Mixing
Need help in solving $int _{ 0 }^{ frac { pi }{ 2 } } int _{ 0 }^{ x }{ { e }^{ sin(y) } } { sin(x)dydx }$
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I have tried to solve $int _{ 0 }^{ frac { pi }{ 2 } } int _{ 0 }^{ x }{ { e }^{ sin(y) } } { sin(x)dydx }$ . I have solved it by using mathematica to evaluate $int { { e }^{ sin(x) }dx } $ but it is turning tedious so hoping to get some insight into solve it in a better way
calculus integration multivariable-calculus definite-integrals
asked Jan 12 at 3:49
SHOURIE MRSSSHOURIE MRSS
363
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add a comment |
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I have tried to solve $int _{ 0 }^{ frac { pi }{ 2 } } int _{ 0 }^{ x }{ { e }^{ sin(y) } } { sin(x)dydx }$ . I have solved it by using mathematica to evaluate $int { { e }^{ sin(x) }dx } $ but it is turning tedious so hoping to get some insight into solve it in a better way
calculus integration multivariable-calculus definite-integrals
asked Jan 12 at 3:49
SHOURIE MRSSSHOURIE MRSS
363
$endgroup$
add a comment |
$begingroup$
I have tried to solve $int _{ 0 }^{ frac { pi }{ 2 } } int _{ 0 }^{ x }{ { e }^{ sin(y) } } { sin(x)dydx }$ . I have solved it by using mathematica to evaluate $int { { e }^{ sin(x) }dx } $ but it is turning tedious so hoping to get some insight into solve it in a better way
calculus integration multivariable-calculus definite-integrals
asked Jan 12 at 3:49
SHOURIE MRSSSHOURIE MRSS
363
$endgroup$
I have tried to solve $int _{ 0 }^{ frac { pi }{ 2 } } int _{ 0 }^{ x }{ { e }^{ sin(y) } } { sin(x)dydx }$ . I have solved it by using mathematica to evaluate $int { { e }^{ sin(x) }dx } $ but it is turning tedious so hoping to get some insight into solve it in a better way
calculus integration multivariable-calculus definite-integrals
calculus integration multivariable-calculus definite-integrals
asked Jan 12 at 3:49
SHOURIE MRSSSHOURIE MRSS
363
asked Jan 12 at 3:49
SHOURIE MRSSSHOURIE MRSS
363
asked Jan 12 at 3:49
SHOURIE MRSSSHOURIE MRSS
363
asked Jan 12 at 3:49
SHOURIE MRSSSHOURIE MRSS
363
asked Jan 12 at 3:49
SHOURIE MRSSSHOURIE MRSS
363
363
add a comment |
add a comment |
1 Answer
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Hint Changing the order of integration you get
$$int _{ 0 }^{ frac { pi }{ 2 } } int _{ 0 }^{ x }{ { e }^{ sin(y) } } { sin(x)dydx }=int _{ 0 }^{ frac { pi }{ 2 } } int _{ y }^{ frac { pi }{ 2 } }{ { e }^{ sin(y) } } { sin(x)dxdy }$$
which is easy to integrate.
answered Jan 12 at 3:54
N. S.N. S.
103k6112208
$endgroup$
$begingroup$
what is the principle behind it?
$endgroup$
– SHOURIE MRSS
Jan 12 at 4:04
$begingroup$
I got the expected answer e-1, but I am still not able to spot the rationale behind the suggestion?
$endgroup$
– SHOURIE MRSS
Jan 12 at 4:20
1
$begingroup$
@SHOURIEMRSS Whenever when you need to integrate a double integral $$int int f(x,y) dy dx$$ and your function is hard/impossible to integrate with respect to $y$ but easy to integrate with respect to $x$ it is natural to change the order of integration.... Start with the easy step, and see what you get, maybe the second integral is also easy....... This is a pretty standard exercise of order of integration.
$endgroup$
– N. S.
Jan 12 at 4:30
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Hint Changing the order of integration you get
$$int _{ 0 }^{ frac { pi }{ 2 } } int _{ 0 }^{ x }{ { e }^{ sin(y) } } { sin(x)dydx }=int _{ 0 }^{ frac { pi }{ 2 } } int _{ y }^{ frac { pi }{ 2 } }{ { e }^{ sin(y) } } { sin(x)dxdy }$$
which is easy to integrate.
answered Jan 12 at 3:54
N. S.N. S.
103k6112208
$endgroup$
$begingroup$
what is the principle behind it?
$endgroup$
– SHOURIE MRSS
Jan 12 at 4:04
$begingroup$
I got the expected answer e-1, but I am still not able to spot the rationale behind the suggestion?
$endgroup$
– SHOURIE MRSS
Jan 12 at 4:20
1
$begingroup$
@SHOURIEMRSS Whenever when you need to integrate a double integral $$int int f(x,y) dy dx$$ and your function is hard/impossible to integrate with respect to $y$ but easy to integrate with respect to $x$ it is natural to change the order of integration.... Start with the easy step, and see what you get, maybe the second integral is also easy....... This is a pretty standard exercise of order of integration.
$endgroup$
– N. S.
