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Integrate without expansion?
Finding $int^1_0 fraclog(1+x)xdx$ without series expansionSimplest way to integrate $int fracu^2+1(u^2-2u-1)^2mathrmdu$How do I integrate $x^frac32e^-x$ from 0 to inf?How to integrate $int_0^pi frac1a-b cos(x) dx$ with calculus tools?Multiple Integration order doesn't agree.Integrate area of function over a tetrahedronHow do I integrate $fracx+1log x$?How to integrate without trigonometric substitutionIntegrate over a triangle in the 2D normal distributionIntegration of an exponential function over a triangle. Numerically stable approximation to the infinite sum?
$begingroup$
I want to evaluate
$$ int_0^1 ( 1 - x^2)^10 dx $$
One way I can do this is by expanding out $(1 - x^2)^10$ term by term, but is there a better way to do this?
integration
asked 8 hours ago
MathStudentMathStudent
443
New contributor
MathStudent is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I want to evaluate
$$ int_0^1 ( 1 - x^2)^10 dx $$
One way I can do this is by expanding out $(1 - x^2)^10$ term by term, but is there a better way to do this?
integration
asked 8 hours ago
MathStudentMathStudent
443
New contributor
MathStudent is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Yes, by coming up with the reduction formula!
$endgroup$
– user209663
8 hours ago
add a comment |
$begingroup$
I want to evaluate
$$ int_0^1 ( 1 - x^2)^10 dx $$
One way I can do this is by expanding out $(1 - x^2)^10$ term by term, but is there a better way to do this?
integration
asked 8 hours ago
MathStudentMathStudent
443
New contributor
MathStudent is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I want to evaluate
$$ int_0^1 ( 1 - x^2)^10 dx $$
One way I can do this is by expanding out $(1 - x^2)^10$ term by term, but is there a better way to do this?
integration
integration
asked 8 hours ago
MathStudentMathStudent
443
New contributor
MathStudent is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
MathStudentMathStudent
443
New contributor
MathStudent is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
MathStudentMathStudent
443
New contributor
MathStudent is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
MathStudentMathStudent
443
asked 8 hours ago
MathStudentMathStudent
443
443
New contributor
MathStudent is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
MathStudent is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Yes, by coming up with the reduction formula!
$endgroup$
– user209663
8 hours ago
add a comment |
$begingroup$
Yes, by coming up with the reduction formula!
$endgroup$
– user209663
8 hours ago
$begingroup$
Yes, by coming up with the reduction formula!
$endgroup$
– user209663
8 hours ago
$begingroup$
Yes, by coming up with the reduction formula!
$endgroup$
– user209663
8 hours ago
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
When you encounter this type of problem, the first thing to do is to come up with a reduction formula. Let us find the reduction formula for $int(k^2 - x^2)^n dx $.
Using integration by parts:
$u = (k-x^2)^n $ and $dv = dx$ then $du = n(k^2 - x^2)^n-1(-2x)dx $ and $v =x$
Thus,
$$int(k^2 - x^2)^n dx = x(k^2 - x^2)^n-1 + 2n int x^2(k^2 - x^2)^n-1 \
= x(k^2 - x^2)^n-1 + 2n int bigg[ k^2(k^2 - x^2)^n-1 - (k^2 - x^2)^n bigg] dx \
= x(k^2 - x^2)^n-1 + 2n int k^2(k^2 - x^2)^n-1 dx - 2n int (k^2 - x^2)^n dx
$$
From here you can see that:
$$2nint(k^2 - x^2)^n dx + int(k^2 - x^2)^n dx = x(k^2 - x^2)^n-1 + 2n int k^2(k^2 - x^2)^n-1 dx - 2n int (k^2 - x^2)^n dx $$
Thus,
$$int(k^2 - x^2)^n dx = fracx(k^2 - x^2)^n-12n+1 + frac2nk^22n+1 int (k^2 - x^2)^n-1 dx$$
For your problem, $k=1$ and $n=10$. Can you take it from here?
answered 8 hours ago
user209663user209663
864411
$endgroup$
$begingroup$
wow! I knew there must be a shorter way. Thank you!
