finding value of integral
$begingroup$
$$int_0^{pi/2} (cos x)^{2011} sin (2013 x) dx $$
The value is given to be $frac{1}{2012}$. I am while solving using by parts, but one term is remaining like $frac{1}{2}$, then $frac{1}{3},$ and so on. Kindly help.
what i am stuck at is following.
$$I(m,n) =int_0^{pi/2} (cos x)^{m} sin (n x) dx $$
$$I(m,n)= frac {1}{m+n} + frac{m}{m+n}I(m-1,n-1)$$
after deriving this i am not able to proceed further
calculus definite-integrals
asked Dec 15 '18 at 8:42
mavericmaveric
88012
$endgroup$
add a comment |
$begingroup$
$$int_0^{pi/2} (cos x)^{2011} sin (2013 x) dx $$
The value is given to be $frac{1}{2012}$. I am while solving using by parts, but one term is remaining like $frac{1}{2}$, then $frac{1}{3},$ and so on. Kindly help.
what i am stuck at is following.
$$I(m,n) =int_0^{pi/2} (cos x)^{m} sin (n x) dx $$
$$I(m,n)= frac {1}{m+n} + frac{m}{m+n}I(m-1,n-1)$$
after deriving this i am not able to proceed further
calculus definite-integrals
asked Dec 15 '18 at 8:42
mavericmaveric
88012
$endgroup$
4
$begingroup$
why downvoting?
$endgroup$
– maveric
Dec 15 '18 at 8:45
3
$begingroup$
In this case, reasons for the down- and closevotes would be helpful. I do not see one.
$endgroup$
– Peter
Dec 15 '18 at 9:20
$begingroup$
What's meant by stuck? You can not derive that or don't know how to utilize that
$endgroup$
– lab bhattacharjee
Dec 15 '18 at 9:37
$begingroup$
i have reached till this step. i am able to proceed further to answer which is 1/2011
$endgroup$
– maveric
Dec 15 '18 at 9:38
$begingroup$
This double recurrence relationship looks a deadend. I have found the general formula $$int_0^{pi/2} (cos x)^{nu-2} sin (nu x) dx=dfrac{1}{nu-1} $$ in a book. It can help to buid a more sympathetic recurrence...
$endgroup$
– Jean Marie
Dec 15 '18 at 9:51
add a comment |
$begingroup$
$$int_0^{pi/2} (cos x)^{2011} sin (2013 x) dx $$
The value is given to be $frac{1}{2012}$. I am while solving using by parts, but one term is remaining like $frac{1}{2}$, then $frac{1}{3},$ and so on. Kindly help.
what i am stuck at is following.
$$I(m,n) =int_0^{pi/2} (cos x)^{m} sin (n x) dx $$
$$I(m,n)= frac {1}{m+n} + frac{m}{m+n}I(m-1,n-1)$$
after deriving this i am not able to proceed further
calculus definite-integrals
asked Dec 15 '18 at 8:42
mavericmaveric
88012
$endgroup$
$$int_0^{pi/2} (cos x)^{2011} sin (2013 x) dx $$
The value is given to be $frac{1}{2012}$. I am while solving using by parts, but one term is remaining like $frac{1}{2}$, then $frac{1}{3},$ and so on. Kindly help.
what i am stuck at is following.
$$I(m,n) =int_0^{pi/2} (cos x)^{m} sin (n x) dx $$
$$I(m,n)= frac {1}{m+n} + frac{m}{m+n}I(m-1,n-1)$$
after deriving this i am not able to proceed further
calculus definite-integrals
calculus definite-integrals
asked Dec 15 '18 at 8:42
mavericmaveric
88012
asked Dec 15 '18 at 8:42
mavericmaveric
88012
edited Dec 15 '18 at 10:26
maveric
asked Dec 15 '18 at 8:42
mavericmaveric
88012
asked Dec 15 '18 at 8:42
mavericmaveric
88012
asked Dec 15 '18 at 8:42
mavericmaveric
88012
88012
4
$begingroup$
why downvoting?
