Mixing
Integration by parts is not working.
0
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I an trying to solve $$int_{-R}^{R}tan^{-1}(sin^2(x))dx$$ with Integration by Parts, by integrating 1 and differentiating our integrand, yielding $$2Rtan^{-1}(sin^2(x))-2int_{-R}^{R}frac{xsin(x)cos(x)}{sin^4(x)+1}$$ and when I plug these into Desmos they yield different results. This probably has something to do with my assumption of the principal value of $tan^{-1}(sin^2(x))$, but I don't know how to fix it.
calculus
asked Jan 14 at 4:22
Milo MosesMilo Moses
113
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show 1 more comment
0
$begingroup$
I an trying to solve $$int_{-R}^{R}tan^{-1}(sin^2(x))dx$$ with Integration by Parts, by integrating 1 and differentiating our integrand, yielding $$2Rtan^{-1}(sin^2(x))-2int_{-R}^{R}frac{xsin(x)cos(x)}{sin^4(x)+1}$$ and when I plug these into Desmos they yield different results. This probably has something to do with my assumption of the principal value of $tan^{-1}(sin^2(x))$, but I don't know how to fix it.
calculus
asked Jan 14 at 4:22
Milo MosesMilo Moses
113
$endgroup$
2
$begingroup$
I don't think it's integral exists in elementary functions.
$endgroup$
– Archis Welankar
Jan 14 at 4:38
$begingroup$
$tiny I=xtan^{-1}(sin^2(x))-frac{1}{4}i(i(text{Li}_2(((1-2i)+2sqrt{-1-i})e^{2ix})+text{Li}_2(((1-2i)-2sqrt{-1-i})e^{2ix}))-i(text{Li}_2(((1+2i)+2sqrt{-1+i})e^{2ix})+text{Li}_2(((1+2i)-2sqrt{-1+i})e^{2ix}))-2log(1+(-1+2i+2sqrt{-1-i})e^{2ix})(x-sin^{-1}(sqrt[4]{-1}))-2log(1+((-1+2i)-2sqrt{-1-i})e^{2ix})(x+sin^{-1}(sqrt[4]{-1}))+2log(1+(-1-2i+2sqrt{-1+i})e^{2ix})(x+sin^{-1}((-1)^{3/4}))+2log(1-((1+2i)+2sqrt{i-1})e^{2ix})(x-sin^{-1}((-1)^{3/4}))-4isin^{-1}((-1)^{3/4})tan^{-1}(frac{2tan(x)}{(i-1)^{3/2}})-4sin^{-1}(frac{1+i}{sqrt{2}})tanh^{-1}(sqrt{-1-i}tan(x)))$
$endgroup$
– Kemono Chen
Jan 14 at 4:58
$begingroup$
where $I=intarctansin^2xdx$. Combining with that the integrand is even, the original integral is not elementary.
$endgroup$
– Kemono Chen
Jan 14 at 5:00
$begingroup$
In your "uv" part, you still have an $x$, inside the $sin.$
$endgroup$
– B. Goddard
Jan 14 at 12:21
$begingroup$
Wait, are you not allowed to do integration by parts on non elementary functions?
$endgroup$
– Milo Moses
Jan 14 at 16:43
|
show 1 more comment
0
0
0
$begingroup$
I an trying to solve $$int_{-R}^{R}tan^{-1}(sin^2(x))dx$$ with Integration by Parts, by integrating 1 and differentiating our integrand, yielding $$2Rtan^{-1}(sin^2(x))-2int_{-R}^{R}frac{xsin(x)cos(x)}{sin^4(x)+1}$$ and when I plug these into Desmos they yield different results. This probably has something to do with my assumption of the principal value of $tan^{-1}(sin^2(x))$, but I don't know how to fix it.
calculus
asked Jan 14 at 4:22
Milo MosesMilo Moses
113
$endgroup$
I an trying to solve $$int_{-R}^{R}tan^{-1}(sin^2(x))dx$$ with Integration by Parts, by integrating 1 and differentiating our integrand, yielding $$2Rtan^{-1}(sin^2(x))-2int_{-R}^{R}frac{xsin(x)cos(x)}{sin^4(x)+1}$$ and when I plug these into Desmos they yield different results. This probably has something to do with my assumption of the principal value of $tan^{-1}(sin^2(x))$, but I don't know how to fix it.
calculus
calculus
asked Jan 14 at 4:22
Milo MosesMilo Moses
113
asked Jan 14 at 4:22
Milo MosesMilo Moses
113
asked Jan 14 at 4:22
Milo MosesMilo Moses
113
asked Jan 14 at 4:22
Milo MosesMilo Moses
113
asked Jan 14 at 4:22
Milo MosesMilo Moses
113
113
2
$begingroup$
I don't think it's integral exists in elementary functions.