$endgroup$
– MathStudent
8 hours ago
$begingroup$
Is this really a shorter way? Evaluating all of the terms of the recursion seems like just as much effort as simply using the binomial expansion.
$endgroup$
– SZN
8 hours ago
$begingroup$
@MathStudent No problem.
$endgroup$
– user209663
8 hours ago
$begingroup$
@SZN Note that this is a definite integral... look at the recursion.... what happen to the first part? This is similar to $int_0^pi sin^n(x) dx$. You also get a recursion... but the trick is the bound of the integral. It makes it easier to evaluate
$endgroup$
– user209663
8 hours ago
$begingroup$
@user209663 what you say is perfectly fine. But by using the binomial expansion I can evaluate the integral in just a few lines without any recursion or argument about bounds.
$endgroup$
– SZN
8 hours ago
add a comment |
$begingroup$
This answer is exactly what the asker is NOT asking for. However, I believe it will be useful to show the asker that expanding out "term-by-term" is not actually that hard.
We have
beginalign
int_0^1 (1-x^2)^10dx &= int_0^1 sum_i=0^10 C(10,i)(-x^2)^idx \
%
& = sum_i=0^10 (-1)^iC(10,i)int_0^1 x^2idx\
%
&= sum_i=0^10 frac(-1)^iC(10,i)2i+1,
endalign
where $$C(n,m) = fracn!m!(n-m)!.$$
answered 8 hours ago
SZNSZN
3,172721
$endgroup$
add a comment |
$begingroup$
An alternate approach is through the beta function. We start by u-substitution and try to make our way towards an integral form of the beta function:
$$beginalign*
int_0^1(1-x^2)^10,mathrm dx &= frac12 int_0^1 (1-u)^10u^-1/2,mathrm du,qquad x=sqrtu \
&= frac12 int_0^1 (1-u)^11-1u^1/2-1,mathrm du\
&= frac12cdotfracGamma(11)Gamma(1/2)Gamma(11+1/2)\
&= frac12cdotfrac10!sqrtpifrac22!4^11cdot11!sqrtpi\
&= frac262144969969.
endalign*$$
answered 8 hours ago
dxdydzdxdydz
8571714
$endgroup$
1
$begingroup$
Upvoted. There's a special place in my heart for using special functions to solve problems that look like they only admit basic methods :-)
$endgroup$
– SZN
8 hours ago
add a comment |
$begingroup$
Similar to dxdydz's answer, let $x=sin(t)$
$$int_0^1 ( 1 - x^2)^10, dx=int_0^frac pi 2cos^21(t),dt$$ and remember that
$$int_0^frac pi 2cos^n(t),dt=fracsqrtpi 2 frac Gamma left(fracn+12right) Gamma
left(fracn+22right)$$
answered 57 mins ago
Claude LeiboviciClaude Leibovici
129k1158139
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
When you encounter this type of problem, the first thing to do is to come up with a reduction formula. Let us find the reduction formula for $int(k^2 - x^2)^n dx $.
Using integration by parts:
$u = (k-x^2)^n $ and $dv = dx$ then $du = n(k^2 - x^2)^n-1(-2x)dx $ and $v =x$
Thus,
$$int(k^2 - x^2)^n dx = x(k^2 - x^2)^n-1 + 2n int x^2(k^2 - x^2)^n-1 \
= x(k^2 - x^2)^n-1 + 2n int bigg[ k^2(k^2 - x^2)^n-1 - (k^2 - x^2)^n bigg] dx \
= x(k^2 - x^2)^n-1 + 2n int k^2(k^2 - x^2)^n-1 dx - 2n int (k^2 - x^2)^n dx
$$
From here you can see that:
$$2nint(k^2 - x^2)^n dx + int(k^2 - x^2)^n dx = x(k^2 - x^2)^n-1 + 2n int k^2(k^2 - x^2)^n-1 dx - 2n int (k^2 - x^2)^n dx $$
Thus,
$$int(k^2 - x^2)^n dx = fracx(k^2 - x^2)^n-12n+1 + frac2nk^22n+1 int (k^2 - x^2)^n-1 dx$$
For your problem, $k=1$ and $n=10$. Can you take it from here?