$endgroup$
– maveric
Dec 15 '18 at 8:45
3
$begingroup$
In this case, reasons for the down- and closevotes would be helpful. I do not see one.
$endgroup$
– Peter
Dec 15 '18 at 9:20
$begingroup$
What's meant by stuck? You can not derive that or don't know how to utilize that
$endgroup$
– lab bhattacharjee
Dec 15 '18 at 9:37
$begingroup$
i have reached till this step. i am able to proceed further to answer which is 1/2011
$endgroup$
– maveric
Dec 15 '18 at 9:38
$begingroup$
This double recurrence relationship looks a deadend. I have found the general formula $$int_0^{pi/2} (cos x)^{nu-2} sin (nu x) dx=dfrac{1}{nu-1} $$ in a book. It can help to buid a more sympathetic recurrence...
$endgroup$
– Jean Marie
Dec 15 '18 at 9:51
add a comment |
4
$begingroup$
why downvoting?
$endgroup$
– maveric
Dec 15 '18 at 8:45
3
$begingroup$
In this case, reasons for the down- and closevotes would be helpful. I do not see one.
$endgroup$
– Peter
Dec 15 '18 at 9:20
$begingroup$
What's meant by stuck? You can not derive that or don't know how to utilize that
$endgroup$
– lab bhattacharjee
Dec 15 '18 at 9:37
$begingroup$
i have reached till this step. i am able to proceed further to answer which is 1/2011
$endgroup$
– maveric
Dec 15 '18 at 9:38
$begingroup$
This double recurrence relationship looks a deadend. I have found the general formula $$int_0^{pi/2} (cos x)^{nu-2} sin (nu x) dx=dfrac{1}{nu-1} $$ in a book. It can help to buid a more sympathetic recurrence...
$endgroup$
– Jean Marie
Dec 15 '18 at 9:51
4
4
$begingroup$
why downvoting?
$endgroup$
– maveric
Dec 15 '18 at 8:45
$begingroup$
why downvoting?
$endgroup$
– maveric
Dec 15 '18 at 8:45
3
3
$begingroup$
In this case, reasons for the down- and closevotes would be helpful. I do not see one.
$endgroup$
– Peter
Dec 15 '18 at 9:20
$begingroup$
In this case, reasons for the down- and closevotes would be helpful. I do not see one.
$endgroup$
– Peter
Dec 15 '18 at 9:20
$begingroup$
What's meant by stuck? You can not derive that or don't know how to utilize that
$endgroup$
– lab bhattacharjee
Dec 15 '18 at 9:37
$begingroup$
What's meant by stuck? You can not derive that or don't know how to utilize that
$endgroup$
– lab bhattacharjee
Dec 15 '18 at 9:37
$begingroup$
i have reached till this step. i am able to proceed further to answer which is 1/2011
$endgroup$
– maveric
Dec 15 '18 at 9:38
$begingroup$
i have reached till this step. i am able to proceed further to answer which is 1/2011
$endgroup$
– maveric
Dec 15 '18 at 9:38
$begingroup$
This double recurrence relationship looks a deadend. I have found the general formula $$int_0^{pi/2} (cos x)^{nu-2} sin (nu x) dx=dfrac{1}{nu-1} $$ in a book. It can help to buid a more sympathetic recurrence...
$endgroup$
– Jean Marie
Dec 15 '18 at 9:51
$begingroup$
This double recurrence relationship looks a deadend. I have found the general formula $$int_0^{pi/2} (cos x)^{nu-2} sin (nu x) dx=dfrac{1}{nu-1} $$ in a book. It can help to buid a more sympathetic recurrence...