Jan 12 at 4:30
add a comment |
$begingroup$
Hint Changing the order of integration you get
$$int _{ 0 }^{ frac { pi }{ 2 } } int _{ 0 }^{ x }{ { e }^{ sin(y) } } { sin(x)dydx }=int _{ 0 }^{ frac { pi }{ 2 } } int _{ y }^{ frac { pi }{ 2 } }{ { e }^{ sin(y) } } { sin(x)dxdy }$$
which is easy to integrate.
answered Jan 12 at 3:54
N. S.N. S.
103k6112208
$endgroup$
$begingroup$
what is the principle behind it?
$endgroup$
– SHOURIE MRSS
Jan 12 at 4:04
$begingroup$
I got the expected answer e-1, but I am still not able to spot the rationale behind the suggestion?
$endgroup$
– SHOURIE MRSS
Jan 12 at 4:20
1
$begingroup$
@SHOURIEMRSS Whenever when you need to integrate a double integral $$int int f(x,y) dy dx$$ and your function is hard/impossible to integrate with respect to $y$ but easy to integrate with respect to $x$ it is natural to change the order of integration.... Start with the easy step, and see what you get, maybe the second integral is also easy....... This is a pretty standard exercise of order of integration.
$endgroup$
– N. S.
Jan 12 at 4:30
add a comment |
$begingroup$
Hint Changing the order of integration you get
$$int _{ 0 }^{ frac { pi }{ 2 } } int _{ 0 }^{ x }{ { e }^{ sin(y) } } { sin(x)dydx }=int _{ 0 }^{ frac { pi }{ 2 } } int _{ y }^{ frac { pi }{ 2 } }{ { e }^{ sin(y) } } { sin(x)dxdy }$$
which is easy to integrate.
answered Jan 12 at 3:54
N. S.N. S.
103k6112208
$endgroup$
Hint Changing the order of integration you get
$$int _{ 0 }^{ frac { pi }{ 2 } } int _{ 0 }^{ x }{ { e }^{ sin(y) } } { sin(x)dydx }=int _{ 0 }^{ frac { pi }{ 2 } } int _{ y }^{ frac { pi }{ 2 } }{ { e }^{ sin(y) } } { sin(x)dxdy }$$
which is easy to integrate.
answered Jan 12 at 3:54
N. S.N. S.
103k6112208
answered Jan 12 at 3:54
N. S.N. S.
103k6112208
answered Jan 12 at 3:54
N. S.N. S.
103k6112208
answered Jan 12 at 3:54
N. S.N. S.
103k6112208
103k6112208
$begingroup$
what is the principle behind it?
$endgroup$
– SHOURIE MRSS
Jan 12 at 4:04
$begingroup$
I got the expected answer e-1, but I am still not able to spot the rationale behind the suggestion?
$endgroup$
– SHOURIE MRSS
Jan 12 at 4:20
1
$begingroup$
@SHOURIEMRSS Whenever when you need to integrate a double integral $$int int f(x,y) dy dx$$ and your function is hard/impossible to integrate with respect to $y$ but easy to integrate with respect to $x$ it is natural to change the order of integration.... Start with the easy step, and see what you get, maybe the second integral is also easy....... This is a pretty standard exercise of order of integration.
$endgroup$
– N. S.
Jan 12 at 4:30
add a comment |
$begingroup$
what is the principle behind it?
$endgroup$
– SHOURIE MRSS
Jan 12 at 4:04
$begingroup$
I got the expected answer e-1, but I am still not able to spot the rationale behind the suggestion?
$endgroup$
– SHOURIE MRSS
Jan 12 at 4:20
1
$begingroup$
@SHOURIEMRSS Whenever when you need to integrate a double integral $$int int f(x,y) dy dx$$ and your function is hard/impossible to integrate with respect to $y$ but easy to integrate with respect to $x$ it is natural to change the order of integration.... Start with the easy step, and see what you get, maybe the second integral is also easy....... This is a pretty standard exercise of order of integration.
$endgroup$
– N. S.
Jan 12 at 4:30
$begingroup$
what is the principle behind it?
$endgroup$
– SHOURIE MRSS
Jan 12 at 4:04
$begingroup$
what is the principle behind it?
$endgroup$
– SHOURIE MRSS
Jan 12 at 4:04
$begingroup$
I got the expected answer e-1, but I am still not able to spot the rationale behind the suggestion?
$endgroup$
– SHOURIE MRSS
Jan 12 at 4:20
$begingroup$
I got the expected answer e-1, but I am still not able to spot the rationale behind the suggestion?
$endgroup$
– SHOURIE MRSS
Jan 12 at 4:20
1
1
$begingroup$
@SHOURIEMRSS Whenever when you need to integrate a double integral $$int int f(x,y) dy dx$$ and your function is hard/impossible to integrate with respect to $y$ but easy to integrate with respect to $x$ it is natural to change the order of integration.... Start with the easy step, and see what you get, maybe the second integral is also easy....... This is a pretty standard exercise of order of integration.
$endgroup$
– N. S.
Jan 12 at 4:30
$begingroup$
@SHOURIEMRSS Whenever when you need to integrate a double integral $$int int f(x,y) dy dx$$ and your function is hard/impossible to integrate with respect to $y$ but easy to integrate with respect to $x$ it is natural to change the order of integration.... Start with the easy step, and see what you get, maybe the second integral is also easy....... This is a pretty standard exercise of order of integration.
$endgroup$
– N. S.
Jan 12 at 4:30
add a comment |
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