answered 8 hours ago
user209663user209663
864411
$endgroup$
$begingroup$
wow! I knew there must be a shorter way. Thank you!
$endgroup$
– MathStudent
8 hours ago
$begingroup$
Is this really a shorter way? Evaluating all of the terms of the recursion seems like just as much effort as simply using the binomial expansion.
$endgroup$
– SZN
8 hours ago
$begingroup$
@MathStudent No problem.
$endgroup$
– user209663
8 hours ago
$begingroup$
@SZN Note that this is a definite integral... look at the recursion.... what happen to the first part? This is similar to $int_0^pi sin^n(x) dx$. You also get a recursion... but the trick is the bound of the integral. It makes it easier to evaluate
$endgroup$
– user209663
8 hours ago
$begingroup$
@user209663 what you say is perfectly fine. But by using the binomial expansion I can evaluate the integral in just a few lines without any recursion or argument about bounds.
$endgroup$
– SZN
8 hours ago
add a comment |
$begingroup$
When you encounter this type of problem, the first thing to do is to come up with a reduction formula. Let us find the reduction formula for $int(k^2 - x^2)^n dx $.
Using integration by parts:
$u = (k-x^2)^n $ and $dv = dx$ then $du = n(k^2 - x^2)^n-1(-2x)dx $ and $v =x$
Thus,
$$int(k^2 - x^2)^n dx = x(k^2 - x^2)^n-1 + 2n int x^2(k^2 - x^2)^n-1 \
= x(k^2 - x^2)^n-1 + 2n int bigg[ k^2(k^2 - x^2)^n-1 - (k^2 - x^2)^n bigg] dx \
= x(k^2 - x^2)^n-1 + 2n int k^2(k^2 - x^2)^n-1 dx - 2n int (k^2 - x^2)^n dx
$$
From here you can see that:
$$2nint(k^2 - x^2)^n dx + int(k^2 - x^2)^n dx = x(k^2 - x^2)^n-1 + 2n int k^2(k^2 - x^2)^n-1 dx - 2n int (k^2 - x^2)^n dx $$
Thus,
$$int(k^2 - x^2)^n dx = fracx(k^2 - x^2)^n-12n+1 + frac2nk^22n+1 int (k^2 - x^2)^n-1 dx$$
For your problem, $k=1$ and $n=10$. Can you take it from here?
answered 8 hours ago
user209663user209663
864411
$endgroup$
$begingroup$
wow! I knew there must be a shorter way. Thank you!
$endgroup$
– MathStudent
8 hours ago
$begingroup$
Is this really a shorter way? Evaluating all of the terms of the recursion seems like just as much effort as simply using the binomial expansion.
$endgroup$
– SZN
8 hours ago
$begingroup$
@MathStudent No problem.
$endgroup$
– user209663
8 hours ago
$begingroup$
@SZN Note that this is a definite integral... look at the recursion.... what happen to the first part? This is similar to $int_0^pi sin^n(x) dx$. You also get a recursion... but the trick is the bound of the integral. It makes it easier to evaluate
$endgroup$
– user209663
8 hours ago
$begingroup$
@user209663 what you say is perfectly fine. But by using the binomial expansion I can evaluate the integral in just a few lines without any recursion or argument about bounds.
$endgroup$
– SZN
8 hours ago
add a comment |
$begingroup$
When you encounter this type of problem, the first thing to do is to come up with a reduction formula. Let us find the reduction formula for $int(k^2 - x^2)^n dx $.
