Evaluating $int frac {3cos x}{(2sin x-5cos x)},dx$
up vote
0
down vote
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$$int dfrac {3cos x}{(2sin x-5cos x)},dx$$
I've been thinking and trying to work this out in quite a few ways:
1)Taking conjugate which actually complicates it further
2)Using half angle formula to convert the expression in terms of tan but I get 2 terms in the integration of which I'm unable to integrate 1 term because I can't seem to be able to make a suitable substitution.
3)Dividing throughout by $cos^2 x$
I find none of these methods to be effective. Please advice.
calculus integration
edited Nov 22 at 14:58
StubbornAtom
5,06011138
asked Nov 22 at 14:48
MadCap
213
add a comment |
up vote
0
down vote
favorite
$$int dfrac {3cos x}{(2sin x-5cos x)},dx$$
I've been thinking and trying to work this out in quite a few ways:
1)Taking conjugate which actually complicates it further
2)Using half angle formula to convert the expression in terms of tan but I get 2 terms in the integration of which I'm unable to integrate 1 term because I can't seem to be able to make a suitable substitution.
3)Dividing throughout by $cos^2 x$
I find none of these methods to be effective. Please advice.
calculus integration
edited Nov 22 at 14:58
StubbornAtom
5,06011138
asked Nov 22 at 14:48
MadCap
213
math.stackexchange.com/questions/3006097/…
– lab bhattacharjee
Nov 22 at 14:55
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$$int dfrac {3cos x}{(2sin x-5cos x)},dx$$
I've been thinking and trying to work this out in quite a few ways:
1)Taking conjugate which actually complicates it further
2)Using half angle formula to convert the expression in terms of tan but I get 2 terms in the integration of which I'm unable to integrate 1 term because I can't seem to be able to make a suitable substitution.
3)Dividing throughout by $cos^2 x$
I find none of these methods to be effective. Please advice.
calculus integration
edited Nov 22 at 14:58
StubbornAtom
5,06011138
asked Nov 22 at 14:48
MadCap
213
$$int dfrac {3cos x}{(2sin x-5cos x)},dx$$
I've been thinking and trying to work this out in quite a few ways:
1)Taking conjugate which actually complicates it further
2)Using half angle formula to convert the expression in terms of tan but I get 2 terms in the integration of which I'm unable to integrate 1 term because I can't seem to be able to make a suitable substitution.
3)Dividing throughout by $cos^2 x$
I find none of these methods to be effective. Please advice.
calculus integration
calculus integration
edited Nov 22 at 14:58
StubbornAtom
5,06011138
asked Nov 22 at 14:48
MadCap
213
edited Nov 22 at 14:58
StubbornAtom
5,06011138
asked Nov 22 at 14:48
MadCap
213
edited Nov 22 at 14:58
StubbornAtom
5,06011138
edited Nov 22 at 14:58
StubbornAtom
5,06011138
edited Nov 22 at 14:58
StubbornAtom
5,06011138
5,06011138
asked Nov 22 at 14:48
MadCap
213
asked Nov 22 at 14:48
MadCap
213
asked Nov 22 at 14:48
MadCap
213
213
math.stackexchange.com/questions/3006097/…
– lab bhattacharjee
Nov 22 at 14:55
add a comment |
math.stackexchange.com/questions/3006097/…
– lab bhattacharjee
Nov 22 at 14:55
math.stackexchange.com/questions/3006097/…
– lab bhattacharjee
Nov 22 at 14:55
math.stackexchange.com/questions/3006097/…
– lab bhattacharjee
Nov 22 at 14:55
add a comment |
3 Answers
3
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3
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Hint. A standard trick is to consider
$$
I=intfrac{sin x}{sin x-5cos x}dx,quad J=intfrac{cos x}{sin x-5cos x}dx
$$ then one may observe that
$$
I-5J=int dx=color{red}?,qquad 5I+J=intfrac{(sin x-5cos x)'}{sin x-5cos x}dx=color{red}?
$$Hope you can take it from here.