$endgroup$
– Jean Marie
Dec 15 '18 at 9:51
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint:
$n=m+2implies$ $$I_{(m,m+2)}=dfrac1{2m+2}+dfrac m{2m+2}I_{(m-1,m+1)}$$
$I_{(0,2)}=1$
$m=1implies$ $$I_{(1,3)}=dfrac14+dfrac14I_{(0,2)}=dfrac12$$
$m=2implies$ $$I_{(2,4)}=dfrac16+dfrac13I_{(1,3)}=dfrac13$$
Use induction for $I_{(m-1,m+1)}=dfrac2{m-1+m+1}$
answered Dec 15 '18 at 9:49
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
$begingroup$
why did you think to replace n= m+2 ? is it because of powers stated? can we solve for any difference?
$endgroup$
– maveric
Dec 15 '18 at 10:13
$begingroup$
@maveric, Yes because of the difference in powers.
$endgroup$
– lab bhattacharjee
Dec 15 '18 at 10:16
$begingroup$
could you ellaborate on induction ?
$endgroup$
– maveric
Dec 15 '18 at 10:27
1
$begingroup$
@If $I_{(m-1,m+1)}=dfrac1m;$ hold true for $m=n$ Establish the same for $m=n+1$
$endgroup$
– lab bhattacharjee
Dec 15 '18 at 10:30
add a comment |
$begingroup$
Note that
$$
begin{align}
I(m,n)
&=int_0^{pi/2}cos^m(x)sin(nx),mathrm{d}xtag1\
&=int_0^{pi/2}cos^{m+1}(x)sin((n-1)x),mathrm{d}x+int_0^{pi/2}cos^m(x)cos((n-1)x)sin(x),mathrm{d}xtag2\
&=I(m+1,n-1)-frac1{m+1}int_0^{pi/2}cos((n-1)x),mathrm{d}cos^{m+1}(x)tag3\
&=I(m+1,n-1)+frac1{m+1}-frac{n-1}{m+1}int_0^{pi/2}cos^{m+1}(x)sin((n-1)x),mathrm{d}xtag4\
&=frac{m-n+2}{m+1},I(m+1,n-1)+frac1{m+1}tag5
end{align}
$$
Explanation:
$(1)$: define $I(m,n)$
$(2)$: $sin(nx)=sin((n-1)x)cos(x)+cos((n-1)x)sin(x)$
$(3)$: prepare to integrate by parts
$(4)$: integrate by parts
$(5)$: simplify
Plug in $m=2011$ and $n=2013$ and see what happens.
However, this gives the answer $frac1{2012}$, which is confirmed by Mathematica.
answered Dec 15 '18 at 10:13
robjohn♦robjohn
268k27309634
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
$begingroup$
Hint:
$n=m+2implies$ $$I_{(m,m+2)}=dfrac1{2m+2}+dfrac m{2m+2}I_{(m-1,m+1)}$$
$I_{(0,2)}=1$
$m=1implies$ $$I_{(1,3)}=dfrac14+dfrac14I_{(0,2)}=dfrac12$$
$m=2implies$ $$I_{(2,4)}=dfrac16+dfrac13I_{(1,3)}=dfrac13$$
Use induction for $I_{(m-1,m+1)}=dfrac2{m-1+m+1}$
answered Dec 15 '18 at 9:49
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
$begingroup$
why did you think to replace n= m+2 ? is it because of powers stated? can we solve for any difference?
$endgroup$
– maveric
Dec 15 '18 at 10:13
$begingroup$
@maveric, Yes because of the difference in powers.
$endgroup$
– lab bhattacharjee
Dec 15 '18 at 10:16
$begingroup$
could you ellaborate on induction ?
