Integral $int frac{cos(x)+sin(2x)}{sin(x)}text{ d}x$












3














I would like to evaluate
$$int dfrac{cos(x)+sin(2x)}{sin(x)}text{ d}xtext{.}$$



This is from Stewart's Calculus text, section 7.2., #19. Please note that I don't have a solutions manual and I am not interested in a complete solution.



I would merely like some guidance on a first step on approaching this one, since I've gotten 9 similar problems correct, and I hit a brick wall on this one. My first thought was perhaps splitting the fraction like so:
$$intcot(x)text{ d}x + intdfrac{sin(2x)}{sin(x)}text{ d}x$$
but this does not look helpful.






calculus integration indefinite-integrals trigonometric-integrals







share|cite|improve this question




















  • 4




    hint : $sin (2x) = 2sin( x) cos (x)$
    – ganeshie8
    Oct 17 '14 at 4:03










  • A tip for solving integrals that have sin or cosine or some trigonometric function is use a bunch of identities
    – Joao
    Oct 17 '14 at 4:05
















3














I would like to evaluate
$$int dfrac{cos(x)+sin(2x)}{sin(x)}text{ d}xtext{.}$$



This is from Stewart's Calculus text, section 7.2., #19. Please note that I don't have a solutions manual and I am not interested in a complete solution.



I would merely like some guidance on a first step on approaching this one, since I've gotten 9 similar problems correct, and I hit a brick wall on this one. My first thought was perhaps splitting the fraction like so:
$$intcot(x)text{ d}x + intdfrac{sin(2x)}{sin(x)}text{ d}x$$
but this does not look helpful.






calculus integration indefinite-integrals trigonometric-integrals







share|cite|improve this question




















  • 4




    hint : $sin (2x) = 2sin( x) cos (x)$
    – ganeshie8
    Oct 17 '14 at 4:03










  • A tip for solving integrals that have sin or cosine or some trigonometric function is use a bunch of identities
    – Joao
    Oct 17 '14 at 4:05














3












3








3







I would like to evaluate
$$int dfrac{cos(x)+sin(2x)}{sin(x)}text{ d}xtext{.}$$



This is from Stewart's Calculus text, section 7.2., #19. Please note that I don't have a solutions manual and I am not interested in a complete solution.



I would merely like some guidance on a first step on approaching this one, since I've gotten 9 similar problems correct, and I hit a brick wall on this one. My first thought was perhaps splitting the fraction like so:
$$intcot(x)text{ d}x + intdfrac{sin(2x)}{sin(x)}text{ d}x$$
but this does not look helpful.






calculus integration indefinite-integrals trigonometric-integrals







share|cite|improve this question















I would like to evaluate
$$int dfrac{cos(x)+sin(2x)}{sin(x)}text{ d}xtext{.}$$



This is from Stewart's Calculus text, section 7.2., #19. Please note that I don't have a solutions manual and I am not interested in a complete solution.



I would merely like some guidance on a first step on approaching this one, since I've gotten 9 similar problems correct, and I hit a brick wall on this one. My first thought was perhaps splitting the fraction like so:
$$intcot(x)text{ d}x + intdfrac{sin(2x)}{sin(x)}text{ d}x$$
but this does not look helpful.






calculus integration indefinite-integrals trigonometric-integrals




calculus integration indefinite-integrals trigonometric-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 27 '18 at 12:15









Martin Sleziak

44.6k8115271




44.6k8115271










asked Oct 17 '14 at 4:02









ClarinetistClarinetist

10.8k42778




10.8k42778








  • 4




    hint : $sin (2x) = 2sin( x) cos (x)$
    – ganeshie8
    Oct 17 '14 at 4:03










  • A tip for solving integrals that have sin or cosine or some trigonometric function is use a bunch of identities
    – Joao
    Oct 17 '14 at 4:05














  • 4




    hint : $sin (2x) = 2sin( x) cos (x)$
    – ganeshie8
    Oct 17 '14 at 4:03










  • A tip for solving integrals that have sin or cosine or some trigonometric function is use a bunch of identities
    – Joao
    Oct 17 '14 at 4:05








4




4




hint : $sin (2x) = 2sin( x) cos (x)$
– ganeshie8
Oct 17 '14 at 4:03




hint : $sin (2x) = 2sin( x) cos (x)$
– ganeshie8
Oct 17 '14 at 4:03












A tip for solving integrals that have sin or cosine or some trigonometric function is use a bunch of identities
– Joao
Oct 17 '14 at 4:05




A tip for solving integrals that have sin or cosine or some trigonometric function is use a bunch of identities
– Joao
Oct 17 '14 at 4:05










