$ int bigl| asin(x)+bcos(x) bigr| {rm d}x = ? $












1












$begingroup$



My friend evaluated this to be
$$
int bigl| asin(x)+bcos(x) bigr| {rm d}x \
= sqrt{a^2+b^2} left(
sin(x-phi)text{sign}(cos(x-phi))
+frac{2}{pi} bigl(x-arctan(tan(x-phi)) bigr)
right) + C
$$
where $phi = arctan left(frac{a}{b} right)$.




My answer instead is much shorter, by simply looking for the intervals of $x$ for which I need to multiply by $-1$, I got



$$
(bsin(x)-acos(x)) cdot text{sign}(asin(x)+bcos(x)) + C
$$



Why does my friend's solution look so big and complex and how did she even come up with the idea to do that?






calculus integration indefinite-integrals absolute-value trigonometric-integrals







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Try $b=1$, $a=0$, then you suggest $sin xcdottext{sign},cos x$ as the antiderivative, but it is not even continuous at some points.
    $endgroup$
    – A.Γ.
    Jan 2 at 16:23






  • 2




    $begingroup$
    What exactly is the point of the $frac{sqrt{cos^2(x-phi)}}{cos(x-phi)}$? It could be replaced by $text{sign}(cos(x-phi))$.
    $endgroup$
    – John Doe
    Jan 2 at 16:23












  • $begingroup$
    @A.Γ. The other solution actually also gives $sin x cdot text{sign} cos x$ in that case
    $endgroup$
    – John Doe
    Jan 2 at 16:27












  • $begingroup$
    @JohnDoe Not really, it has the extra term with $arctan$ that shifts the function in different intervals by a multiple of $2$ to make it continuous.
    $endgroup$
    – A.Γ.
    Jan 2 at 16:43












  • $begingroup$
    @A.Γ. Ah yes, you're right.
    $endgroup$
    – John Doe
    Jan 2 at 16:44
















1












$begingroup$



My friend evaluated this to be
$$
int bigl| asin(x)+bcos(x) bigr| {rm d}x \
= sqrt{a^2+b^2} left(
sin(x-phi)text{sign}(cos(x-phi))
+frac{2}{pi} bigl(x-arctan(tan(x-phi)) bigr)
right) + C
$$
where $phi = arctan left(frac{a}{b} right)$.




My answer instead is much shorter, by simply looking for the intervals of $x$ for which I need to multiply by $-1$, I got



$$
(bsin(x)-acos(x)) cdot text{sign}(asin(x)+bcos(x)) + C
$$



Why does my friend's solution look so big and complex and how did she even come up with the idea to do that?






calculus integration indefinite-integrals absolute-value trigonometric-integrals







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Try $b=1$, $a=0$, then you suggest $sin xcdottext{sign},cos x$ as the antiderivative, but it is not even continuous at some points.
    $endgroup$
    – A.Γ.
    Jan 2 at 16:23






  • 2




    $begingroup$
    What exactly is the point of the $frac{sqrt{cos^2(x-phi)}}{cos(x-phi)}$? It could be replaced by $text{sign}(cos(x-phi))$.
    $endgroup$
    – John Doe
    Jan 2 at 16:23












  • $begingroup$
    @A.Γ. The other solution actually also gives $sin x cdot text{sign} cos x$ in that case
    $endgroup$
    – John Doe
    Jan 2 at 16:27












  • $begingroup$
    @JohnDoe Not really, it has the extra term with $arctan$ that shifts the function in different intervals by a multiple of $2$ to make it continuous.
    $endgroup$
    – A.Γ.
    Jan 2 at 16:43












  • $begingroup$
    @A.Γ. Ah yes, you're right.
    $endgroup$
    – John Doe
    Jan 2 at 16:44














1












1








1





$begingroup$



My friend evaluated this to be
$$
int bigl| asin(x)+bcos(x) bigr| {rm d}x \
= sqrt{a^2+b^2} left(
sin(x-phi)text{sign}(cos(x-phi))
+frac{2}{pi} bigl(x-arctan(tan(x-phi)) bigr)
right) + C
$$
where $phi = arctan left(frac{a}{b} right)$.




My answer instead is much shorter, by simply looking for the intervals of $x$ for which I need to multiply by $-1$, I got



$$
(bsin(x)-acos(x)) cdot text{sign}(asin(x)+bcos(x)) + C
$$



Why does my friend's solution look so big and complex and how did she even come up with the idea to do that?






calculus integration indefinite-integrals absolute-value trigonometric-integrals







share|cite|improve this question











$endgroup$





My friend evaluated this to be
$$
int bigl| asin(x)+bcos(x) bigr| {rm d}x \
= sqrt{a^2+b^2} left(
sin(x-phi)text{sign}(cos(x-phi))
+frac{2}{pi} bigl(x-arctan(tan(x-phi)) bigr)
right) + C
$$
where $phi = arctan left(frac{a}{b} right)$.




