Evaluating $ int frac1{5S_{8} - 9S_{10} + 7S_{12} - 2S_{14}} {rm d}x$, where $S_n = sin^n(x) + cos^n(x)$












5












$begingroup$



Prove that



$$ int frac1{5S_{8} - 9S_{10} + 7S_{12} - 2S_{14}} {rm d}x = 2x - arctan left( frac{tan2x}{2 + tan^22x} right) + C$$



where $S_n = sin^n(x) + cos^n(x)$.




Even differentiating the right doesn't end up with anything close to that monster integrand.



$$ frac{{rm d}}{{rm d}x} left( 2x - arctan left( frac{tan2x}{2 + tan^22x} right) + C right)
\ = 2 + frac{(tan^22x+1)(2tan^22x-4)}{tan^42x+5tan^22x+4}
\ = frac{ 4 } {sin^42x+ 5sin^22xcos^22x + 4cos^42x}
\ = frac1{sin^8x + sin^2xcos^6x +cos^2xsin^6x + cos^8x} $$






integration trigonometry indefinite-integrals trigonometric-integrals







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    The denominator in your final expression is $S_6$
    $endgroup$
    – Empy2
    Dec 13 '18 at 12:14






  • 2




    $begingroup$
    $S_0=2,S_2=1$ $$S_{m+2}=sin^mx(1-cos^2x)+cos^mx(1-sin^2x)=S_m-sin^2xcos^2xS_{m-2}$$
    $endgroup$
    – lab bhattacharjee
    Dec 13 '18 at 14:38
















5












$begingroup$



Prove that



$$ int frac1{5S_{8} - 9S_{10} + 7S_{12} - 2S_{14}} {rm d}x = 2x - arctan left( frac{tan2x}{2 + tan^22x} right) + C$$



where $S_n = sin^n(x) + cos^n(x)$.




Even differentiating the right doesn't end up with anything close to that monster integrand.



$$ frac{{rm d}}{{rm d}x} left( 2x - arctan left( frac{tan2x}{2 + tan^22x} right) + C right)
\ = 2 + frac{(tan^22x+1)(2tan^22x-4)}{tan^42x+5tan^22x+4}
\ = frac{ 4 } {sin^42x+ 5sin^22xcos^22x + 4cos^42x}
\ = frac1{sin^8x + sin^2xcos^6x +cos^2xsin^6x + cos^8x} $$






integration trigonometry indefinite-integrals trigonometric-integrals







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    The denominator in your final expression is $S_6$
    $endgroup$
    – Empy2
    Dec 13 '18 at 12:14






  • 2




    $begingroup$
    $S_0=2,S_2=1$ $$S_{m+2}=sin^mx(1-cos^2x)+cos^mx(1-sin^2x)=S_m-sin^2xcos^2xS_{m-2}$$
    $endgroup$
    – lab bhattacharjee
    Dec 13 '18 at 14:38














5












5








5


1



$begingroup$



Prove that



$$ int frac1{5S_{8} - 9S_{10} + 7S_{12} - 2S_{14}} {rm d}x = 2x - arctan left( frac{tan2x}{2 + tan^22x} right) + C$$



where $S_n = sin^n(x) + cos^n(x)$.




Even differentiating the right doesn't end up with anything close to that monster integrand.



$$ frac{{rm d}}{{rm d}x} left( 2x - arctan left( frac{tan2x}{2 + tan^22x} right) + C right)
\ = 2 + frac{(tan^22x+1)(2tan^22x-4)}{tan^42x+5tan^22x+4}
\ = frac{ 4 } {sin^42x+ 5sin^22xcos^22x + 4cos^42x}
\ = frac1{sin^8x + sin^2xcos^6x +cos^2xsin^6x + cos^8x} $$






integration trigonometry indefinite-integrals trigonometric-integrals







share|cite|improve this question











$endgroup$





Prove that



$$ int frac1{5S_{8} - 9S_{10} + 7S_{12} - 2S_{14}} {rm d}x = 2x - arctan left( frac{tan2x}{2 + tan^22x} right) + C$$



where $S_n = sin^n(x) + cos^n(x)$.