1 Answer
1






active

oldest

votes


















7














Substituting using the double-angle identity
$$sin (2x) = 2 sin x cos x$$
will transform the integrand into an expression that involves only $sin x$ and $cos x$, which suggests a particular substitution.






share|cite|improve this answer

















  • 1




    You don’t even need a substitution: after using the double-angle formula, separate the integrand into two fractions, each easily integrated.
    – Lubin
    Oct 17 '14 at 4:12






  • 1




    @Lubin Well, the integral of $cot x$ is not that easy to remember. I usually recognise the integrand as of the form $frac{(sin x)'}{sin x}$, but that's basically equivalent to substituting $u = sin x$.
    – Deepak
    Oct 17 '14 at 4:14












  • @Lubin You're quite right, of course, but when a student first encounters this type of integral they may well have only seen integrands involving $sin$ and $cos$ and in particular may not yet have seen how to handle $int cot x ,dx$.
    – Travis
    Oct 17 '14 at 4:18










  • @Travis - Thank you for your suggestion. I'm a recent math graduate who's studying for the Math GRE Subject Test, and I want to keep memorization to a minimum, so I appreciate yor answer.
    – Clarinetist
    Oct 17 '14 at 4:19






  • 2




    @Lubin I think it is well-known to everyone who has learned calculus, but it may not be obvious to people who are only just learning about trig integrals.
    – Travis
    Oct 17 '14 at 4:36











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









7














Substituting using the double-angle identity
$$sin (2x) = 2 sin x cos x$$
will transform the integrand into an expression that involves only $sin x$ and $cos x$, which suggests a particular substitution.






share|cite|improve this answer

















  • 1




    You don’t even need a substitution: after using the double-angle formula, separate the integrand into two fractions, each easily integrated.
    – Lubin
    Oct 17 '14 at 4:12






  • 1




    @Lubin Well, the integral of $cot x$ is not that easy to remember. I usually recognise the integrand as of the form $frac{(sin x)'}{sin x}$, but that's basically equivalent to substituting $u = sin x$.
    – Deepak
    Oct 17 '14 at 4:14












  • @Lubin You're quite right, of course, but when a student first encounters this type of integral they may well have only seen integrands involving $sin$ and $cos$ and in particular may not yet have seen how to handle $int cot x ,dx$.
    – Travis
    Oct 17 '14 at 4:18










  • @Travis - Thank you for your suggestion. I'm a recent math graduate who's studying for the Math GRE Subject Test, and I want to keep memorization to a minimum, so I appreciate yor answer.
    – Clarinetist
    Oct 17 '14 at 4:19






  • 2




    @Lubin I think it is well-known to everyone who has learned calculus, but it may not be obvious to people who are only just learning about trig integrals.
    – Travis
    Oct 17 '14 at 4:36
















7














Substituting using the double-angle identity
$$sin (2x) = 2 sin x cos x$$
will transform the integrand into an expression that involves only $sin x$ and $cos x$, which suggests a particular substitution.






share|cite|improve this answer

















  • 1




    You don’t even need a substitution: after using the double-angle formula, separate the integrand into two fractions, each easily integrated.
    – Lubin
    Oct 17 '14 at 4:12






  • 1




    @Lubin Well, the integral of $cot x$ is not that easy to remember. I usually recognise the integrand as of the form $frac{(sin x)'}{sin x}$, but that's basically equivalent to substituting $u = sin x$.
    – Deepak
    Oct 17 '14 at 4:14












  • @Lubin You're quite right, of course, but when a student first encounters this type of integral they may well have only seen integrands involving $sin$ and $cos$ and in particular may not yet have seen how to handle $int cot x ,dx$.
    – Travis
    Oct 17 '14 at 4:18










  • @Travis - Thank you for your suggestion. I'm a recent math graduate who's studying for the Math GRE Subject Test, and I want to keep memorization to a minimum, so I appreciate yor answer.
    – Clarinetist
    Oct 17 '14 at 4:19






  • 2




    @Lubin I think it is well-known to everyone who has learned calculus, but it may not be obvious to people who are only just learning about trig integrals.
    – Travis
    Oct 17 '14 at 4:36














7












7








7






Substituting using the double-angle identity
$$sin (2x) = 2 sin x cos x$$
will transform the integrand into an expression that involves only $sin x$ and $cos x$, which suggests a particular substitution.






share|cite|improve this answer












Substituting using the double-angle identity
$$sin (2x) = 2 sin x cos x$$
will transform the integrand into an expression that involves only $sin x$ and $cos x$, which suggests a particular substitution.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Oct 17 '14 at 4:04









