Mixing
trigonometric integral with tangent
$begingroup$
$$intbiggl( sqrt[3]{frac{sin x+1}{cos x}}+sqrt[3]frac{sin x-1}{cos x}biggr)frac{1}{cos^2x},dx ,;xinBigl(-frac{pi}{2},frac{pi}{2}Bigr)$$ Can somebody give some tips about how can I evaluate this integral,please? I observed that $frac{1}{cos^2x}$ is the derivative of tangent, but I don't know how to do it using the substitution $t=tan x$ because there will be $frac{1}{cos x}$ inside and I think there is another method.
calculus integration
edited Jan 12 at 19:52
Bernard
121k740116
asked Jan 12 at 19:49
GaboruGaboru
877
$endgroup$
add a comment |
$begingroup$
$$intbiggl( sqrt[3]{frac{sin x+1}{cos x}}+sqrt[3]frac{sin x-1}{cos x}biggr)frac{1}{cos^2x},dx ,;xinBigl(-frac{pi}{2},frac{pi}{2}Bigr)$$ Can somebody give some tips about how can I evaluate this integral,please? I observed that $frac{1}{cos^2x}$ is the derivative of tangent, but I don't know how to do it using the substitution $t=tan x$ because there will be $frac{1}{cos x}$ inside and I think there is another method.
calculus integration
edited Jan 12 at 19:52
Bernard
121k740116
asked Jan 12 at 19:49
GaboruGaboru
877
$endgroup$
$begingroup$
Are you trying to evaluate the integral with the limits $int_{-pi/2}^{pi/2}$, or the indefinite integral?
$endgroup$
– John Doe
Jan 12 at 20:09
$begingroup$
Maybe it should be $frac{1+sin{x}}{cos^2x}$ instead $frac{1}{cos^2x}$?
$endgroup$
– Michael Rozenberg
Jan 12 at 20:11
$begingroup$
Notice, that, when you substitute $x + pi$ for $x$ in the integrand, you get the same function, so according to wikipedia, a sensible change of variable will be $tan$.
$endgroup$
– Viktor Glombik
Jan 12 at 20:17
add a comment |
$begingroup$
$$intbiggl( sqrt[3]{frac{sin x+1}{cos x}}+sqrt[3]frac{sin x-1}{cos x}biggr)frac{1}{cos^2x},dx ,;xinBigl(-frac{pi}{2},frac{pi}{2}Bigr)$$ Can somebody give some tips about how can I evaluate this integral,please? I observed that $frac{1}{cos^2x}$ is the derivative of tangent, but I don't know how to do it using the substitution $t=tan x$ because there will be $frac{1}{cos x}$ inside and I think there is another method.
calculus integration
edited Jan 12 at 19:52
Bernard
121k740116
asked Jan 12 at 19:49
GaboruGaboru
877
$endgroup$
$$intbiggl( sqrt[3]{frac{sin x+1}{cos x}}+sqrt[3]frac{sin x-1}{cos x}biggr)frac{1}{cos^2x},dx ,;xinBigl(-frac{pi}{2},frac{pi}{2}Bigr)$$ Can somebody give some tips about how can I evaluate this integral,please? I observed that $frac{1}{cos^2x}$ is the derivative of tangent, but I don't know how to do it using the substitution $t=tan x$ because there will be $frac{1}{cos x}$ inside and I think there is another method.
calculus integration
calculus integration
edited Jan 12 at 19:52
Bernard
121k740116
asked Jan 12 at 19:49
GaboruGaboru
877
edited Jan 12 at 19:52
Bernard
121k740116
asked Jan 12 at 19:49
GaboruGaboru
877
edited Jan 12 at 19:52
Bernard
121k740116
edited Jan 12 at 19:52
Bernard
121k740116
edited Jan 12 at 19:52
Bernard
121k740116
121k740116
asked Jan 12 at 19:49
GaboruGaboru
877
asked Jan 12 at 19:49
GaboruGaboru
877
asked Jan 12 at 19:49
GaboruGaboru
877
877
$begingroup$
Are you trying to evaluate the integral with the limits $int_{-pi/2}^{pi/2}$, or the indefinite integral?