My answer instead is much shorter, by simply looking for the intervals of $x$ for which I need to multiply by $-1$, I got



$$
(bsin(x)-acos(x)) cdot text{sign}(asin(x)+bcos(x)) + C
$$



Why does my friend's solution look so big and complex and how did she even come up with the idea to do that?






calculus integration indefinite-integrals absolute-value trigonometric-integrals




calculus integration indefinite-integrals absolute-value trigonometric-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 14:29







Mint

















asked Jan 2 at 16:14









MintMint

5301417




5301417








  • 1




    $begingroup$
    Try $b=1$, $a=0$, then you suggest $sin xcdottext{sign},cos x$ as the antiderivative, but it is not even continuous at some points.
    $endgroup$
    – A.Γ.
    Jan 2 at 16:23






  • 2




    $begingroup$
    What exactly is the point of the $frac{sqrt{cos^2(x-phi)}}{cos(x-phi)}$? It could be replaced by $text{sign}(cos(x-phi))$.
    $endgroup$
    – John Doe
    Jan 2 at 16:23












  • $begingroup$
    @A.Γ. The other solution actually also gives $sin x cdot text{sign} cos x$ in that case
    $endgroup$
    – John Doe
    Jan 2 at 16:27












  • $begingroup$
    @JohnDoe Not really, it has the extra term with $arctan$ that shifts the function in different intervals by a multiple of $2$ to make it continuous.
    $endgroup$
    – A.Γ.
    Jan 2 at 16:43












  • $begingroup$
    @A.Γ. Ah yes, you're right.
    $endgroup$
    – John Doe
    Jan 2 at 16:44














  • 1




    $begingroup$
    Try $b=1$, $a=0$, then you suggest $sin xcdottext{sign},cos x$ as the antiderivative, but it is not even continuous at some points.
    $endgroup$
    – A.Γ.
    Jan 2 at 16:23






  • 2




    $begingroup$
    What exactly is the point of the $frac{sqrt{cos^2(x-phi)}}{cos(x-phi)}$? It could be replaced by $text{sign}(cos(x-phi))$.
    $endgroup$
    – John Doe
    Jan 2 at 16:23












  • $begingroup$
    @A.Γ. The other solution actually also gives $sin x cdot text{sign} cos x$ in that case
    $endgroup$
    – John Doe
    Jan 2 at 16:27












  • $begingroup$
    @JohnDoe Not really, it has the extra term with $arctan$ that shifts the function in different intervals by a multiple of $2$ to make it continuous.
    $endgroup$
    – A.Γ.
    Jan 2 at 16:43












  • $begingroup$
    @A.Γ. Ah yes, you're right.
    $endgroup$
    – John Doe
    Jan 2 at 16:44








1




1




$begingroup$
Try $b=1$, $a=0$, then you suggest $sin xcdottext{sign},cos x$ as the antiderivative, but it is not even continuous at some points.
$endgroup$
– A.Γ.
Jan 2 at 16:23




$begingroup$
Try $b=1$, $a=0$, then you suggest $sin xcdottext{sign},cos x$ as the antiderivative, but it is not even continuous at some points.
$endgroup$
– A.Γ.
Jan 2 at 16:23




2




2




$begingroup$
What exactly is the point of the $frac{sqrt{cos^2(x-phi)}}{cos(x-phi)}$? It could be replaced by $text{sign}(cos(x-phi))$.
$endgroup$
– John Doe
Jan 2 at 16:23






$begingroup$
What exactly is the point of the $frac{sqrt{cos^2(x-phi)}}{cos(x-phi)}$? It could be replaced by $text{sign}(cos(x-phi))$.
$endgroup$
– John Doe
Jan 2 at 16:23














$begingroup$
@A.Γ. The other solution actually also gives $sin x cdot text{sign} cos x$ in that case
$endgroup$
– John Doe
Jan 2 at 16:27






$begingroup$
@A.Γ. The other solution actually also gives $sin x cdot text{sign} cos x$ in that case
$endgroup$
– John Doe
Jan 2 at 16:27














$begingroup$
@JohnDoe Not really, it has the extra term with $arctan$ that shifts the function in different intervals by a multiple of $2$ to make it continuous.
$endgroup$
– A.Γ.
Jan 2 at 16:43






$begingroup$
@JohnDoe Not really, it has the extra term with $arctan$ that shifts the function in different intervals by a multiple of $2$ to make it continuous.
$endgroup$
– A.Γ.
Jan 2 at 16:43














