqrt{(4+3 sinfrac{2}{x})}}}biggr),dx ,;xin Bigl(0,inftyBigr)$$
I used the substitution $u=frac{1}{x}$ and I got $$-intbiggl({frac{sin u}{sqrt{(4+3 sin2u)}}}biggr),du$$
Can somebody give me some tips about what should I do next, please?
real-analysis calculus trigonometry indefinite-integrals substitution
edited Jan 16 at 21:24
Bernard
121k740116
asked Jan 16 at 20:19
GaboruGaboru
1047
$endgroup$
add a comment |
$begingroup$
I have the next integral:
$$intbiggl({frac{sin frac{1}{x}}{x^2sqrt{(4+3 sinfrac{2}{x})}}}biggr),dx ,;xin Bigl(0,inftyBigr)$$
I used the substitution $u=frac{1}{x}$ and I got $$-intbiggl({frac{sin u}{sqrt{(4+3 sin2u)}}}biggr),du$$
Can somebody give me some tips about what should I do next, please?
real-analysis calculus trigonometry indefinite-integrals substitution
edited Jan 16 at 21:24
Bernard
121k740116
asked Jan 16 at 20:19
GaboruGaboru
1047
$endgroup$
add a comment |
$begingroup$
I have the next integral:
$$intbiggl({frac{sin frac{1}{x}}{x^2sqrt{(4+3 sinfrac{2}{x})}}}biggr),dx ,;xin Bigl(0,inftyBigr)$$
I used the substitution $u=frac{1}{x}$ and I got $$-intbiggl({frac{sin u}{sqrt{(4+3 sin2u)}}}biggr),du$$
Can somebody give me some tips about what should I do next, please?
real-analysis calculus trigonometry indefinite-integrals substitution
edited Jan 16 at 21:24
Bernard
121k740116
asked Jan 16 at 20:19
GaboruGaboru
1047
$endgroup$
I have the next integral:
$$intbiggl({frac{sin frac{1}{x}}{x^2sqrt{(4+3 sinfrac{2}{x})}}}biggr),dx ,;xin Bigl(0,inftyBigr)$$
I used the substitution $u=frac{1}{x}$ and I got $$-intbiggl({frac{sin u}{sqrt{(4+3 sin2u)}}}biggr),du$$
Can somebody give me some tips about what should I do next, please?
real-analysis calculus trigonometry indefinite-integrals substitution
real-analysis calculus trigonometry indefinite-integrals substitution
edited Jan 16 at 21:24
Bernard
121k740116
asked Jan 16 at 20:19
GaboruGaboru
1047
edited Jan 16 at 21:24
Bernard
121k740116
asked Jan 16 at 20:19
GaboruGaboru
1047
edited Jan 16 at 21:24
Bernard
121k740116
edited Jan 16 at 21:24
Bernard
121k740116
edited Jan 16 at 21:24
Bernard
121k740116
121k740116
asked Jan 16 at 20:19
GaboruGaboru
1047
asked Jan 16 at 20:19
GaboruGaboru
1047
asked Jan 16 at 20:19
GaboruGaboru
1047
1047
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The substitution $u=1/x$ yields $dx=-frac{1}{u^2},du$, so the integral becomes
$$
intfrac{-sin u}{sqrt{4+3sin2u}},du=
intfrac{-sin u}{sqrt{4+3sin2u}},du
$$
This can be improved by setting $u=pi/4-v$, so we get
$$
frac{1}{sqrt{2}}intfrac{cos v-sin v}{sqrt{4+3cos2v}},dv=
frac{1}{sqrt{2}}biggl(
intfrac{cos v}{sqrt{7-6sin^2v}},dv
-intfrac{sin v}{sqrt{6cos^2v+1}},dv
biggr)
$$
that you should be able to manage.
answered Jan 16 at 21:14
egregegreg
182k1486204
$endgroup$
add a comment |
$begingroup$
As $(sin vpmcos v)^2=1pmsin2v$
$$intdfrac{2sin v dv}{f(sin2v)}=intdfrac{(sin v-cos v) dv}{f(sin2v)}+intdfrac{(sin v+cos v) dv}{f(sin2v)}=I+J$$ where $f(sin2v)$ is a function of $sin2v$
As $displaystyleint(sin v-cos v)=-(sin v+cos v)+C,$ set $sin v+cos v=y$ for
$$I=intdfrac{(sin v-cos v) dv}{f((sin v+cos v)^2-1)}$$
Similarly, set $sin v-cos v=z$ for $$J=intdfrac{(sin v+cos v) dv}{f(1-(sin v-cos v)^2)}$$
answered Jan 17 at 7:55
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
The substitution $u=1/x$ yields $dx=-frac{1}{u^2},du$, so the integral becomes
$$
intfrac{-sin u}{sqrt{4+3sin2u}},du=
intfrac{-sin u}{sqrt{4+3sin2u}},du
$$
This can be improved by setting $u=pi/4-v$, so we get
$$
frac{1}{sqrt{2}}intfrac{cos v-sin v}{sqrt{4+3cos2v}},dv=
frac{1}{sqrt{2}}biggl(
intfrac{cos v}{sqrt{7-6sin^2v}},dv
-intfrac{sin v}{sqrt{6cos^2v+1}},dv
biggr)
$$
that you should be able to manage.
answered Jan 16 at 21:14
egregegreg
182k1486204
$endgroup$
add a comment |
$begingroup$
The substitution $u=1/x$ yields $dx=-frac{1}{u^2},du$, so the integral becomes
$$
intfrac{-sin u}{sqrt{4+3sin2u}},du=
intfrac{-sin u}{sqrt{4+3sin2u}},du
$$
This can be improved by setting $u=pi/4-v$, so we get
$$
frac{1}{sqrt{2}}intfrac{cos v-sin v}{sqrt{4+3cos2v}},dv=
frac{1}{sqrt{2}}biggl(
intfrac{cos v}{sqrt{7-6sin^2v}},dv
-intfrac{sin v}{sqrt{6cos^2v+1}},dv
biggr)
$$
that you should be able to manage.