$endgroup$
– John Doe
Jan 12 at 20:09
$begingroup$
Maybe it should be $frac{1+sin{x}}{cos^2x}$ instead $frac{1}{cos^2x}$?
$endgroup$
– Michael Rozenberg
Jan 12 at 20:11
$begingroup$
Notice, that, when you substitute $x + pi$ for $x$ in the integrand, you get the same function, so according to wikipedia, a sensible change of variable will be $tan$.
$endgroup$
– Viktor Glombik
Jan 12 at 20:17
add a comment |
$begingroup$
Are you trying to evaluate the integral with the limits $int_{-pi/2}^{pi/2}$, or the indefinite integral?
$endgroup$
– John Doe
Jan 12 at 20:09
$begingroup$
Maybe it should be $frac{1+sin{x}}{cos^2x}$ instead $frac{1}{cos^2x}$?
$endgroup$
– Michael Rozenberg
Jan 12 at 20:11
$begingroup$
Notice, that, when you substitute $x + pi$ for $x$ in the integrand, you get the same function, so according to wikipedia, a sensible change of variable will be $tan$.
$endgroup$
– Viktor Glombik
Jan 12 at 20:17
$begingroup$
Are you trying to evaluate the integral with the limits $int_{-pi/2}^{pi/2}$, or the indefinite integral?
$endgroup$
– John Doe
Jan 12 at 20:09
$begingroup$
Are you trying to evaluate the integral with the limits $int_{-pi/2}^{pi/2}$, or the indefinite integral?
$endgroup$
– John Doe
Jan 12 at 20:09
$begingroup$
Maybe it should be $frac{1+sin{x}}{cos^2x}$ instead $frac{1}{cos^2x}$?
$endgroup$
– Michael Rozenberg
Jan 12 at 20:11
$begingroup$
Maybe it should be $frac{1+sin{x}}{cos^2x}$ instead $frac{1}{cos^2x}$?
$endgroup$
– Michael Rozenberg
Jan 12 at 20:11
$begingroup$
Notice, that, when you substitute $x + pi$ for $x$ in the integrand, you get the same function, so according to wikipedia, a sensible change of variable will be $tan$.
$endgroup$
– Viktor Glombik
Jan 12 at 20:17
$begingroup$
Notice, that, when you substitute $x + pi$ for $x$ in the integrand, you get the same function, so according to wikipedia, a sensible change of variable will be $tan$.
$endgroup$
– Viktor Glombik
Jan 12 at 20:17
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
I will use integration by parts twice. First we write
begin{align}
I &= int left (sqrt[3]{frac{sin x + 1}{cos x}} + sqrt[3]{frac{sin x - 1}{cos x}} right ) sec^2 x , dx\
&= int big{(}sqrt[3]{tan x + sec x} + sqrt[3]{tan x - sec x} big{)} sec^2 x , dx.
end{align}
If we let
$$f(x) = sqrt[3]{tan x + sec x} + sqrt[3]{tan x - sec x}$$
and
$$g(x) = sqrt[3]{tan x + sec x} - sqrt[3]{tan x - sec x},$$
then obsereve that
$$f'(x) = frac{1}{3} sec x cdot g(x) quad text{and} quad g'(x) = frac{1}{3} sec x cdot f(x).$$
So for the integral we have
begin{align}
I &= int sec^2 x cdot f(x) , dx\
&= f(x) cdot tan x - int tan x cdot f'(x) , dx qquad text{(by parts)}\
&= f(x) cdot tan x - frac{1}{3} int sec x tan x cdot g(x) , dx \
&= f(x) cdot tan x - frac{1}{3} sec x cdot g(x) + frac{1}{3} int sec x cdot g'(x) , dx quad text{(by parts)}\
&= f(x) cdot tan x - frac{1}{3} sec x cdot g(x) + frac{1}{9} int sec^2 x cdot f(x) , dx\
&= f(x) cdot tan x - frac{1}{3} sec x cdot g(x) + frac{1}{9} I,
end{align}
giving
$$I = frac{9}{8} tan x cdot f(x) - frac{3}{8} sec x cdot g(x) + C$$
or
begin{align}
I &= frac{9}{8} tan x big{(} sqrt[3]{tan x + sec x} + sqrt[3]{tan x - sec x} big{)}\
& qquad - frac{3}{8} sec x big{(} sqrt[3]{tan x + sec x} - sqrt[3]{tan x - sec x} big{)} + C.
end{align}
answered Jan 13 at 9:48
omegadotomegadot
5,8372728
$endgroup$
add a comment |
$begingroup$
First note that if $f$ is a function with antiderivative $F$, then the antiderivative of its inverse function $f^{-1}(x)$ is $F( f^{-1}(x))-xf^{-1}(x)$, which can be easily shown by differentiating.