Using integration by parts:
$u = (k-x^2)^n $ and $dv = dx$ then $du = n(k^2 - x^2)^n-1(-2x)dx $ and $v =x$
Thus,
$$int(k^2 - x^2)^n dx = x(k^2 - x^2)^n-1 + 2n int x^2(k^2 - x^2)^n-1 \
= x(k^2 - x^2)^n-1 + 2n int bigg[ k^2(k^2 - x^2)^n-1 - (k^2 - x^2)^n bigg] dx \
= x(k^2 - x^2)^n-1 + 2n int k^2(k^2 - x^2)^n-1 dx - 2n int (k^2 - x^2)^n dx
$$
From here you can see that:
$$2nint(k^2 - x^2)^n dx + int(k^2 - x^2)^n dx = x(k^2 - x^2)^n-1 + 2n int k^2(k^2 - x^2)^n-1 dx - 2n int (k^2 - x^2)^n dx $$
Thus,
$$int(k^2 - x^2)^n dx = fracx(k^2 - x^2)^n-12n+1 + frac2nk^22n+1 int (k^2 - x^2)^n-1 dx$$
For your problem, $k=1$ and $n=10$. Can you take it from here?
answered 8 hours ago
user209663user209663
864411
$endgroup$
When you encounter this type of problem, the first thing to do is to come up with a reduction formula. Let us find the reduction formula for $int(k^2 - x^2)^n dx $.
Using integration by parts:
$u = (k-x^2)^n $ and $dv = dx$ then $du = n(k^2 - x^2)^n-1(-2x)dx $ and $v =x$
Thus,
$$int(k^2 - x^2)^n dx = x(k^2 - x^2)^n-1 + 2n int x^2(k^2 - x^2)^n-1 \
= x(k^2 - x^2)^n-1 + 2n int bigg[ k^2(k^2 - x^2)^n-1 - (k^2 - x^2)^n bigg] dx \
= x(k^2 - x^2)^n-1 + 2n int k^2(k^2 - x^2)^n-1 dx - 2n int (k^2 - x^2)^n dx
$$
From here you can see that:
$$2nint(k^2 - x^2)^n dx + int(k^2 - x^2)^n dx = x(k^2 - x^2)^n-1 + 2n int k^2(k^2 - x^2)^n-1 dx - 2n int (k^2 - x^2)^n dx $$
Thus,
$$int(k^2 - x^2)^n dx = fracx(k^2 - x^2)^n-12n+1 + frac2nk^22n+1 int (k^2 - x^2)^n-1 dx$$
For your problem, $k=1$ and $n=10$. Can you take it from here?
answered 8 hours ago
user209663user209663
864411
answered 8 hours ago
user209663user209663
864411
answered 8 hours ago
user209663user209663
864411
answered 8 hours ago
user209663user209663
864411
864411
$begingroup$
wow! I knew there must be a shorter way. Thank you!
$endgroup$
– MathStudent
8 hours ago
$begingroup$
Is this really a shorter way? Evaluating all of the terms of the recursion seems like just as much effort as simply using the binomial expansion.
$endgroup$
– SZN
8 hours ago
$begingroup$
@MathStudent No problem.
$endgroup$
– user209663
8 hours ago
$begingroup$
@SZN Note that this is a definite integral... look at the recursion.... what happen to the first part? This is similar to $int_0^pi sin^n(x) dx$. You also get a recursion... but the trick is the bound of the integral. It makes it easier to evaluate
$endgroup$
– user209663
8 hours ago
$begingroup$
@user209663 what you say is perfectly fine. But by using the binomial expansion I can evaluate the integral in just a few lines without any recursion or argument about bounds.
$endgroup$
– SZN
8 hours ago
add a comment |
$begingroup$
wow! I knew there must be a shorter way. Thank you!