TravisTravis

59.8k767146




59.8k767146








  • 1




    You don’t even need a substitution: after using the double-angle formula, separate the integrand into two fractions, each easily integrated.
    – Lubin
    Oct 17 '14 at 4:12






  • 1




    @Lubin Well, the integral of $cot x$ is not that easy to remember. I usually recognise the integrand as of the form $frac{(sin x)'}{sin x}$, but that's basically equivalent to substituting $u = sin x$.
    – Deepak
    Oct 17 '14 at 4:14












  • @Lubin You're quite right, of course, but when a student first encounters this type of integral they may well have only seen integrands involving $sin$ and $cos$ and in particular may not yet have seen how to handle $int cot x ,dx$.
    – Travis
    Oct 17 '14 at 4:18










  • @Travis - Thank you for your suggestion. I'm a recent math graduate who's studying for the Math GRE Subject Test, and I want to keep memorization to a minimum, so I appreciate yor answer.
    – Clarinetist
    Oct 17 '14 at 4:19






  • 2




    @Lubin I think it is well-known to everyone who has learned calculus, but it may not be obvious to people who are only just learning about trig integrals.
    – Travis
    Oct 17 '14 at 4:36














  • 1




    You don’t even need a substitution: after using the double-angle formula, separate the integrand into two fractions, each easily integrated.
    – Lubin
    Oct 17 '14 at 4:12






  • 1




    @Lubin Well, the integral of $cot x$ is not that easy to remember. I usually recognise the integrand as of the form $frac{(sin x)'}{sin x}$, but that's basically equivalent to substituting $u = sin x$.
    – Deepak
    Oct 17 '14 at 4:14












  • @Lubin You're quite right, of course, but when a student first encounters this type of integral they may well have only seen integrands involving $sin$ and $cos$ and in particular may not yet have seen how to handle $int cot x ,dx$.
    – Travis
    Oct 17 '14 at 4:18










  • @Travis - Thank you for your suggestion. I'm a recent math graduate who's studying for the Math GRE Subject Test, and I want to keep memorization to a minimum, so I appreciate yor answer.
    – Clarinetist
    Oct 17 '14 at 4:19






  • 2




    @Lubin I think it is well-known to everyone who has learned calculus, but it may not be obvious to people who are only just learning about trig integrals.
    – Travis
    Oct 17 '14 at 4:36








1




1




You don’t even need a substitution: after using the double-angle formula, separate the integrand into two fractions, each easily integrated.
– Lubin
Oct 17 '14 at 4:12




You don’t even need a substitution: after using the double-angle formula, separate the integrand into two fractions, each easily integrated.
– Lubin
Oct 17 '14 at 4:12




1




1




@Lubin Well, the integral of $cot x$ is not that easy to remember. I usually recognise the integrand as of the form $frac{(sin x)'}{sin x}$, but that's basically equivalent to substituting $u = sin x$.
– Deepak
Oct 17 '14 at 4:14






@Lubin Well, the integral of $cot x$ is not that easy to remember. I usually recognise the integrand as of the form $frac{(sin x)'}{sin x}$, but that's basically equivalent to substituting $u = sin x$.
– Deepak
Oct 17 '14 at 4:14














@Lubin You're quite right, of course, but when a student first encounters this type of integral they may well have only seen integrands involving $sin$ and $cos$ and in particular may not yet have seen how to handle $int cot x ,dx$.
– Travis
Oct 17 '14 at 4:18




@Lubin You're quite right, of course, but when a student first encounters this type of integral they may well have only seen integrands involving $sin$ and $cos$ and in particular may not yet have seen how to handle $int cot x ,dx$.
– Travis
Oct 17 '14 at 4:18












@Travis - Thank you for your suggestion. I'm a recent math graduate who's studying for the Math GRE Subject Test, and I want to keep memorization to a minimum, so I appreciate yor answer.
– Clarinetist
Oct 17 '14 at 4:19




@Travis - Thank you for your suggestion. I'm a recent math graduate who's studying for the Math GRE Subject Test, and I want to keep memorization to a minimum, so I appreciate yor answer.
– Clarinetist
Oct 17 '14 at 4:19




2




2




@Lubin I think it is well-known to everyone who has learned calculus, but it may not be obvious to people who are only just learning about trig integrals.
– Travis
Oct 17 '14 at 4:36




@Lubin I think it is well-known to everyone who has learned calculus, but it may not be obvious to people who are only just learning about trig integrals.
– Travis
Oct 17 '14 at 4:36


















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