Now let's consider your integral.
Making a trigonometric substitution allows us to consider instead the following integrand:
$$s(x)=sqrt[3]{x+sqrt{x^2+1}}+sqrt[3]{x-sqrt{x^2+1}}$$
Notice that $s$ satisfies the property
$$s=sqrt[3]{2x-3s}$$
or
$$s^3+3s=2x$$
If $t$ is the inverse function of $s$ so that $s(t(x))=x$ and $t(s(x))=x$, we have from the above equation that
$$2t=x^3+3x$$
Thus, we may use the formula mentioned at the beginning of the post to solve your integral. The antiderivative of $t$ with respect to $x$ is
$$T(x)=frac{x^4+6x^2}{8}$$
and so the antiderivative of $s$ is given by
$$S(x)=T(s(x))-xs(x)$$
where $T$ and $s$ are defined above. This yields the desired result, and now all you have to do is undo the trig substitution at the beginning, and the value of your integral is
$$T(s(tan(x)))-tan(x)s(tan(x))$$
answered Jan 12 at 20:17
FrpzzdFrpzzd
23k841109
$endgroup$
$begingroup$
Is the approach taken here a known method to approach a certain family(ies) of integrals? (similar to say the Half Tangent substitution method for rational trigonometric functions)
$endgroup$
– DavidG
Jan 13 at 0:00
1
$begingroup$
@DavidG Not that I know of. I just worked it out from scratch.
$endgroup$
– Frpzzd
Jan 13 at 0:36
$begingroup$
How did you get that expresion of s(x) ?
$endgroup$
– Gaboru
Jan 13 at 1:36
add a comment |
$begingroup$
Hint: Substitute $$sin(x)=frac{2t}{1+t^2}$$
$$cos(x)=frac{1-t^2}{1+t^2}$$
$$dx=frac{2dt}{1+t^2}$$
answered Jan 12 at 20:19
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.2k42865
$endgroup$
1
$begingroup$
NONONONONO it makes things wayyyyy worse
$endgroup$
– clathratus
Jan 12 at 20:49
$begingroup$
This gives $$intleft(sqrt[3]frac{1+t}{1-t}+sqrt[3]frac{t-1}{t+1}right)frac{2(1+t^2)}{(1-t^2)^2},dt$$...what then?
$endgroup$
– John Doe
Jan 12 at 21:27
$begingroup$
Now i would substitute the roots.
$endgroup$
– Dr. Sonnhard Graubner
Jan 12 at 21:35
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I will use integration by parts twice. First we write
begin{align}
I &= int left (sqrt[3]{frac{sin x + 1}{cos x}} + sqrt[3]{frac{sin x - 1}{cos x}} right ) sec^2 x , dx\
&= int big{(}sqrt[3]{tan x + sec x} + sqrt[3]{tan x - sec x} big{)} sec^2 x , dx.
end{align}
If we let
$$f(x) = sqrt[3]{tan x + sec x} + sqrt[3]{tan x - sec x}$$
and
$$g(x) = sqrt[3]{tan x + sec x} - sqrt[3]{tan x - sec x},$$
then obsereve that
$$f'(x) = frac{1}{3} sec x cdot g(x) quad text{and} quad g'(x) = frac{1}{3} sec x cdot f(x).$$
So for the integral we have
begin{align}
I &= int sec^2 x cdot f(x) , dx\
&= f(x) cdot tan x - int tan x cdot f'(x) , dx qquad text{(by parts)}\
&= f(x) cdot tan x - frac{1}{3} int sec x tan x cdot g(x) , dx \
&= f(x) cdot tan x - frac{1}{3} sec x cdot g(x) + frac{1}{3} int sec x cdot g'(x) , dx quad text{(by parts)}\
&= f(x) cdot tan x - frac{1}{3} sec x cdot g(x) + frac{1}{9} int sec^2 x cdot f(x) , dx\
&= f(x) cdot tan x - frac{1}{3} sec x cdot g(x) + frac{1}{9} I,
end{align}
giving
$$I = frac{9}{8} tan x cdot f(x) - frac{3}{8} sec x cdot g(x) + C$$
or
begin{align}
I &= frac{9}{8} tan x big{(} sqrt[3]{tan x + sec x} + sqrt[3]{tan x - sec x} big{)}\
& qquad - frac{3}{8} sec x big{(} sqrt[3]{tan x + sec x} - sqrt[3]{tan x - sec x} big{)} + C.