$endgroup$
– Archis Welankar
Jan 14 at 4:38
$begingroup$
$tiny I=xtan^{-1}(sin^2(x))-frac{1}{4}i(i(text{Li}_2(((1-2i)+2sqrt{-1-i})e^{2ix})+text{Li}_2(((1-2i)-2sqrt{-1-i})e^{2ix}))-i(text{Li}_2(((1+2i)+2sqrt{-1+i})e^{2ix})+text{Li}_2(((1+2i)-2sqrt{-1+i})e^{2ix}))-2log(1+(-1+2i+2sqrt{-1-i})e^{2ix})(x-sin^{-1}(sqrt[4]{-1}))-2log(1+((-1+2i)-2sqrt{-1-i})e^{2ix})(x+sin^{-1}(sqrt[4]{-1}))+2log(1+(-1-2i+2sqrt{-1+i})e^{2ix})(x+sin^{-1}((-1)^{3/4}))+2log(1-((1+2i)+2sqrt{i-1})e^{2ix})(x-sin^{-1}((-1)^{3/4}))-4isin^{-1}((-1)^{3/4})tan^{-1}(frac{2tan(x)}{(i-1)^{3/2}})-4sin^{-1}(frac{1+i}{sqrt{2}})tanh^{-1}(sqrt{-1-i}tan(x)))$
$endgroup$
– Kemono Chen
Jan 14 at 4:58
$begingroup$
where $I=intarctansin^2xdx$. Combining with that the integrand is even, the original integral is not elementary.
$endgroup$
– Kemono Chen
Jan 14 at 5:00
$begingroup$
In your "uv" part, you still have an $x$, inside the $sin.$
$endgroup$
– B. Goddard
Jan 14 at 12:21
$begingroup$
Wait, are you not allowed to do integration by parts on non elementary functions?
$endgroup$
– Milo Moses
Jan 14 at 16:43
|
show 1 more comment
2
$begingroup$
I don't think it's integral exists in elementary functions.
$endgroup$
– Archis Welankar
Jan 14 at 4:38
$begingroup$
$tiny I=xtan^{-1}(sin^2(x))-frac{1}{4}i(i(text{Li}_2(((1-2i)+2sqrt{-1-i})e^{2ix})+text{Li}_2(((1-2i)-2sqrt{-1-i})e^{2ix}))-i(text{Li}_2(((1+2i)+2sqrt{-1+i})e^{2ix})+text{Li}_2(((1+2i)-2sqrt{-1+i})e^{2ix}))-2log(1+(-1+2i+2sqrt{-1-i})e^{2ix})(x-sin^{-1}(sqrt[4]{-1}))-2log(1+((-1+2i)-2sqrt{-1-i})e^{2ix})(x+sin^{-1}(sqrt[4]{-1}))+2log(1+(-1-2i+2sqrt{-1+i})e^{2ix})(x+sin^{-1}((-1)^{3/4}))+2log(1-((1+2i)+2sqrt{i-1})e^{2ix})(x-sin^{-1}((-1)^{3/4}))-4isin^{-1}((-1)^{3/4})tan^{-1}(frac{2tan(x)}{(i-1)^{3/2}})-4sin^{-1}(frac{1+i}{sqrt{2}})tanh^{-1}(sqrt{-1-i}tan(x)))$
$endgroup$
– Kemono Chen
Jan 14 at 4:58
$begingroup$
where $I=intarctansin^2xdx$. Combining with that the integrand is even, the original integral is not elementary.
$endgroup$
– Kemono Chen
Jan 14 at 5:00
$begingroup$
In your "uv" part, you still have an $x$, inside the $sin.$
$endgroup$
– B. Goddard
Jan 14 at 12:21
$begingroup$
Wait, are you not allowed to do integration by parts on non elementary functions?
$endgroup$
– Milo Moses
Jan 14 at 16:43
2
2
$begingroup$
I don't think it's integral exists in elementary functions.
$endgroup$
– Archis Welankar
Jan 14 at 4:38
$begingroup$
I don't think it's integral exists in elementary functions.
