How this indefinite integral can be performed? [closed]

-1

$$
intfrac{Asin(x) - cos(x)}{sin(x) - A cos(x)},dx
$$

How can we check the existence of this integral?

Can this function be integrated using substitution? If yes, then what substitution should be made?

edited Dec 26 '18 at 17:53

egreg

186k1486208

asked Dec 26 '18 at 16:20

Vibin Narayanan

164

closed as off-topic by RRL, Lord_Farin, Saad, user91500, José Carlos Santos Dec 27 '18 at 12:04

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Lord_Farin, Saad, user91500, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment |

-1

$$
intfrac{Asin(x) - cos(x)}{sin(x) - A cos(x)},dx
$$

How can we check the existence of this integral?

Can this function be integrated using substitution? If yes, then what substitution should be made?

edited Dec 26 '18 at 17:53

egreg

186k1486208

asked Dec 26 '18 at 16:20

Vibin Narayanan

164

closed as off-topic by RRL, Lord_Farin, Saad, user91500, José Carlos Santos Dec 27 '18 at 12:04

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Lord_Farin, Saad, user91500, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment |

-1

$$
intfrac{Asin(x) - cos(x)}{sin(x) - A cos(x)},dx
$$

How can we check the existence of this integral?

Can this function be integrated using substitution? If yes, then what substitution should be made?

edited Dec 26 '18 at 17:53

egreg

186k1486208

asked Dec 26 '18 at 16:20

Vibin Narayanan

164

$$
intfrac{Asin(x) - cos(x)}{sin(x) - A cos(x)},dx
$$

How can we check the existence of this integral?

Can this function be integrated using substitution? If yes, then what substitution should be made?

calculus

edited Dec 26 '18 at 17:53

egreg

186k1486208

asked Dec 26 '18 at 16:20

Vibin Narayanan

164

edited Dec 26 '18 at 17:53

egreg

186k1486208

asked Dec 26 '18 at 16:20

Vibin Narayanan

164

edited Dec 26 '18 at 17:53

egreg

186k1486208

edited Dec 26 '18 at 17:53

egreg

186k1486208

edited Dec 26 '18 at 17:53

egreg

186k1486208

asked Dec 26 '18 at 16:20

Vibin Narayanan

164

asked Dec 26 '18 at 16:20

Vibin Narayanan

164

asked Dec 26 '18 at 16:20

Vibin Narayanan

164

closed as off-topic by RRL, Lord_Farin, Saad, user91500, José Carlos Santos Dec 27 '18 at 12:04

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Lord_Farin, Saad, user91500, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by RRL, Lord_Farin, Saad, user91500, José Carlos Santos Dec 27 '18 at 12:04

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Lord_Farin, Saad, user91500, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment |

2 Answers
2

active

oldest

votes

Or, if for some constant $a,b$, we can express

$cfrac {A sin(x) - cos(x)} {sin(x) - A cos(x)}$

$cfrac {a( sin(x) - A cos(x))-b(cos(x)+Asin(x))} {sin(x) - A cos(x)}$

$=a-b cfrac {mathrm{d}(sin(x)-Acos(x))/mathrm{d}x} {sin(x) - A cos(x)}$

Then, it can be easily separated and integrated.

we have

$a-bA=A$

$aA+b=1$

$implies a=cfrac {2A}{1+A^2}, b=cfrac {1-A^2}{1+A^2}$

hence $displaystyle int cfrac {A sin(x) - cos(x)} {sin(x) - A cos(x)} mathrm{d} x=cfrac {2A}{1+A^2} x-cfrac {1-A^2}{1+A^2}ln left|sin x-Acos xright|+C$

edited Dec 26 '18 at 17:38

answered Dec 26 '18 at 16:32

Lance

64239

add a comment |

Hint: Use $$sin(x)=frac{2t}{1+t^2}$$
$$cos(x)=frac{1-t^2}{1+t^2}$$
$$dx=frac{2dt}{1+t^2}$$
the so-called Weierstrass substitution.

answered Dec 26 '18 at 16:25

Dr. Sonnhard Graubner

78.9k42867

add a comment |

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

Or, if for some constant $a,b$, we can express

$cfrac {A sin(x) - cos(x)} {sin(x) - A cos(x)}$

$cfrac {a( sin(x) - A cos(x))-b(cos(x)+Asin(x))} {sin(x) - A cos(x)}$

$=a-b cfrac {mathrm{d}(sin(x)-Acos(x))/mathrm{d}x} {sin(x) - A cos(x)}$

Then, it can be easily separated and integrated.