end{align}
answered Jan 13 at 9:48
omegadotomegadot
5,8372728
$endgroup$
add a comment |
$begingroup$
I will use integration by parts twice. First we write
begin{align}
I &= int left (sqrt[3]{frac{sin x + 1}{cos x}} + sqrt[3]{frac{sin x - 1}{cos x}} right ) sec^2 x , dx\
&= int big{(}sqrt[3]{tan x + sec x} + sqrt[3]{tan x - sec x} big{)} sec^2 x , dx.
end{align}
If we let
$$f(x) = sqrt[3]{tan x + sec x} + sqrt[3]{tan x - sec x}$$
and
$$g(x) = sqrt[3]{tan x + sec x} - sqrt[3]{tan x - sec x},$$
then obsereve that
$$f'(x) = frac{1}{3} sec x cdot g(x) quad text{and} quad g'(x) = frac{1}{3} sec x cdot f(x).$$
So for the integral we have
begin{align}
I &= int sec^2 x cdot f(x) , dx\
&= f(x) cdot tan x - int tan x cdot f'(x) , dx qquad text{(by parts)}\
&= f(x) cdot tan x - frac{1}{3} int sec x tan x cdot g(x) , dx \
&= f(x) cdot tan x - frac{1}{3} sec x cdot g(x) + frac{1}{3} int sec x cdot g'(x) , dx quad text{(by parts)}\
&= f(x) cdot tan x - frac{1}{3} sec x cdot g(x) + frac{1}{9} int sec^2 x cdot f(x) , dx\
&= f(x) cdot tan x - frac{1}{3} sec x cdot g(x) + frac{1}{9} I,
end{align}
giving
$$I = frac{9}{8} tan x cdot f(x) - frac{3}{8} sec x cdot g(x) + C$$
or
begin{align}
I &= frac{9}{8} tan x big{(} sqrt[3]{tan x + sec x} + sqrt[3]{tan x - sec x} big{)}\
& qquad - frac{3}{8} sec x big{(} sqrt[3]{tan x + sec x} - sqrt[3]{tan x - sec x} big{)} + C.
end{align}
answered Jan 13 at 9:48
omegadotomegadot
5,8372728
$endgroup$
add a comment |
$begingroup$
I will use integration by parts twice. First we write
begin{align}
I &= int left (sqrt[3]{frac{sin x + 1}{cos x}} + sqrt[3]{frac{sin x - 1}{cos x}} right ) sec^2 x , dx\
&= int big{(}sqrt[3]{tan x + sec x} + sqrt[3]{tan x - sec x} big{)} sec^2 x , dx.
end{align}
If we let
$$f(x) = sqrt[3]{tan x + sec x} + sqrt[3]{tan x - sec x}$$
and
$$g(x) = sqrt[3]{tan x + sec x} - sqrt[3]{tan x - sec x},$$
then obsereve that
$$f'(x) = frac{1}{3} sec x cdot g(x) quad text{and} quad g'(x) = frac{1}{3} sec x cdot f(x).$$
So for the integral we have
begin{align}
I &= int sec^2 x cdot f(x) , dx\
&= f(x) cdot tan x - int tan x cdot f'(x) , dx qquad text{(by parts)}\
&= f(x) cdot tan x - frac{1}{3} int sec x tan x cdot g(x) , dx \
&= f(x) cdot tan x - frac{1}{3} sec x cdot g(x) + frac{1}{3} int sec x cdot g'(x) , dx quad text{(by parts)}\
&= f(x) cdot tan x - frac{1}{3} sec x cdot g(x) + frac{1}{9} int sec^2 x cdot f(x) , dx\
&= f(x) cdot tan x - frac{1}{3} sec x cdot g(x) + frac{1}{9} I,
end{align}
giving
$$I = frac{9}{8} tan x cdot f(x) - frac{3}{8} sec x cdot g(x) + C$$
or
begin{align}
I &= frac{9}{8} tan x big{(} sqrt[3]{tan x + sec x} + sqrt[3]{tan x - sec x} big{)}\
& qquad - frac{3}{8} sec x big{(} sqrt[3]{tan x + sec x} - sqrt[3]{tan x - sec x} big{)} + C.
