What method has to be used to integrate this? Seems to be non-integrable, but is. Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Integral of $int frac1sqrtx(1-x) dx $Evaluate $intfraccos xsin x + cos x,textdx$.Definite integration problem (trig).Differentiating a function and using the result to calculate the indefinite integral of another.How do you evaluate an integral with an absolute value?Evaluate $int_0^4 fracln xsqrt4x-x^2 ,mathrm dx$Anti-derivative of continuous function $frac12+sin x$Evaluating the integral: $intlimits_0^inftyleft(fracsin(ax)xright)^2 dx , a neq 0 $The indefinite integral $intfracoperatornameLi_2(x)1+sqrtx,dx$: what is the strategy to get such indefinite integralHow to evaluate $int_pi/2^pi sqrt1 - frac12cos^2 x + sin x sin 2x ;mathrmdx$
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What method has to be used to integrate this? Seems to be non-integrable, but is.
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Integral of $int frac1sqrtx(1-x) dx $Evaluate $intfraccos xsin x + cos x,textdx$.Definite integration problem (trig).Differentiating a function and using the result to calculate the indefinite integral of another.How do you evaluate an integral with an absolute value?Evaluate $int_0^4 fracln xsqrt4x-x^2 ,mathrm dx$Anti-derivative of continuous function $frac12+sin x$Evaluating the integral: $intlimits_0^inftyleft(fracsin(ax)xright)^2 dx , a neq 0 $The indefinite integral $intfracoperatornameLi_2(x)1+sqrtx,dx$: what is the strategy to get such indefinite integralHow to evaluate $int_pi/2^pi sqrt1 - frac12cos^2 x + sin x sin 2x ;mathrmdx$
$begingroup$
I'm doing an special assignment for my calculus class and I have to integrate the following in order to obtain the distance of a certain epicycloid from $0$ to $2pi$. I don't believe the specific epicycloid is relevant to the question. Heres the integral:
$$4 int_0^fracpi2 sqrt(-5sin(t)+5sin(5t))^2 + (5cos(t)-5cos(5t))^2 ; mathrmdt $$
Wolfram Alpha tells me the result is $40$, however when I try to apply Barrow to the indefinite integral, I get the result is undefined. This makes sense, since the indefinite integral is
$$ frac-5 cos(2 t) sqrtsin^2(2 t)sin(2 t)$$
and the sine of $2 cdot fracpi2$ and the sine of $0$ is, obviously, $0$. So that's undefined, can't be solved. However Wolfram comes up with $40$ and I've got no idea how.
calculus integration
edited Apr 8 at 21:32
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asked Apr 8 at 20:54
JereJere
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add a comment |
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I'm doing an special assignment for my calculus class and I have to integrate the following in order to obtain the distance of a certain epicycloid from $0$ to $2pi$. I don't believe the specific epicycloid is relevant to the question. Heres the integral:
$$4 int_0^fracpi2 sqrt(-5sin(t)+5sin(5t))^2 + (5cos(t)-5cos(5t))^2 ; mathrmdt $$
Wolfram Alpha tells me the result is $40$, however when I try to apply Barrow to the indefinite integral, I get the result is undefined. This makes sense, since the indefinite integral is
$$ frac-5 cos(2 t) sqrtsin^2(2 t)sin(2 t)$$
and the sine of $2 cdot fracpi2$ and the sine of $0$ is, obviously, $0$. So that's undefined, can't be solved. However Wolfram comes up with $40$ and I've got no idea how.
calculus integration
edited Apr 8 at 21:32
user
6,57011031
asked Apr 8 at 20:54
JereJere
11
New contributor
Jere is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Why not factor out the $5$s immediately?