$endgroup$
– MathStudent
8 hours ago
$begingroup$
Is this really a shorter way? Evaluating all of the terms of the recursion seems like just as much effort as simply using the binomial expansion.
$endgroup$
– SZN
8 hours ago
$begingroup$
@MathStudent No problem.
$endgroup$
– user209663
8 hours ago
$begingroup$
@SZN Note that this is a definite integral... look at the recursion.... what happen to the first part? This is similar to $int_0^pi sin^n(x) dx$. You also get a recursion... but the trick is the bound of the integral. It makes it easier to evaluate
$endgroup$
– user209663
8 hours ago
$begingroup$
@user209663 what you say is perfectly fine. But by using the binomial expansion I can evaluate the integral in just a few lines without any recursion or argument about bounds.
$endgroup$
– SZN
8 hours ago
$begingroup$
wow! I knew there must be a shorter way. Thank you!
$endgroup$
– MathStudent
8 hours ago
$begingroup$
wow! I knew there must be a shorter way. Thank you!
$endgroup$
– MathStudent
8 hours ago
$begingroup$
Is this really a shorter way? Evaluating all of the terms of the recursion seems like just as much effort as simply using the binomial expansion.
$endgroup$
– SZN
8 hours ago
$begingroup$
Is this really a shorter way? Evaluating all of the terms of the recursion seems like just as much effort as simply using the binomial expansion.
$endgroup$
– SZN
8 hours ago
$begingroup$
@MathStudent No problem.
$endgroup$
– user209663
8 hours ago
$begingroup$
@MathStudent No problem.
$endgroup$
– user209663
8 hours ago
$begingroup$
@SZN Note that this is a definite integral... look at the recursion.... what happen to the first part? This is similar to $int_0^pi sin^n(x) dx$. You also get a recursion... but the trick is the bound of the integral. It makes it easier to evaluate
$endgroup$
– user209663
8 hours ago
$begingroup$
@SZN Note that this is a definite integral... look at the recursion.... what happen to the first part? This is similar to $int_0^pi sin^n(x) dx$. You also get a recursion... but the trick is the bound of the integral. It makes it easier to evaluate
$endgroup$
– user209663
8 hours ago
$begingroup$
@user209663 what you say is perfectly fine. But by using the binomial expansion I can evaluate the integral in just a few lines without any recursion or argument about bounds.
$endgroup$
– SZN
8 hours ago
$begingroup$
@user209663 what you say is perfectly fine. But by using the binomial expansion I can evaluate the integral in just a few lines without any recursion or argument about bounds.
$endgroup$
– SZN
8 hours ago
add a comment |
$begingroup$
This answer is exactly what the asker is NOT asking for. However, I believe it will be useful to show the asker that expanding out "term-by-term" is not actually that hard.
We have
beginalign
int_0^1 (1-x^2)^10dx &= int_0^1 sum_i=0^10 C(10,i)(-x^2)^idx \
%
& = sum_i=0^10 (-1)^iC(10,i)int_0^1 x^2idx\
%
&= sum_i=0^10 frac(-1)^iC(10,i)2i+1,
endalign
where $$C(n,m) = fracn!m!(n-m)!.$$
answered 8 hours ago
SZNSZN
3,172721
$endgroup$
add a comment |
$begingroup$
This answer is exactly what the asker is NOT asking for. However, I believe it will be useful to show the asker that expanding out "term-by-term" is not actually that hard.
We have
beginalign
int_0^1 (1-x^2)^10dx &= int_0^1 sum_i=0^10 C(10,i)(-x^2)^idx \
%
& = sum_i=0^10 (-1)^iC(10,i)int_0^1 x^2idx\
%
&= sum_i=0^10 frac(-1)^iC(10,i)2i+1,
endalign
where $$C(n,m) = fracn!m!(n-m)!.$$
answered 8 hours ago
SZNSZN
3,172721
$endgroup$
add a comment |
$begingroup$
This answer is exactly what the asker is NOT asking for. However, I believe it will be useful to show the asker that expanding out "term-by-term" is not actually that hard.