Even differentiating the right doesn't end up with anything close to that monster integrand.



$$ frac{{rm d}}{{rm d}x} left( 2x - arctan left( frac{tan2x}{2 + tan^22x} right) + C right)
\ = 2 + frac{(tan^22x+1)(2tan^22x-4)}{tan^42x+5tan^22x+4}
\ = frac{ 4 } {sin^42x+ 5sin^22xcos^22x + 4cos^42x}
\ = frac1{sin^8x + sin^2xcos^6x +cos^2xsin^6x + cos^8x} $$






integration trigonometry indefinite-integrals trigonometric-integrals




integration trigonometry 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 Dec 13 '18 at 13:23









Blue

48.6k870156




48.6k870156










asked Dec 13 '18 at 12:08









MintMint

5311417




5311417








  • 1




    $begingroup$
    The denominator in your final expression is $S_6$
    $endgroup$
    – Empy2
    Dec 13 '18 at 12:14






  • 2




    $begingroup$
    $S_0=2,S_2=1$ $$S_{m+2}=sin^mx(1-cos^2x)+cos^mx(1-sin^2x)=S_m-sin^2xcos^2xS_{m-2}$$
    $endgroup$
    – lab bhattacharjee
    Dec 13 '18 at 14:38














  • 1




    $begingroup$
    The denominator in your final expression is $S_6$
    $endgroup$
    – Empy2
    Dec 13 '18 at 12:14






  • 2




    $begingroup$
    $S_0=2,S_2=1$ $$S_{m+2}=sin^mx(1-cos^2x)+cos^mx(1-sin^2x)=S_m-sin^2xcos^2xS_{m-2}$$
    $endgroup$
    – lab bhattacharjee
    Dec 13 '18 at 14:38








1




1




$begingroup$
The denominator in your final expression is $S_6$
$endgroup$
– Empy2
Dec 13 '18 at 12:14




$begingroup$
The denominator in your final expression is $S_6$
$endgroup$
– Empy2
Dec 13 '18 at 12:14




2




2




$begingroup$
$S_0=2,S_2=1$ $$S_{m+2}=sin^mx(1-cos^2x)+cos^mx(1-sin^2x)=S_m-sin^2xcos^2xS_{m-2}$$
$endgroup$
– lab bhattacharjee
Dec 13 '18 at 14:38




$begingroup$
$S_0=2,S_2=1$ $$S_{m+2}=sin^mx(1-cos^2x)+cos^mx(1-sin^2x)=S_m-sin^2xcos^2xS_{m-2}$$
$endgroup$
– lab bhattacharjee
Dec 13 '18 at 14:38










1 Answer
1






active

oldest

votes


















5












$begingroup$

Let $c=cos2x$. Then applying the double angle identities,
$$sin^2x=frac{1-cos2x}2,quadcos^2x=frac{1+cos2x}2,$$
to each occurrence of $sin^nx$ and $cos^nx$, we get
$$begin{cases}
S_8=2^{-3}(1+6c^2+c^4)\
S_{10}=2^{-4}(1+10c^2+5c^4)\
S_{12}=2^{-5}(1+15c^2+15c^4+c^6)\
S_{14}=2^{-6}(1+21c^2+35c^4+7c^6)
end{cases}$$
Combining these expressions to match the denominator of the integrand, we find that it reduces to
$$5S_8-9S_{10}+7S_{12}-2S_{14}=2^{-2}(1+3c^2)$$
so that
$$intfrac{mathrm dx}{5S_8-9S_{10}+7S_{12}-2S_{14}}=intfrac4{1+3cos^22x},mathrm dx=intfrac8{5+3cos4x},mathrm dx$$
The antiderivative is $-arctan(2cot2x)+C$, which I don't think should be too hard to reconcile with the given solution.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    The pure elegance and beauty of this answer is incredible. I commend you with a meaningless upvote.
    $endgroup$
    – clathratus
    Dec 14 '18 at 16:20











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









5












$begingroup$

Let $c=cos2x$. Then applying the double angle identities,
$$sin^2x=frac{1-cos2x}2,quadcos^2x=frac{1+cos2x}2,$$
to each occurrence of $sin^nx$ and $cos^nx$, we get
$$begin{cases}
S_8=2^{-3}(1+6c^2+c^4)\
S_{10}=2^{-4}(1+10c^2+5c^4)\
S_{12}=2^{-5}(1+15c^2+15c^4+c^6)\
S_{14}=2^{-6}(1+21c^2+35c^4+7c^6)
end{cases}$$
Combining these expressions to match the denominator of the integrand, we find that it reduces to
$$5S_8-9S_{10}+7S_{12}-2S_{14}=2^{-2}(1+3c^2)$$
so that
$$intfrac{mathrm dx}{5S_8-9S_{10}+7S_{12}-2S_{14}}=intfrac4{1+3cos^22x},mathrm dx=intfrac8{5+3cos4x},mathrm dx$$
The antiderivative is $-arctan(2cot2x)+C$, which I don't think should be too hard to reconcile with the given solution.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    The pure elegance and beauty of this answer is incredible. I commend you with a meaningless upvote.
    $endgroup$
    – clathratus
    Dec 14 '18 at 16:20
