$endgroup$
– maveric
Dec 15 '18 at 10:27
1
$begingroup$
@If $I_{(m-1,m+1)}=dfrac1m;$ hold true for $m=n$ Establish the same for $m=n+1$
$endgroup$
– lab bhattacharjee
Dec 15 '18 at 10:30
add a comment |
$begingroup$
Hint:
$n=m+2implies$ $$I_{(m,m+2)}=dfrac1{2m+2}+dfrac m{2m+2}I_{(m-1,m+1)}$$
$I_{(0,2)}=1$
$m=1implies$ $$I_{(1,3)}=dfrac14+dfrac14I_{(0,2)}=dfrac12$$
$m=2implies$ $$I_{(2,4)}=dfrac16+dfrac13I_{(1,3)}=dfrac13$$
Use induction for $I_{(m-1,m+1)}=dfrac2{m-1+m+1}$
answered Dec 15 '18 at 9:49
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
$begingroup$
why did you think to replace n= m+2 ? is it because of powers stated? can we solve for any difference?
$endgroup$
– maveric
Dec 15 '18 at 10:13
$begingroup$
@maveric, Yes because of the difference in powers.
$endgroup$
– lab bhattacharjee
Dec 15 '18 at 10:16
$begingroup$
could you ellaborate on induction ?
$endgroup$
– maveric
Dec 15 '18 at 10:27
1
$begingroup$
@If $I_{(m-1,m+1)}=dfrac1m;$ hold true for $m=n$ Establish the same for $m=n+1$
$endgroup$
– lab bhattacharjee
Dec 15 '18 at 10:30
add a comment |
$begingroup$
Hint:
$n=m+2implies$ $$I_{(m,m+2)}=dfrac1{2m+2}+dfrac m{2m+2}I_{(m-1,m+1)}$$
$I_{(0,2)}=1$
$m=1implies$ $$I_{(1,3)}=dfrac14+dfrac14I_{(0,2)}=dfrac12$$
$m=2implies$ $$I_{(2,4)}=dfrac16+dfrac13I_{(1,3)}=dfrac13$$
Use induction for $I_{(m-1,m+1)}=dfrac2{m-1+m+1}$
answered Dec 15 '18 at 9:49
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
Hint:
$n=m+2implies$ $$I_{(m,m+2)}=dfrac1{2m+2}+dfrac m{2m+2}I_{(m-1,m+1)}$$
$I_{(0,2)}=1$
$m=1implies$ $$I_{(1,3)}=dfrac14+dfrac14I_{(0,2)}=dfrac12$$
$m=2implies$ $$I_{(2,4)}=dfrac16+dfrac13I_{(1,3)}=dfrac13$$
Use induction for $I_{(m-1,m+1)}=dfrac2{m-1+m+1}$
answered Dec 15 '18 at 9:49
lab bhattacharjeelab bhattacharjee
226k15157275
answered Dec 15 '18 at 9:49
lab bhattacharjeelab bhattacharjee
226k15157275
answered Dec 15 '18 at 9:49
lab bhattacharjeelab bhattacharjee
226k15157275
answered Dec 15 '18 at 9:49
lab bhattacharjeelab bhattacharjee
226k15157275
226k15157275
$begingroup$
why did you think to replace n= m+2 ? is it because of powers stated? can we solve for any difference?
$endgroup$
– maveric
Dec 15 '18 at 10:13
$begingroup$
@maveric, Yes because of the difference in powers.
$endgroup$
– lab bhattacharjee
Dec 15 '18 at 10:16
$begingroup$
could you ellaborate on induction ?
$endgroup$
– maveric
Dec 15 '18 at 10:27
1
$begingroup$
@If $I_{(m-1,m+1)}=dfrac1m;$ hold true for $m=n$ Establish the same for $m=n+1$
$endgroup$
– lab bhattacharjee
Dec 15 '18 at 10:30
add a comment |
$begingroup$
why did you think to replace n= m+2 ? is it because of powers stated? can we solve for any difference?
$endgroup$
– maveric
Dec 15 '18 at 10:13
$begingroup$
@maveric, Yes because of the difference in powers.
$endgroup$
– lab bhattacharjee
Dec 15 '18 at 10:16
$begingroup$
could you ellaborate on induction ?