end{align}
answered Jan 13 at 9:48
omegadotomegadot
5,8372728
$endgroup$
I will use integration by parts twice. First we write
begin{align}
I &= int left (sqrt[3]{frac{sin x + 1}{cos x}} + sqrt[3]{frac{sin x - 1}{cos x}} right ) sec^2 x , dx\
&= int big{(}sqrt[3]{tan x + sec x} + sqrt[3]{tan x - sec x} big{)} sec^2 x , dx.
end{align}
If we let
$$f(x) = sqrt[3]{tan x + sec x} + sqrt[3]{tan x - sec x}$$
and
$$g(x) = sqrt[3]{tan x + sec x} - sqrt[3]{tan x - sec x},$$
then obsereve that
$$f'(x) = frac{1}{3} sec x cdot g(x) quad text{and} quad g'(x) = frac{1}{3} sec x cdot f(x).$$
So for the integral we have
begin{align}
I &= int sec^2 x cdot f(x) , dx\
&= f(x) cdot tan x - int tan x cdot f'(x) , dx qquad text{(by parts)}\
&= f(x) cdot tan x - frac{1}{3} int sec x tan x cdot g(x) , dx \
&= f(x) cdot tan x - frac{1}{3} sec x cdot g(x) + frac{1}{3} int sec x cdot g'(x) , dx quad text{(by parts)}\
&= f(x) cdot tan x - frac{1}{3} sec x cdot g(x) + frac{1}{9} int sec^2 x cdot f(x) , dx\
&= f(x) cdot tan x - frac{1}{3} sec x cdot g(x) + frac{1}{9} I,
end{align}
giving
$$I = frac{9}{8} tan x cdot f(x) - frac{3}{8} sec x cdot g(x) + C$$
or
begin{align}
I &= frac{9}{8} tan x big{(} sqrt[3]{tan x + sec x} + sqrt[3]{tan x - sec x} big{)}\
& qquad - frac{3}{8} sec x big{(} sqrt[3]{tan x + sec x} - sqrt[3]{tan x - sec x} big{)} + C.
end{align}
answered Jan 13 at 9:48
omegadotomegadot
5,8372728
answered Jan 13 at 9:48
omegadotomegadot
5,8372728
answered Jan 13 at 9:48
omegadotomegadot
5,8372728
answered Jan 13 at 9:48
omegadotomegadot
5,8372728
5,8372728
add a comment |
add a comment |
$begingroup$
First note that if $f$ is a function with antiderivative $F$, then the antiderivative of its inverse function $f^{-1}(x)$ is $F( f^{-1}(x))-xf^{-1}(x)$, which can be easily shown by differentiating.
Now let's consider your integral.
Making a trigonometric substitution allows us to consider instead the following integrand:
$$s(x)=sqrt[3]{x+sqrt{x^2+1}}+sqrt[3]{x-sqrt{x^2+1}}$$
Notice that $s$ satisfies the property
$$s=sqrt[3]{2x-3s}$$
or
$$s^3+3s=2x$$
If $t$ is the inverse function of $s$ so that $s(t(x))=x$ and $t(s(x))=x$, we have from the above equation that
$$2t=x^3+3x$$
Thus, we may use the formula mentioned at the beginning of the post to solve your integral. The antiderivative of $t$ with respect to $x$ is
$$T(x)=frac{x^4+6x^2}{8}$$
and so the antiderivative of $s$ is given by
$$S(x)=T(s(x))-xs(x)$$
where $T$ and $s$ are defined above. This yields the desired result, and now all you have to do is undo the trig substitution at the beginning, and the value of your integral is
$$T(s(tan(x)))-tan(x)s(tan(x))$$
answered Jan 12 at 20:17
FrpzzdFrpzzd
23k841109
$endgroup$
$begingroup$
Is the approach taken here a known method to approach a certain family(ies) of integrals? (similar to say the Half Tangent substitution method for rational trigonometric functions)
$endgroup$
– DavidG
Jan 13 at 0:00
1
$begingroup$
@DavidG Not that I know of. I just worked it out from scratch.