5












$begingroup$

Let $c=cos2x$. Then applying the double angle identities,
$$sin^2x=frac{1-cos2x}2,quadcos^2x=frac{1+cos2x}2,$$
to each occurrence of $sin^nx$ and $cos^nx$, we get
$$begin{cases}
S_8=2^{-3}(1+6c^2+c^4)\
S_{10}=2^{-4}(1+10c^2+5c^4)\
S_{12}=2^{-5}(1+15c^2+15c^4+c^6)\
S_{14}=2^{-6}(1+21c^2+35c^4+7c^6)
end{cases}$$
Combining these expressions to match the denominator of the integrand, we find that it reduces to
$$5S_8-9S_{10}+7S_{12}-2S_{14}=2^{-2}(1+3c^2)$$
so that
$$intfrac{mathrm dx}{5S_8-9S_{10}+7S_{12}-2S_{14}}=intfrac4{1+3cos^22x},mathrm dx=intfrac8{5+3cos4x},mathrm dx$$
The antiderivative is $-arctan(2cot2x)+C$, which I don't think should be too hard to reconcile with the given solution.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    The pure elegance and beauty of this answer is incredible. I commend you with a meaningless upvote.
    $endgroup$
    – clathratus
    Dec 14 '18 at 16:20














5












5








5





$begingroup$

Let $c=cos2x$. Then applying the double angle identities,
$$sin^2x=frac{1-cos2x}2,quadcos^2x=frac{1+cos2x}2,$$
to each occurrence of $sin^nx$ and $cos^nx$, we get
$$begin{cases}
S_8=2^{-3}(1+6c^2+c^4)\
S_{10}=2^{-4}(1+10c^2+5c^4)\
S_{12}=2^{-5}(1+15c^2+15c^4+c^6)\
S_{14}=2^{-6}(1+21c^2+35c^4+7c^6)
end{cases}$$
Combining these expressions to match the denominator of the integrand, we find that it reduces to
$$5S_8-9S_{10}+7S_{12}-2S_{14}=2^{-2}(1+3c^2)$$
so that
$$intfrac{mathrm dx}{5S_8-9S_{10}+7S_{12}-2S_{14}}=intfrac4{1+3cos^22x},mathrm dx=intfrac8{5+3cos4x},mathrm dx$$
The antiderivative is $-arctan(2cot2x)+C$, which I don't think should be too hard to reconcile with the given solution.






share|cite|improve this answer









$endgroup$



Let $c=cos2x$. Then applying the double angle identities,
$$sin^2x=frac{1-cos2x}2,quadcos^2x=frac{1+cos2x}2,$$
to each occurrence of $sin^nx$ and $cos^nx$, we get
$$begin{cases}
S_8=2^{-3}(1+6c^2+c^4)\
S_{10}=2^{-4}(1+10c^2+5c^4)\
S_{12}=2^{-5}(1+15c^2+15c^4+c^6)\
S_{14}=2^{-6}(1+21c^2+35c^4+7c^6)
end{cases}$$
Combining these expressions to match the denominator of the integrand, we find that it reduces to
$$5S_8-9S_{10}+7S_{12}-2S_{14}=2^{-2}(1+3c^2)$$
so that
$$intfrac{mathrm dx}{5S_8-9S_{10}+7S_{12}-2S_{14}}=intfrac4{1+3cos^22x},mathrm dx=intfrac8{5+3cos4x},mathrm dx$$
The antiderivative is $-arctan(2cot2x)+C$, which I don't think should be too hard to reconcile with the given solution.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 13 '18 at 17:17









user170231user170231

4,15811429




4,15811429








  • 1




    $begingroup$
    The pure elegance and beauty of this answer is incredible. I commend you with a meaningless upvote.
    $endgroup$
    – clathratus
    Dec 14 '18 at 16:20














  • 1




    $begingroup$
    The pure elegance and beauty of this answer is incredible. I commend you with a meaningless upvote.
    $endgroup$
    – clathratus
    Dec 14 '18 at 16:20








1




1




$begingroup$
The pure elegance and beauty of this answer is incredible. I commend you with a meaningless upvote.
$endgroup$
– clathratus
Dec 14 '18 at 16:20




$begingroup$
The pure elegance and beauty of this answer is incredible. I commend you with a meaningless upvote.
$endgroup$
– clathratus
Dec 14 '18 at 16:20


















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