$endgroup$
– maveric
Dec 15 '18 at 10:27
1
$begingroup$
@If $I_{(m-1,m+1)}=dfrac1m;$ hold true for $m=n$ Establish the same for $m=n+1$
$endgroup$
– lab bhattacharjee
Dec 15 '18 at 10:30
$begingroup$
why did you think to replace n= m+2 ? is it because of powers stated? can we solve for any difference?
$endgroup$
– maveric
Dec 15 '18 at 10:13
$begingroup$
why did you think to replace n= m+2 ? is it because of powers stated? can we solve for any difference?
$endgroup$
– maveric
Dec 15 '18 at 10:13
$begingroup$
@maveric, Yes because of the difference in powers.
$endgroup$
– lab bhattacharjee
Dec 15 '18 at 10:16
$begingroup$
@maveric, Yes because of the difference in powers.
$endgroup$
– lab bhattacharjee
Dec 15 '18 at 10:16
$begingroup$
could you ellaborate on induction ?
$endgroup$
– maveric
Dec 15 '18 at 10:27
$begingroup$
could you ellaborate on induction ?
$endgroup$
– maveric
Dec 15 '18 at 10:27
1
1
$begingroup$
@If $I_{(m-1,m+1)}=dfrac1m;$ hold true for $m=n$ Establish the same for $m=n+1$
$endgroup$
– lab bhattacharjee
Dec 15 '18 at 10:30
$begingroup$
@If $I_{(m-1,m+1)}=dfrac1m;$ hold true for $m=n$ Establish the same for $m=n+1$
$endgroup$
– lab bhattacharjee
Dec 15 '18 at 10:30
add a comment |
$begingroup$
Note that
$$
begin{align}
I(m,n)
&=int_0^{pi/2}cos^m(x)sin(nx),mathrm{d}xtag1\
&=int_0^{pi/2}cos^{m+1}(x)sin((n-1)x),mathrm{d}x+int_0^{pi/2}cos^m(x)cos((n-1)x)sin(x),mathrm{d}xtag2\
&=I(m+1,n-1)-frac1{m+1}int_0^{pi/2}cos((n-1)x),mathrm{d}cos^{m+1}(x)tag3\
&=I(m+1,n-1)+frac1{m+1}-frac{n-1}{m+1}int_0^{pi/2}cos^{m+1}(x)sin((n-1)x),mathrm{d}xtag4\
&=frac{m-n+2}{m+1},I(m+1,n-1)+frac1{m+1}tag5
end{align}
$$
Explanation:
$(1)$: define $I(m,n)$
$(2)$: $sin(nx)=sin((n-1)x)cos(x)+cos((n-1)x)sin(x)$
$(3)$: prepare to integrate by parts
$(4)$: integrate by parts
$(5)$: simplify
Plug in $m=2011$ and $n=2013$ and see what happens.
However, this gives the answer $frac1{2012}$, which is confirmed by Mathematica.
answered Dec 15 '18 at 10:13
robjohn♦robjohn
268k27309634
$endgroup$
add a comment |
$begingroup$
Note that
$$
begin{align}
I(m,n)
&=int_0^{pi/2}cos^m(x)sin(nx),mathrm{d}xtag1\
&=int_0^{pi/2}cos^{m+1}(x)sin((n-1)x),mathrm{d}x+int_0^{pi/2}cos^m(x)cos((n-1)x)sin(x),mathrm{d}xtag2\
&=I(m+1,n-1)-frac1{m+1}int_0^{pi/2}cos((n-1)x),mathrm{d}cos^{m+1}(x)tag3\
&=I(m+1,n-1)+frac1{m+1}-frac{n-1}{m+1}int_0^{pi/2}cos^{m+1}(x)sin((n-1)x),mathrm{d}xtag4\
&=frac{m-n+2}{m+1},I(m+1,n-1)+frac1{m+1}tag5
end{align}
$$
Explanation:
$(1)$: define $I(m,n)$
$(2)$: $sin(nx)=sin((n-1)x)cos(x)+cos((n-1)x)sin(x)$
$(3)$: prepare to integrate by parts
$(4)$: integrate by parts
$(5)$: simplify
Plug in $m=2011$ and $n=2013$ and see what happens.