$endgroup$
– David G. Stork
Apr 8 at 21:13
add a comment |
$begingroup$
I'm doing an special assignment for my calculus class and I have to integrate the following in order to obtain the distance of a certain epicycloid from $0$ to $2pi$. I don't believe the specific epicycloid is relevant to the question. Heres the integral:
$$4 int_0^fracpi2 sqrt(-5sin(t)+5sin(5t))^2 + (5cos(t)-5cos(5t))^2 ; mathrmdt $$
Wolfram Alpha tells me the result is $40$, however when I try to apply Barrow to the indefinite integral, I get the result is undefined. This makes sense, since the indefinite integral is
$$ frac-5 cos(2 t) sqrtsin^2(2 t)sin(2 t)$$
and the sine of $2 cdot fracpi2$ and the sine of $0$ is, obviously, $0$. So that's undefined, can't be solved. However Wolfram comes up with $40$ and I've got no idea how.
calculus integration
edited Apr 8 at 21:32
user
6,57011031
asked Apr 8 at 20:54
JereJere
11
New contributor
Jere is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I'm doing an special assignment for my calculus class and I have to integrate the following in order to obtain the distance of a certain epicycloid from $0$ to $2pi$. I don't believe the specific epicycloid is relevant to the question. Heres the integral:
$$4 int_0^fracpi2 sqrt(-5sin(t)+5sin(5t))^2 + (5cos(t)-5cos(5t))^2 ; mathrmdt $$
Wolfram Alpha tells me the result is $40$, however when I try to apply Barrow to the indefinite integral, I get the result is undefined. This makes sense, since the indefinite integral is
$$ frac-5 cos(2 t) sqrtsin^2(2 t)sin(2 t)$$
and the sine of $2 cdot fracpi2$ and the sine of $0$ is, obviously, $0$. So that's undefined, can't be solved. However Wolfram comes up with $40$ and I've got no idea how.
calculus integration
calculus integration
edited Apr 8 at 21:32
user
6,57011031
asked Apr 8 at 20:54
JereJere
11
New contributor
Jere is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Apr 8 at 21:32
user
6,57011031
asked Apr 8 at 20:54
JereJere
11
New contributor
Jere is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Apr 8 at 21:32
user
6,57011031
edited Apr 8 at 21:32
user
6,57011031
edited Apr 8 at 21:32
user
6,57011031
6,57011031
asked Apr 8 at 20:54
JereJere
11
New contributor
Jere is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 8 at 20:54
JereJere
11
asked Apr 8 at 20:54
JereJere
11
11
New contributor
Jere is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Jere is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Jere is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Why not factor out the $5$s immediately?
$endgroup$
– David G. Stork
Apr 8 at 21:13
add a comment |
$begingroup$
Why not factor out the $5$s immediately?
$endgroup$
– David G. Stork
Apr 8 at 21:13
$begingroup$
Why not factor out the $5$s immediately?
$endgroup$
– David G. Stork
Apr 8 at 21:13
$begingroup$
Why not factor out the $5$s immediately?
$endgroup$
– David G. Stork
Apr 8 at 21:13
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
When in doubt, plot:
The integral is indeed $40$.
Factor the $5$ out of the integrand and note that
$$sqrt(- sin (t) + sin (5 t))^2 + (cos (t) - cos (5 t))^2 = sin^2 (2 t)$$
(The integral of $sin^2 (x)$ is straightforward.)
That the denominator vanishes at the endpoints of integration doesn't mean the full result does.
answered Apr 8 at 21:17
David G. StorkDavid G. Stork
12.2k41836
$endgroup$
add a comment |
$begingroup$
Whenever I see $sqrtsin^2(2t)$ in a context where $sin(2t)ge 0$, I always like to replace it with $sin(2t)$. This might help!
Having said that, if your indefinite integral is correct, you get the answer $-40$, which doesn't look right.
answered Apr 8 at 21:23
TonyKTonyK
44.1k358137
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add a comment |
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$begingroup$
When in doubt, plot:
The integral is indeed $40$.
Factor the $5$ out of the integrand and note that
$$sqrt(- sin (t) + sin (5 t))^2 + (cos (t) - cos (5 t))^2 = sin^2 (2 t)$$
(The integral of $sin^2 (x)$ is straightforward.)