answered Jan 16 at 21:14
egregegreg
182k1486204
$endgroup$
add a comment |
$begingroup$
The substitution $u=1/x$ yields $dx=-frac{1}{u^2},du$, so the integral becomes
$$
intfrac{-sin u}{sqrt{4+3sin2u}},du=
intfrac{-sin u}{sqrt{4+3sin2u}},du
$$
This can be improved by setting $u=pi/4-v$, so we get
$$
frac{1}{sqrt{2}}intfrac{cos v-sin v}{sqrt{4+3cos2v}},dv=
frac{1}{sqrt{2}}biggl(
intfrac{cos v}{sqrt{7-6sin^2v}},dv
-intfrac{sin v}{sqrt{6cos^2v+1}},dv
biggr)
$$
that you should be able to manage.
answered Jan 16 at 21:14
egregegreg
182k1486204
$endgroup$
The substitution $u=1/x$ yields $dx=-frac{1}{u^2},du$, so the integral becomes
$$
intfrac{-sin u}{sqrt{4+3sin2u}},du=
intfrac{-sin u}{sqrt{4+3sin2u}},du
$$
This can be improved by setting $u=pi/4-v$, so we get
$$
frac{1}{sqrt{2}}intfrac{cos v-sin v}{sqrt{4+3cos2v}},dv=
frac{1}{sqrt{2}}biggl(
intfrac{cos v}{sqrt{7-6sin^2v}},dv
-intfrac{sin v}{sqrt{6cos^2v+1}},dv
biggr)
$$
that you should be able to manage.
answered Jan 16 at 21:14
egregegreg
182k1486204
answered Jan 16 at 21:14
egregegreg
182k1486204
answered Jan 16 at 21:14
egregegreg
182k1486204
answered Jan 16 at 21:14
egregegreg
182k1486204
182k1486204
add a comment |
add a comment |
$begingroup$
As $(sin vpmcos v)^2=1pmsin2v$
$$intdfrac{2sin v dv}{f(sin2v)}=intdfrac{(sin v-cos v) dv}{f(sin2v)}+intdfrac{(sin v+cos v) dv}{f(sin2v)}=I+J$$ where $f(sin2v)$ is a function of $sin2v$
As $displaystyleint(sin v-cos v)=-(sin v+cos v)+C,$ set $sin v+cos v=y$ for
$$I=intdfrac{(sin v-cos v) dv}{f((sin v+cos v)^2-1)}$$
Similarly, set $sin v-cos v=z$ for $$J=intdfrac{(sin v+cos v) dv}{f(1-(sin v-cos v)^2)}$$
answered Jan 17 at 7:55
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
add a comment |
$begingroup$
As $(sin vpmcos v)^2=1pmsin2v$
$$intdfrac{2sin v dv}{f(sin2v)}=intdfrac{(sin v-cos v) dv}{f(sin2v)}+intdfrac{(sin v+cos v) dv}{f(sin2v)}=I+J$$ where $f(sin2v)$ is a function of $sin2v$
As $displaystyleint(sin v-cos v)=-(sin v+cos v)+C,$ set $sin v+cos v=y$ for
$$I=intdfrac{(sin v-cos v) dv}{f((sin v+cos v)^2-1)}$$
Similarly, set $sin v-cos v=z$ for $$J=intdfrac{(sin v+cos v) dv}{f(1-(sin v-cos v)^2)}$$
answered Jan 17 at 7:55
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
add a comment |
$begingroup$
As $(sin vpmcos v)^2=1pmsin2v$
$$intdfrac{2sin v dv}{f(sin2v)}=intdfrac{(sin v-cos v) dv}{f(sin2v)}+intdfrac{(sin v+cos v) dv}{f(sin2v)}=I+J$$ where $f(sin2v)$ is a function of $sin2v$
As $displaystyleint(sin v-cos v)=-(sin v+cos v)+C,$ set $sin v+cos v=y$ for
$$I=intdfrac{(sin v-cos v) dv}{f((sin v+cos v)^2-1)}$$
Similarly, set $sin v-cos v=z$ for $$J=intdfrac{(sin v+cos v) dv}{f(1-(sin v-cos v)^2)}$$
answered Jan 17 at 7:55
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
As $(sin vpmcos v)^2=1pmsin2v$
$$intdfrac{2sin v dv}{f(sin2v)}=intdfrac{(sin v-cos v) dv}{f(sin2v)}+intdfrac{(sin v+cos v) dv}{f(sin2v)}=I+J$$ where $f(sin2v)$ is a function of $sin2v$
As $displaystyleint(sin v-cos v)=-(sin v+cos v)+C,$ set $sin v+cos v=y$ for
$$I=intdfrac{(sin v-cos v) dv}{f((sin v+cos v)^2-1)}$$
Similarly, set $sin v-cos v=z$ for $$J=intdfrac{(sin v+cos v) dv}{f(1-(sin v-cos v)^2)}$$
answered Jan 17 at 7:55
lab bhattacharjeelab bhattacharjee
226k15157275
answered Jan 17 at 7:55
lab bhattacharjeelab bhattacharjee
226k15157275
answered Jan 17 at 7:55
lab bhattacharjeelab bhattacharjee
226k15157275
answered Jan 17 at 7:55
lab bhattacharjeelab bhattacharjee
226k15157275
226k15157275
add a comment |
add a comment |
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