$endgroup$
– Archis Welankar
Jan 14 at 4:38
$begingroup$
$tiny I=xtan^{-1}(sin^2(x))-frac{1}{4}i(i(text{Li}_2(((1-2i)+2sqrt{-1-i})e^{2ix})+text{Li}_2(((1-2i)-2sqrt{-1-i})e^{2ix}))-i(text{Li}_2(((1+2i)+2sqrt{-1+i})e^{2ix})+text{Li}_2(((1+2i)-2sqrt{-1+i})e^{2ix}))-2log(1+(-1+2i+2sqrt{-1-i})e^{2ix})(x-sin^{-1}(sqrt[4]{-1}))-2log(1+((-1+2i)-2sqrt{-1-i})e^{2ix})(x+sin^{-1}(sqrt[4]{-1}))+2log(1+(-1-2i+2sqrt{-1+i})e^{2ix})(x+sin^{-1}((-1)^{3/4}))+2log(1-((1+2i)+2sqrt{i-1})e^{2ix})(x-sin^{-1}((-1)^{3/4}))-4isin^{-1}((-1)^{3/4})tan^{-1}(frac{2tan(x)}{(i-1)^{3/2}})-4sin^{-1}(frac{1+i}{sqrt{2}})tanh^{-1}(sqrt{-1-i}tan(x)))$
$endgroup$
– Kemono Chen
Jan 14 at 4:58
$begingroup$
$tiny I=xtan^{-1}(sin^2(x))-frac{1}{4}i(i(text{Li}_2(((1-2i)+2sqrt{-1-i})e^{2ix})+text{Li}_2(((1-2i)-2sqrt{-1-i})e^{2ix}))-i(text{Li}_2(((1+2i)+2sqrt{-1+i})e^{2ix})+text{Li}_2(((1+2i)-2sqrt{-1+i})e^{2ix}))-2log(1+(-1+2i+2sqrt{-1-i})e^{2ix})(x-sin^{-1}(sqrt[4]{-1}))-2log(1+((-1+2i)-2sqrt{-1-i})e^{2ix})(x+sin^{-1}(sqrt[4]{-1}))+2log(1+(-1-2i+2sqrt{-1+i})e^{2ix})(x+sin^{-1}((-1)^{3/4}))+2log(1-((1+2i)+2sqrt{i-1})e^{2ix})(x-sin^{-1}((-1)^{3/4}))-4isin^{-1}((-1)^{3/4})tan^{-1}(frac{2tan(x)}{(i-1)^{3/2}})-4sin^{-1}(frac{1+i}{sqrt{2}})tanh^{-1}(sqrt{-1-i}tan(x)))$
$endgroup$
– Kemono Chen
Jan 14 at 4:58
$begingroup$
where $I=intarctansin^2xdx$. Combining with that the integrand is even, the original integral is not elementary.
$endgroup$
– Kemono Chen
Jan 14 at 5:00
$begingroup$
where $I=intarctansin^2xdx$. Combining with that the integrand is even, the original integral is not elementary.
$endgroup$
– Kemono Chen
Jan 14 at 5:00
$begingroup$
In your "uv" part, you still have an $x$, inside the $sin.$
$endgroup$
– B. Goddard
Jan 14 at 12:21
$begingroup$
In your "uv" part, you still have an $x$, inside the $sin.$
$endgroup$
– B. Goddard
Jan 14 at 12:21
$begingroup$
Wait, are you not allowed to do integration by parts on non elementary functions?
$endgroup$
– Milo Moses
Jan 14 at 16:43
$begingroup$
Wait, are you not allowed to do integration by parts on non elementary functions?
$endgroup$
– Milo Moses
Jan 14 at 16:43
|
show 1 more comment
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2
$begingroup$
I don't think it's integral exists in elementary functions.
$endgroup$
– Archis Welankar
Jan 14 at 4:38
$begingroup$
$tiny I=xtan^{-1}(sin^2(x))-frac{1}{4}i(i(text{Li}_2(((1-2i)+2sqrt{-1-i})e^{2ix})+text{Li}_2(((1-2i)-2sqrt{-1-i})e^{2ix}))-i(text{Li}_2(((1+2i)+2sqrt{-1+i})e^{2ix})+text{Li}_2(((1+2i)-2sqrt{-1+i})e^{2ix}))-2log(1+(-1+2i+2sqrt{-1-i})e^{2ix})(x-sin^{-1}(sqrt[4]{-1}))-2log(1+((-1+2i)-2sqrt{-1-i})e^{2ix})(x+sin^{-1}(sqrt[4]{-1}))+2log(1+(-1-2i+2sqrt{-1+i})e^{2ix})(x+sin^{-1}((-1)^{3/4}))+2log(1-((1+2i)+2sqrt{i-1})e^{2ix})(x-sin^{-1}((-1)^{3/4}))-4isin^{-1}((-1)^{3/4})tan^{-1}(frac{2tan(x)}{(i-1)^{3/2}})-4sin^{-1}(frac{1+i}{sqrt{2}})tanh^{-1}(sqrt{-1-i}tan(x)))$
$endgroup$
– Kemono Chen
Jan 14 at 4:58
$begingroup$
where $I=intarctansin^2xdx$. Combining with that the integrand is even, the original integral is not elementary.
$endgroup$
– Kemono Chen
Jan 14 at 5:00
$begingroup$
In your "uv" part, you still have an $x$, inside the $sin.$
$endgroup$
– B. Goddard
Jan 14 at 12:21
$begingroup$
Wait, are you not allowed to do integration by parts on non elementary functions?
$endgroup$
– Milo Moses
Jan 14 at 16:43