we have

$a-bA=A$

$aA+b=1$

$implies a=cfrac {2A}{1+A^2}, b=cfrac {1-A^2}{1+A^2}$

hence $displaystyle int cfrac {A sin(x) - cos(x)} {sin(x) - A cos(x)} mathrm{d} x=cfrac {2A}{1+A^2} x-cfrac {1-A^2}{1+A^2}ln left|sin x-Acos xright|+C$

edited Dec 26 '18 at 17:38

answered Dec 26 '18 at 16:32

Lance

64239

add a comment |

Or, if for some constant $a,b$, we can express

$cfrac {A sin(x) - cos(x)} {sin(x) - A cos(x)}$

$cfrac {a( sin(x) - A cos(x))-b(cos(x)+Asin(x))} {sin(x) - A cos(x)}$

$=a-b cfrac {mathrm{d}(sin(x)-Acos(x))/mathrm{d}x} {sin(x) - A cos(x)}$

Then, it can be easily separated and integrated.

we have

$a-bA=A$

$aA+b=1$

$implies a=cfrac {2A}{1+A^2}, b=cfrac {1-A^2}{1+A^2}$

hence $displaystyle int cfrac {A sin(x) - cos(x)} {sin(x) - A cos(x)} mathrm{d} x=cfrac {2A}{1+A^2} x-cfrac {1-A^2}{1+A^2}ln left|sin x-Acos xright|+C$

edited Dec 26 '18 at 17:38

answered Dec 26 '18 at 16:32

Lance

64239

add a comment |

Or, if for some constant $a,b$, we can express

$cfrac {A sin(x) - cos(x)} {sin(x) - A cos(x)}$

$cfrac {a( sin(x) - A cos(x))-b(cos(x)+Asin(x))} {sin(x) - A cos(x)}$

$=a-b cfrac {mathrm{d}(sin(x)-Acos(x))/mathrm{d}x} {sin(x) - A cos(x)}$

Then, it can be easily separated and integrated.

we have

$a-bA=A$

$aA+b=1$

$implies a=cfrac {2A}{1+A^2}, b=cfrac {1-A^2}{1+A^2}$

hence $displaystyle int cfrac {A sin(x) - cos(x)} {sin(x) - A cos(x)} mathrm{d} x=cfrac {2A}{1+A^2} x-cfrac {1-A^2}{1+A^2}ln left|sin x-Acos xright|+C$

edited Dec 26 '18 at 17:38

answered Dec 26 '18 at 16:32

Lance

64239

Or, if for some constant $a,b$, we can express

$cfrac {A sin(x) - cos(x)} {sin(x) - A cos(x)}$

$cfrac {a( sin(x) - A cos(x))-b(cos(x)+Asin(x))} {sin(x) - A cos(x)}$

$=a-b cfrac {mathrm{d}(sin(x)-Acos(x))/mathrm{d}x} {sin(x) - A cos(x)}$

Then, it can be easily separated and integrated.

we have

$a-bA=A$

$aA+b=1$

$implies a=cfrac {2A}{1+A^2}, b=cfrac {1-A^2}{1+A^2}$

hence $displaystyle int cfrac {A sin(x) - cos(x)} {sin(x) - A cos(x)} mathrm{d} x=cfrac {2A}{1+A^2} x-cfrac {1-A^2}{1+A^2}ln left|sin x-Acos xright|+C$

edited Dec 26 '18 at 17:38

answered Dec 26 '18 at 16:32

Lance

64239

edited Dec 26 '18 at 17:38

answered Dec 26 '18 at 16:32

Lance

64239

answered Dec 26 '18 at 16:32

Lance

64239

answered Dec 26 '18 at 16:32

Lance

64239

add a comment |

Hint: Use $$sin(x)=frac{2t}{1+t^2}$$
$$cos(x)=frac{1-t^2}{1+t^2}$$
$$dx=frac{2dt}{1+t^2}$$
the so-called Weierstrass substitution.

answered Dec 26 '18 at 16:25

Dr. Sonnhard Graubner

78.9k42867

add a comment |

Hint: Use $$sin(x)=frac{2t}{1+t^2}$$
$$cos(x)=frac{1-t^2}{1+t^2}$$
$$dx=frac{2dt}{1+t^2}$$
the so-called Weierstrass substitution.

answered Dec 26 '18 at 16:25

Dr. Sonnhard Graubner

78.9k42867

add a comment |

Hint: Use $$sin(x)=frac{2t}{1+t^2}$$
$$cos(x)=frac{1-t^2}{1+t^2}$$
$$dx=frac{2dt}{1+t^2}$$
the so-called Weierstrass substitution.

answered Dec 26 '18 at 16:25

Dr. Sonnhard Graubner

78.9k42867

Hint: Use $$sin(x)=frac{2t}{1+t^2}$$
$$cos(x)=frac{1-t^2}{1+t^2}$$
$$dx=frac{2dt}{1+t^2}$$
the so-called Weierstrass substitution.

answered Dec 26 '18 at 16:25

Dr. Sonnhard Graubner

78.9k42867

answered Dec 26 '18 at 16:25

Dr. Sonnhard Graubner

78.9k42867

answered Dec 26 '18 at 16:25

Dr. Sonnhard Graubner

78.9k42867

answered Dec 26 '18 at 16:25

Dr. Sonnhard Graubner

78.9k42867

add a comment |

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