$endgroup$
– Frpzzd
Jan 13 at 0:36
$begingroup$
How did you get that expresion of s(x) ?
$endgroup$
– Gaboru
Jan 13 at 1:36
add a comment |
$begingroup$
First note that if $f$ is a function with antiderivative $F$, then the antiderivative of its inverse function $f^{-1}(x)$ is $F( f^{-1}(x))-xf^{-1}(x)$, which can be easily shown by differentiating.
Now let's consider your integral.
Making a trigonometric substitution allows us to consider instead the following integrand:
$$s(x)=sqrt[3]{x+sqrt{x^2+1}}+sqrt[3]{x-sqrt{x^2+1}}$$
Notice that $s$ satisfies the property
$$s=sqrt[3]{2x-3s}$$
or
$$s^3+3s=2x$$
If $t$ is the inverse function of $s$ so that $s(t(x))=x$ and $t(s(x))=x$, we have from the above equation that
$$2t=x^3+3x$$
Thus, we may use the formula mentioned at the beginning of the post to solve your integral. The antiderivative of $t$ with respect to $x$ is
$$T(x)=frac{x^4+6x^2}{8}$$
and so the antiderivative of $s$ is given by
$$S(x)=T(s(x))-xs(x)$$
where $T$ and $s$ are defined above. This yields the desired result, and now all you have to do is undo the trig substitution at the beginning, and the value of your integral is
$$T(s(tan(x)))-tan(x)s(tan(x))$$
answered Jan 12 at 20:17
FrpzzdFrpzzd
23k841109
$endgroup$
$begingroup$
Is the approach taken here a known method to approach a certain family(ies) of integrals? (similar to say the Half Tangent substitution method for rational trigonometric functions)
$endgroup$
– DavidG
Jan 13 at 0:00
1
$begingroup$
@DavidG Not that I know of. I just worked it out from scratch.
$endgroup$
– Frpzzd
Jan 13 at 0:36
$begingroup$
How did you get that expresion of s(x) ?
$endgroup$
– Gaboru
Jan 13 at 1:36
add a comment |
$begingroup$
First note that if $f$ is a function with antiderivative $F$, then the antiderivative of its inverse function $f^{-1}(x)$ is $F( f^{-1}(x))-xf^{-1}(x)$, which can be easily shown by differentiating.
Now let's consider your integral.
Making a trigonometric substitution allows us to consider instead the following integrand:
$$s(x)=sqrt[3]{x+sqrt{x^2+1}}+sqrt[3]{x-sqrt{x^2+1}}$$
Notice that $s$ satisfies the property
$$s=sqrt[3]{2x-3s}$$
or
$$s^3+3s=2x$$
If $t$ is the inverse function of $s$ so that $s(t(x))=x$ and $t(s(x))=x$, we have from the above equation that
$$2t=x^3+3x$$
Thus, we may use the formula mentioned at the beginning of the post to solve your integral. The antiderivative of $t$ with respect to $x$ is
$$T(x)=frac{x^4+6x^2}{8}$$
and so the antiderivative of $s$ is given by
$$S(x)=T(s(x))-xs(x)$$
where $T$ and $s$ are defined above. This yields the desired result, and now all you have to do is undo the trig substitution at the beginning, and the value of your integral is
$$T(s(tan(x)))-tan(x)s(tan(x))$$
answered Jan 12 at 20:17
FrpzzdFrpzzd
23k841109
$endgroup$
First note that if $f$ is a function with antiderivative $F$, then the antiderivative of its inverse function $f^{-1}(x)$ is $F( f^{-1}(x))-xf^{-1}(x)$, which can be easily shown by differentiating.
Now let's consider your integral.