answered Nov 22 at 14:55
Olivier Oloa
107k17175293
add a comment |
up vote
1
down vote
Use:
$3 cos x = lambda (2sin x - 5cos x) + mu(2cos x + 5 sin x) $
Putting $x = 0$ and $x = pi/2 $ and solving the two equations we get:
$lambda = -dfrac {15}{29} $ and $mu = dfrac 6{29}$
So we have:
$$int frac {3cos x}{(2sin x-5cos x)},dx
\ = int dfrac{lambda(2sin x - 5 cos x)+ mu(2cos x + 5 sin x)}{2sin x - 5 cos x} \ = int lambda dx + muint dfrac{2cos x+ 5sin x}{2 sin x - 5 cos x}dx \ =lambda x + mu ln (2sin x - 5cos x) + c = -dfrac{15}{29} x + dfrac{6}{29} ln (2sin x - 5cos x) + c$$
answered Nov 22 at 14:56
Abcd
2,93221131
add a comment |
up vote
0
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Hint:
By shifting the argument, you can turn the denominator to $sqrt{29}cos x$. Then the numerator becomes of the form $acos x+bsin x$, and the fraction is easy to integrate.
answered Nov 22 at 15:04
Yves Daoust
123k668219
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Hint. A standard trick is to consider
$$
I=intfrac{sin x}{sin x-5cos x}dx,quad J=intfrac{cos x}{sin x-5cos x}dx
$$ then one may observe that
$$
I-5J=int dx=color{red}?,qquad 5I+J=intfrac{(sin x-5cos x)'}{sin x-5cos x}dx=color{red}?
$$Hope you can take it from here.
answered Nov 22 at 14:55
Olivier Oloa
107k17175293
add a comment |
up vote
3
down vote
Hint. A standard trick is to consider
$$
I=intfrac{sin x}{sin x-5cos x}dx,quad J=intfrac{cos x}{sin x-5cos x}dx
$$ then one may observe that
$$
I-5J=int dx=color{red}?,qquad 5I+J=intfrac{(sin x-5cos x)'}{sin x-5cos x}dx=color{red}?
$$Hope you can take it from here.
answered Nov 22 at 14:55
Olivier Oloa
107k17175293
add a comment |
up vote
3
down vote
up vote
3
down vote
Hint. A standard trick is to consider
$$
I=intfrac{sin x}{sin x-5cos x}dx,quad J=intfrac{cos x}{sin x-5cos x}dx
$$ then one may observe that
$$
I-5J=int dx=color{red}?,qquad 5I+J=intfrac{(sin x-5cos x)'}{sin x-5cos x}dx=color{red}?
$$Hope you can take it from here.
answered Nov 22 at 14:55
Olivier Oloa
107k17175293
Hint. A standard trick is to consider
$$
I=intfrac{sin x}{sin x-5cos x}dx,quad J=intfrac{cos x}{sin x-5cos x}dx
$$ then one may observe that
$$
I-5J=int dx=color{red}?,qquad 5I+J=intfrac{(sin x-5cos x)'}{sin x-5cos x}dx=color{red}?
$$Hope you can take it from here.
answered Nov 22 at 14:55
Olivier Oloa
107k17175293
answered Nov 22 at 14:55
Olivier Oloa
107k17175293
answered Nov 22 at 14:55
Olivier Oloa
107k17175293
answered Nov 22 at 14:55
Olivier Oloa
107k17175293
107k17175293
add a comment |
add a comment |
up vote
1
down vote
Use:
$3 cos x = lambda (2sin x - 5cos x) + mu(2cos x + 5 sin x) $
Putting $x = 0$ and $x = pi/2 $ and solving the two equations we get:
$lambda = -dfrac {15}{29} $ and $mu = dfrac 6{29}$
So we have:
$$int frac {3cos x}{(2sin x-5cos x)},dx
\ = int dfrac{lambda(2sin x - 5 cos x)+ mu(2cos x + 5 sin x)}{2sin x - 5 cos x} \ = int lambda dx + muint dfrac{2cos x+ 5sin x}{2 sin x - 5 cos x}dx \ =lambda x + mu ln (2sin x - 5cos x) + c = -dfrac{15}{29} x + dfrac{6}{29} ln (2sin x - 5cos x) + c$$
answered Nov 22 at 14:56
Abcd
2,93221131
add a comment |
up vote
1
down vote
Use:
$3 cos x = lambda (2sin x - 5cos x) + mu(2cos x + 5 sin x) $
Putting $x = 0$ and $x = pi/2 $ and solving the two equations we get:
$lambda = -dfrac {15}{29} $ and $mu = dfrac 6{29}$
So we have:
$$int frac {3cos x}{(2sin x-5cos x)},dx
\ = int dfrac{lambda(2sin x - 5 cos x)+ mu(2cos x + 5 sin x)}{2sin x - 5 cos x} \ = int lambda dx + muint dfrac{2cos x+ 5sin x}{2 sin x - 5 cos x}dx \ =lambda x + mu ln (2sin x - 5cos x) + c = -dfrac{15}{29} x + dfrac{6}{29} ln (2sin x - 5cos x) + c$$
answered Nov 22 at 14:56
Abcd
2,93221131
add a comment |
up vote
1
down vote
up vote
1
down vote
Use:
$3 cos x = lambda (2sin x - 5cos x) + mu(2cos x + 5 sin x) $
Putting $x = 0$ and $x = pi/2 $ and solving the two equations we get:
$lambda = -dfrac {15}{29} $ and $mu = dfrac 6{29}$
So we have:
$$int frac {3cos x}{(2sin x-5cos x)},dx
\ = int dfrac{lambda(2sin x - 5 cos x)+ mu(2cos x + 5 sin x)}{2sin x - 5 cos x} \ = int lambda dx + muint dfrac{2cos x+ 5sin x}{2 sin x - 5 cos x}dx \ =lambda x + mu ln (2sin x - 5cos x) + c = -dfrac{15}{29} x + dfrac{6}{29} ln (2sin x - 5cos x) + c$$
answered Nov 22 at 14:56
Abcd
2,93221131
Use:
$3 cos x = lambda (2sin x - 5cos x) + mu(2cos x + 5 sin x) $
Putting $x = 0$ and $x = pi/2 $ and solving the two equations we get:
$lambda = -dfrac {15}{29} $ and $mu = dfrac 6{29}$
So we have:
$$int frac {3cos x}{(2sin x-5cos x)},dx
\ = int dfrac{lambda(2sin x - 5 cos x)+ mu(2cos x + 5 sin x)}{2sin x - 5 cos x} \ = int lambda dx + muint dfrac{2cos x+ 5sin x}{2 sin x - 5 cos x}dx \ =lambda x + mu ln (2sin x - 5cos x) + c = -dfrac{15}{29} x + dfrac{6}{29} ln (2sin x - 5cos x) + c$$
answered Nov 22 at 14:56
Abcd
2,93221131
edited Nov 22 at 15:07
answered Nov 22 at 14:56
Abcd
2,93221131
answered Nov 22 at 14:56
Abcd
2,93221131
answered Nov 22 at 14:56
Abcd
2,93221131
2,93221131
add a comment |
add a comment |
up vote
0
down vote
Hint:
By shifting the argument, you can turn the denominator to $sqrt{29}cos x$. Then the numerator becomes of the form $acos x+bsin x$, and the fraction is easy to integrate.
answered Nov 22 at 15:04
Yves Daoust
123k668219
add a comment |
up vote
0
down vote
Hint:
By shifting the argument, you can turn the denominator to $sqrt{29}cos x$. Then the numerator becomes of the form $acos x+bsin x$, and the fraction is easy to integrate.
answered Nov 22 at 15:04
Yves Daoust
123k668219
add a comment |
up vote
0
down vote
up vote
0
down vote
Hint:
By shifting the argument, you can turn the denominator to $sqrt{29}cos x$. Then the numerator becomes of the form $acos x+bsin x$, and the fraction is easy to integrate.
answered Nov 22 at 15:04
Yves Daoust
123k668219
Hint:
By shifting the argument, you can turn the denominator to $sqrt{29}cos x$. Then the numerator becomes of the form $acos x+bsin x$, and the fraction is easy to integrate.
answered Nov 22 at 15:04
Yves Daoust
123k668219
answered Nov 22 at 15:04
Yves Daoust
123k668219
answered Nov 22 at 15:04
Yves Daoust
123k668219
answered Nov 22 at 15:04
Yves Daoust
123k668219
123k668219
add a comment |
add a comment |
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math.stackexchange.com/questions/3006097/…
– lab bhattacharjee
Nov 22 at 14:55