We have
beginalign
int_0^1 (1-x^2)^10dx &= int_0^1 sum_i=0^10 C(10,i)(-x^2)^idx \
%
& = sum_i=0^10 (-1)^iC(10,i)int_0^1 x^2idx\
%
&= sum_i=0^10 frac(-1)^iC(10,i)2i+1,
endalign
where $$C(n,m) = fracn!m!(n-m)!.$$
answered 8 hours ago
SZNSZN
3,172721
$endgroup$
This answer is exactly what the asker is NOT asking for. However, I believe it will be useful to show the asker that expanding out "term-by-term" is not actually that hard.
We have
beginalign
int_0^1 (1-x^2)^10dx &= int_0^1 sum_i=0^10 C(10,i)(-x^2)^idx \
%
& = sum_i=0^10 (-1)^iC(10,i)int_0^1 x^2idx\
%
&= sum_i=0^10 frac(-1)^iC(10,i)2i+1,
endalign
where $$C(n,m) = fracn!m!(n-m)!.$$
answered 8 hours ago
SZNSZN
3,172721
answered 8 hours ago
SZNSZN
3,172721
answered 8 hours ago
SZNSZN
3,172721
answered 8 hours ago
SZNSZN
3,172721
3,172721
add a comment |
add a comment |
$begingroup$
An alternate approach is through the beta function. We start by u-substitution and try to make our way towards an integral form of the beta function:
$$beginalign*
int_0^1(1-x^2)^10,mathrm dx &= frac12 int_0^1 (1-u)^10u^-1/2,mathrm du,qquad x=sqrtu \
&= frac12 int_0^1 (1-u)^11-1u^1/2-1,mathrm du\
&= frac12cdotfracGamma(11)Gamma(1/2)Gamma(11+1/2)\
&= frac12cdotfrac10!sqrtpifrac22!4^11cdot11!sqrtpi\
&= frac262144969969.
endalign*$$
answered 8 hours ago
dxdydzdxdydz
8571714
$endgroup$
1
$begingroup$
Upvoted. There's a special place in my heart for using special functions to solve problems that look like they only admit basic methods :-)
$endgroup$
– SZN
8 hours ago
add a comment |
$begingroup$
An alternate approach is through the beta function. We start by u-substitution and try to make our way towards an integral form of the beta function:
$$beginalign*
int_0^1(1-x^2)^10,mathrm dx &= frac12 int_0^1 (1-u)^10u^-1/2,mathrm du,qquad x=sqrtu \
&= frac12 int_0^1 (1-u)^11-1u^1/2-1,mathrm du\
&= frac12cdotfracGamma(11)Gamma(1/2)Gamma(11+1/2)\
&= frac12cdotfrac10!sqrtpifrac22!4^11cdot11!sqrtpi\
&= frac262144969969.
endalign*$$
answered 8 hours ago
dxdydzdxdydz
8571714
$endgroup$
1
$begingroup$
Upvoted. There's a special place in my heart for using special functions to solve problems that look like they only admit basic methods :-)
$endgroup$
– SZN
8 hours ago
add a comment |
$begingroup$
An alternate approach is through the beta function. We start by u-substitution and try to make our way towards an integral form of the beta function:
$$beginalign*
int_0^1(1-x^2)^10,mathrm dx &= frac12 int_0^1 (1-u)^10u^-1/2,mathrm du,qquad x=sqrtu \
&= frac12 int_0^1 (1-u)^11-1u^1/2-1,mathrm du\
&= frac12cdotfracGamma(11)Gamma(1/2)Gamma(11+1/2)\
&= frac12cdotfrac10!sqrtpifrac22!4^11cdot11!sqrtpi\
&= frac262144969969.