That the denominator vanishes at the endpoints of integration doesn't mean the full result does.
answered Apr 8 at 21:17
David G. StorkDavid G. Stork
12.2k41836
$endgroup$
add a comment |
$begingroup$
When in doubt, plot:
The integral is indeed $40$.
Factor the $5$ out of the integrand and note that
$$sqrt(- sin (t) + sin (5 t))^2 + (cos (t) - cos (5 t))^2 = sin^2 (2 t)$$
(The integral of $sin^2 (x)$ is straightforward.)
That the denominator vanishes at the endpoints of integration doesn't mean the full result does.
answered Apr 8 at 21:17
David G. StorkDavid G. Stork
12.2k41836
$endgroup$
add a comment |
$begingroup$
When in doubt, plot:
The integral is indeed $40$.
Factor the $5$ out of the integrand and note that
$$sqrt(- sin (t) + sin (5 t))^2 + (cos (t) - cos (5 t))^2 = sin^2 (2 t)$$
(The integral of $sin^2 (x)$ is straightforward.)
That the denominator vanishes at the endpoints of integration doesn't mean the full result does.
answered Apr 8 at 21:17
David G. StorkDavid G. Stork
12.2k41836
$endgroup$
When in doubt, plot:
The integral is indeed $40$.
Factor the $5$ out of the integrand and note that
$$sqrt(- sin (t) + sin (5 t))^2 + (cos (t) - cos (5 t))^2 = sin^2 (2 t)$$
(The integral of $sin^2 (x)$ is straightforward.)
That the denominator vanishes at the endpoints of integration doesn't mean the full result does.
answered Apr 8 at 21:17
David G. StorkDavid G. Stork
12.2k41836
edited Apr 8 at 21:22
answered Apr 8 at 21:17
David G. StorkDavid G. Stork
12.2k41836
answered Apr 8 at 21:17
David G. StorkDavid G. Stork
12.2k41836
answered Apr 8 at 21:17
David G. StorkDavid G. Stork
12.2k41836
12.2k41836
add a comment |
add a comment |
$begingroup$
Whenever I see $sqrtsin^2(2t)$ in a context where $sin(2t)ge 0$, I always like to replace it with $sin(2t)$. This might help!
Having said that, if your indefinite integral is correct, you get the answer $-40$, which doesn't look right.
answered Apr 8 at 21:23
TonyKTonyK
44.1k358137
$endgroup$
add a comment |
$begingroup$
Whenever I see $sqrtsin^2(2t)$ in a context where $sin(2t)ge 0$, I always like to replace it with $sin(2t)$. This might help!
Having said that, if your indefinite integral is correct, you get the answer $-40$, which doesn't look right.
answered Apr 8 at 21:23
TonyKTonyK
44.1k358137
$endgroup$
add a comment |
$begingroup$
Whenever I see $sqrtsin^2(2t)$ in a context where $sin(2t)ge 0$, I always like to replace it with $sin(2t)$. This might help!
Having said that, if your indefinite integral is correct, you get the answer $-40$, which doesn't look right.
answered Apr 8 at 21:23
TonyKTonyK
44.1k358137
$endgroup$
Whenever I see $sqrtsin^2(2t)$ in a context where $sin(2t)ge 0$, I always like to replace it with $sin(2t)$. This might help!
Having said that, if your indefinite integral is correct, you get the answer $-40$, which doesn't look right.
answered Apr 8 at 21:23
TonyKTonyK
44.1k358137
answered Apr 8 at 21:23
TonyKTonyK
44.1k358137
answered Apr 8 at 21:23
TonyKTonyK
44.1k358137
answered Apr 8 at 21:23
TonyKTonyK
44.1k358137
44.1k358137
add a comment |
add a comment |
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$begingroup$
Why not factor out the $5$s immediately?
$endgroup$
– David G. Stork
Apr 8 at 21:13