Making a trigonometric substitution allows us to consider instead the following integrand:
$$s(x)=sqrt[3]{x+sqrt{x^2+1}}+sqrt[3]{x-sqrt{x^2+1}}$$
Notice that $s$ satisfies the property
$$s=sqrt[3]{2x-3s}$$
or
$$s^3+3s=2x$$
If $t$ is the inverse function of $s$ so that $s(t(x))=x$ and $t(s(x))=x$, we have from the above equation that
$$2t=x^3+3x$$
Thus, we may use the formula mentioned at the beginning of the post to solve your integral. The antiderivative of $t$ with respect to $x$ is
$$T(x)=frac{x^4+6x^2}{8}$$
and so the antiderivative of $s$ is given by
$$S(x)=T(s(x))-xs(x)$$
where $T$ and $s$ are defined above. This yields the desired result, and now all you have to do is undo the trig substitution at the beginning, and the value of your integral is
$$T(s(tan(x)))-tan(x)s(tan(x))$$
answered Jan 12 at 20:17
FrpzzdFrpzzd
23k841109
answered Jan 12 at 20:17
FrpzzdFrpzzd
23k841109
answered Jan 12 at 20:17
FrpzzdFrpzzd
23k841109
answered Jan 12 at 20:17
FrpzzdFrpzzd
23k841109
23k841109
$begingroup$
Is the approach taken here a known method to approach a certain family(ies) of integrals? (similar to say the Half Tangent substitution method for rational trigonometric functions)
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– DavidG
Jan 13 at 0:00
1
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@DavidG Not that I know of. I just worked it out from scratch.
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– Frpzzd
Jan 13 at 0:36
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How did you get that expresion of s(x) ?
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– Gaboru
Jan 13 at 1:36
add a comment |
$begingroup$
Is the approach taken here a known method to approach a certain family(ies) of integrals? (similar to say the Half Tangent substitution method for rational trigonometric functions)
$endgroup$
– DavidG
Jan 13 at 0:00
1
$begingroup$
@DavidG Not that I know of. I just worked it out from scratch.
$endgroup$
– Frpzzd
Jan 13 at 0:36
$begingroup$
How did you get that expresion of s(x) ?
$endgroup$
– Gaboru
Jan 13 at 1:36
$begingroup$
Is the approach taken here a known method to approach a certain family(ies) of integrals? (similar to say the Half Tangent substitution method for rational trigonometric functions)
$endgroup$
– DavidG
Jan 13 at 0:00
$begingroup$
Is the approach taken here a known method to approach a certain family(ies) of integrals? (similar to say the Half Tangent substitution method for rational trigonometric functions)
$endgroup$
– DavidG
Jan 13 at 0:00
1
1
$begingroup$
@DavidG Not that I know of. I just worked it out from scratch.
$endgroup$
– Frpzzd
Jan 13 at 0:36
$begingroup$
@DavidG Not that I know of. I just worked it out from scratch.
$endgroup$
– Frpzzd
Jan 13 at 0:36
$begingroup$
How did you get that expresion of s(x) ?
$endgroup$
– Gaboru
Jan 13 at 1:36
$begingroup$
How did you get that expresion of s(x) ?
$endgroup$
– Gaboru
Jan 13 at 1:36
add a comment |
$begingroup$
Hint: Substitute $$sin(x)=frac{2t}{1+t^2}$$
$$cos(x)=frac{1-t^2}{1+t^2}$$
$$dx=frac{2dt}{1+t^2}$$
answered Jan 12 at 20:19
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.2k42865
$endgroup$
1
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NONONONONO it makes things wayyyyy worse
$endgroup$
– clathratus
Jan 12 at 20:49
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This gives $$intleft(sqrt[3]frac{1+t}{1-t}+sqrt[3]frac{t-1}{t+1}right)frac{2(1+t^2)}{(1-t^2)^2},dt$$...what then?
$endgroup$
– John Doe
Jan 12 at 21:27
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Now i would substitute the roots.