endalign*$$
answered 8 hours ago
dxdydzdxdydz
8571714
$endgroup$
An alternate approach is through the beta function. We start by u-substitution and try to make our way towards an integral form of the beta function:
$$beginalign*
int_0^1(1-x^2)^10,mathrm dx &= frac12 int_0^1 (1-u)^10u^-1/2,mathrm du,qquad x=sqrtu \
&= frac12 int_0^1 (1-u)^11-1u^1/2-1,mathrm du\
&= frac12cdotfracGamma(11)Gamma(1/2)Gamma(11+1/2)\
&= frac12cdotfrac10!sqrtpifrac22!4^11cdot11!sqrtpi\
&= frac262144969969.
endalign*$$
answered 8 hours ago
dxdydzdxdydz
8571714
answered 8 hours ago
dxdydzdxdydz
8571714
answered 8 hours ago
dxdydzdxdydz
8571714
answered 8 hours ago
dxdydzdxdydz
8571714
8571714
1
$begingroup$
Upvoted. There's a special place in my heart for using special functions to solve problems that look like they only admit basic methods :-)
$endgroup$
– SZN
8 hours ago
add a comment |
1
$begingroup$
Upvoted. There's a special place in my heart for using special functions to solve problems that look like they only admit basic methods :-)
$endgroup$
– SZN
8 hours ago
1
1
$begingroup$
Upvoted. There's a special place in my heart for using special functions to solve problems that look like they only admit basic methods :-)
$endgroup$
– SZN
8 hours ago
$begingroup$
Upvoted. There's a special place in my heart for using special functions to solve problems that look like they only admit basic methods :-)
$endgroup$
– SZN
8 hours ago
add a comment |
$begingroup$
Similar to dxdydz's answer, let $x=sin(t)$
$$int_0^1 ( 1 - x^2)^10, dx=int_0^frac pi 2cos^21(t),dt$$ and remember that
$$int_0^frac pi 2cos^n(t),dt=fracsqrtpi 2 frac Gamma left(fracn+12right) Gamma
left(fracn+22right)$$
answered 57 mins ago
Claude LeiboviciClaude Leibovici
129k1158139
$endgroup$
add a comment |
$begingroup$
Similar to dxdydz's answer, let $x=sin(t)$
$$int_0^1 ( 1 - x^2)^10, dx=int_0^frac pi 2cos^21(t),dt$$ and remember that
$$int_0^frac pi 2cos^n(t),dt=fracsqrtpi 2 frac Gamma left(fracn+12right) Gamma
left(fracn+22right)$$
answered 57 mins ago
Claude LeiboviciClaude Leibovici
129k1158139
$endgroup$
add a comment |
$begingroup$
Similar to dxdydz's answer, let $x=sin(t)$
$$int_0^1 ( 1 - x^2)^10, dx=int_0^frac pi 2cos^21(t),dt$$ and remember that
$$int_0^frac pi 2cos^n(t),dt=fracsqrtpi 2 frac Gamma left(fracn+12right) Gamma
left(fracn+22right)$$
answered 57 mins ago
Claude LeiboviciClaude Leibovici
129k1158139
$endgroup$
Similar to dxdydz's answer, let $x=sin(t)$
$$int_0^1 ( 1 - x^2)^10, dx=int_0^frac pi 2cos^21(t),dt$$ and remember that
$$int_0^frac pi 2cos^n(t),dt=fracsqrtpi 2 frac Gamma left(fracn+12right) Gamma
left(fracn+22right)$$
answered 57 mins ago
Claude LeiboviciClaude Leibovici
129k1158139
answered 57 mins ago
Claude LeiboviciClaude Leibovici
129k1158139
answered 57 mins ago
Claude LeiboviciClaude Leibovici
129k1158139
answered 57 mins ago
Claude LeiboviciClaude Leibovici
129k1158139
129k1158139
add a comment |
add a comment |
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$begingroup$
Yes, by coming up with the reduction formula!
$endgroup$
– user209663
8 hours ago