$begingroup$
@A.Γ. Ah yes, you're right.
$endgroup$
– John Doe
Jan 2 at 16:44




$begingroup$
@A.Γ. Ah yes, you're right.
$endgroup$
– John Doe
Jan 2 at 16:44










1 Answer
1






active

oldest

votes


















3












$begingroup$

She is probably starting by replacing the $asin x + bcos x$ and rewriting in terms of a function with an amplitude and phase. One way to do this is to multiply by $1$ and draw a right triangle with angle $phi$, adjacent side length $b$, and opposite side length $a$. Then one has



$$begin{aligned} left|asin x + bcos xright| &= sqrt{a^{2}+b^{2}}left|frac{a}{sqrt{a^{2}+b^{2}}},sin x + frac{b}{sqrt{a^{2}+b^{2}}},cos xright| \
&= sqrt{a^{2}+b^{2}}left|sinphisin x + cosphicos xright| \
&= sqrt{a^{2}+b^{2}}left|cos(x-phi)right|.end{aligned}$$



This procedure is motivated in physics/engineering as a more elegant way of writing solutions to the harmonic oscillator equation, as amplitude and phase both have physical interpretations and are often easier to determine given initial conditions.






share|cite|improve this answer









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    1 Answer
    1






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    active

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    active

    oldest

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    3












    $begingroup$

    She is probably starting by replacing the $asin x + bcos x$ and rewriting in terms of a function with an amplitude and phase. One way to do this is to multiply by $1$ and draw a right triangle with angle $phi$, adjacent side length $b$, and opposite side length $a$. Then one has



    $$begin{aligned} left|asin x + bcos xright| &= sqrt{a^{2}+b^{2}}left|frac{a}{sqrt{a^{2}+b^{2}}},sin x + frac{b}{sqrt{a^{2}+b^{2}}},cos xright| \
    &= sqrt{a^{2}+b^{2}}left|sinphisin x + cosphicos xright| \
    &= sqrt{a^{2}+b^{2}}left|cos(x-phi)right|.end{aligned}$$



    This procedure is motivated in physics/engineering as a more elegant way of writing solutions to the harmonic oscillator equation, as amplitude and phase both have physical interpretations and are often easier to determine given initial conditions.






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      She is probably starting by replacing the $asin x + bcos x$ and rewriting in terms of a function with an amplitude and phase. One way to do this is to multiply by $1$ and draw a right triangle with angle $phi$, adjacent side length $b$, and opposite side length $a$. Then one has



      $$begin{aligned} left|asin x + bcos xright| &= sqrt{a^{2}+b^{2}}left|frac{a}{sqrt{a^{2}+b^{2}}},sin x + frac{b}{sqrt{a^{2}+b^{2}}},cos xright| \
      &= sqrt{a^{2}+b^{2}}left|sinphisin x + cosphicos xright| \
      &= sqrt{a^{2}+b^{2}}left|cos(x-phi)right|.end{aligned}$$



      This procedure is motivated in physics/engineering as a more elegant way of writing solutions to the harmonic oscillator equation, as amplitude and phase both have physical interpretations and are often easier to determine given initial conditions.






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        She is probably starting by replacing the $asin x + bcos x$ and rewriting in terms of a function with an amplitude and phase. One way to do this is to multiply by $1$ and draw a right triangle with angle $phi$, adjacent side length $b$, and opposite side length $a$. Then one has



        $$begin{aligned} left|asin x + bcos xright| &= sqrt{a^{2}+b^{2}}left|frac{a}{sqrt{a^{2}+b^{2}}},sin x + frac{b}{sqrt{a^{2}+b^{2}}},cos xright| \
        &= sqrt{a^{2}+b^{2}}left|sinphisin x + cosphicos xright| \
        &= sqrt{a^{2}+b^{2}}left|cos(x-phi)right|.end{aligned}$$



        This procedure is motivated in physics/engineering as a more elegant way of writing solutions to the harmonic oscillator equation, as amplitude and phase both have physical interpretations and are often easier to determine given initial conditions.






        share|cite|improve this answer









        $endgroup$



        She is probably starting by replacing the $asin x + bcos x$ and rewriting in terms of a function with an amplitude and phase. One way to do this is to multiply by $1$ and draw a right triangle with angle $phi$, adjacent side length $b$, and opposite side length $a$. Then one has



        $$begin{aligned} left|asin x + bcos xright| &= sqrt{a^{2}+b^{2}}left|frac{a}{sqrt{a^{2}+b^{2}}},sin x + frac{b}{sqrt{a^{2}+b^{2}}},cos xright| \
        &= sqrt{a^{2}+b^{2}}left|sinphisin x + cosphicos xright| \
        &= sqrt{a^{2}+b^{2}}left|cos(x-phi)right|.end{aligned}$$



        This procedure is motivated in physics/engineering as a more elegant way of writing solutions to the harmonic oscillator equation, as amplitude and phase both have physical interpretations and are often easier to determine given initial conditions.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 2 at 18:33









        IninterrompueIninterrompue

        63519




        63519






























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