$endgroup$
– Dr. Sonnhard Graubner
Jan 12 at 21:35
add a comment |
$begingroup$
Hint: Substitute $$sin(x)=frac{2t}{1+t^2}$$
$$cos(x)=frac{1-t^2}{1+t^2}$$
$$dx=frac{2dt}{1+t^2}$$
answered Jan 12 at 20:19
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.2k42865
$endgroup$
1
$begingroup$
NONONONONO it makes things wayyyyy worse
$endgroup$
– clathratus
Jan 12 at 20:49
$begingroup$
This gives $$intleft(sqrt[3]frac{1+t}{1-t}+sqrt[3]frac{t-1}{t+1}right)frac{2(1+t^2)}{(1-t^2)^2},dt$$...what then?
$endgroup$
– John Doe
Jan 12 at 21:27
$begingroup$
Now i would substitute the roots.
$endgroup$
– Dr. Sonnhard Graubner
Jan 12 at 21:35
add a comment |
$begingroup$
Hint: Substitute $$sin(x)=frac{2t}{1+t^2}$$
$$cos(x)=frac{1-t^2}{1+t^2}$$
$$dx=frac{2dt}{1+t^2}$$
answered Jan 12 at 20:19
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.2k42865
$endgroup$
Hint: Substitute $$sin(x)=frac{2t}{1+t^2}$$
$$cos(x)=frac{1-t^2}{1+t^2}$$
$$dx=frac{2dt}{1+t^2}$$
answered Jan 12 at 20:19
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.2k42865
answered Jan 12 at 20:19
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.2k42865
answered Jan 12 at 20:19
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.2k42865
answered Jan 12 at 20:19
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.2k42865
75.2k42865
1
$begingroup$
NONONONONO it makes things wayyyyy worse
$endgroup$
– clathratus
Jan 12 at 20:49
$begingroup$
This gives $$intleft(sqrt[3]frac{1+t}{1-t}+sqrt[3]frac{t-1}{t+1}right)frac{2(1+t^2)}{(1-t^2)^2},dt$$...what then?
$endgroup$
– John Doe
Jan 12 at 21:27
$begingroup$
Now i would substitute the roots.
$endgroup$
– Dr. Sonnhard Graubner
Jan 12 at 21:35
add a comment |
1
$begingroup$
NONONONONO it makes things wayyyyy worse
$endgroup$
– clathratus
Jan 12 at 20:49
$begingroup$
This gives $$intleft(sqrt[3]frac{1+t}{1-t}+sqrt[3]frac{t-1}{t+1}right)frac{2(1+t^2)}{(1-t^2)^2},dt$$...what then?
$endgroup$
– John Doe
Jan 12 at 21:27
$begingroup$
Now i would substitute the roots.
$endgroup$
– Dr. Sonnhard Graubner
Jan 12 at 21:35
1
1
$begingroup$
NONONONONO it makes things wayyyyy worse
$endgroup$
– clathratus
Jan 12 at 20:49
$begingroup$
NONONONONO it makes things wayyyyy worse
$endgroup$
– clathratus
Jan 12 at 20:49
$begingroup$
This gives $$intleft(sqrt[3]frac{1+t}{1-t}+sqrt[3]frac{t-1}{t+1}right)frac{2(1+t^2)}{(1-t^2)^2},dt$$...what then?
$endgroup$
– John Doe
Jan 12 at 21:27
$begingroup$
This gives $$intleft(sqrt[3]frac{1+t}{1-t}+sqrt[3]frac{t-1}{t+1}right)frac{2(1+t^2)}{(1-t^2)^2},dt$$...what then?
$endgroup$
– John Doe
Jan 12 at 21:27
$begingroup$
Now i would substitute the roots.
$endgroup$
– Dr. Sonnhard Graubner
Jan 12 at 21:35
$begingroup$
Now i would substitute the roots.
$endgroup$
– Dr. Sonnhard Graubner
Jan 12 at 21:35
add a comment |
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$begingroup$
Are you trying to evaluate the integral with the limits $int_{-pi/2}^{pi/2}$, or the indefinite integral?
$endgroup$
– John Doe
Jan 12 at 20:09
$begingroup$
Maybe it should be $frac{1+sin{x}}{cos^2x}$ instead $frac{1}{cos^2x}$?
$endgroup$
– Michael Rozenberg
Jan 12 at 20:11
$begingroup$
Notice, that, when you substitute $x + pi$ for $x$ in the integrand, you get the same function, so according to wikipedia, a sensible change of variable will be $tan$.
$endgroup$
– Viktor Glombik
Jan 12 at 20:17