However, this gives the answer $frac1{2012}$, which is confirmed by Mathematica.
answered Dec 15 '18 at 10:13
robjohn♦robjohn
268k27309634
$endgroup$
add a comment |
$begingroup$
Note that
$$
begin{align}
I(m,n)
&=int_0^{pi/2}cos^m(x)sin(nx),mathrm{d}xtag1\
&=int_0^{pi/2}cos^{m+1}(x)sin((n-1)x),mathrm{d}x+int_0^{pi/2}cos^m(x)cos((n-1)x)sin(x),mathrm{d}xtag2\
&=I(m+1,n-1)-frac1{m+1}int_0^{pi/2}cos((n-1)x),mathrm{d}cos^{m+1}(x)tag3\
&=I(m+1,n-1)+frac1{m+1}-frac{n-1}{m+1}int_0^{pi/2}cos^{m+1}(x)sin((n-1)x),mathrm{d}xtag4\
&=frac{m-n+2}{m+1},I(m+1,n-1)+frac1{m+1}tag5
end{align}
$$
Explanation:
$(1)$: define $I(m,n)$
$(2)$: $sin(nx)=sin((n-1)x)cos(x)+cos((n-1)x)sin(x)$
$(3)$: prepare to integrate by parts
$(4)$: integrate by parts
$(5)$: simplify
Plug in $m=2011$ and $n=2013$ and see what happens.
However, this gives the answer $frac1{2012}$, which is confirmed by Mathematica.
answered Dec 15 '18 at 10:13
robjohn♦robjohn
268k27309634
$endgroup$
Note that
$$
begin{align}
I(m,n)
&=int_0^{pi/2}cos^m(x)sin(nx),mathrm{d}xtag1\
&=int_0^{pi/2}cos^{m+1}(x)sin((n-1)x),mathrm{d}x+int_0^{pi/2}cos^m(x)cos((n-1)x)sin(x),mathrm{d}xtag2\
&=I(m+1,n-1)-frac1{m+1}int_0^{pi/2}cos((n-1)x),mathrm{d}cos^{m+1}(x)tag3\
&=I(m+1,n-1)+frac1{m+1}-frac{n-1}{m+1}int_0^{pi/2}cos^{m+1}(x)sin((n-1)x),mathrm{d}xtag4\
&=frac{m-n+2}{m+1},I(m+1,n-1)+frac1{m+1}tag5
end{align}
$$
Explanation:
$(1)$: define $I(m,n)$
$(2)$: $sin(nx)=sin((n-1)x)cos(x)+cos((n-1)x)sin(x)$
$(3)$: prepare to integrate by parts
$(4)$: integrate by parts
$(5)$: simplify
Plug in $m=2011$ and $n=2013$ and see what happens.
However, this gives the answer $frac1{2012}$, which is confirmed by Mathematica.
answered Dec 15 '18 at 10:13
robjohn♦robjohn
268k27309634
edited Dec 15 '18 at 10:21
answered Dec 15 '18 at 10:13
robjohn♦robjohn
268k27309634
answered Dec 15 '18 at 10:13
robjohn♦robjohn
268k27309634
answered Dec 15 '18 at 10:13
robjohn♦robjohn
268k27309634
268k27309634
add a comment |
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4
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why downvoting?
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– maveric
Dec 15 '18 at 8:45
3
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In this case, reasons for the down- and closevotes would be helpful. I do not see one.
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– Peter
Dec 15 '18 at 9:20
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What's meant by stuck? You can not derive that or don't know how to utilize that
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– lab bhattacharjee
Dec 15 '18 at 9:37
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i have reached till this step. i am able to proceed further to answer which is 1/2011
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– maveric
Dec 15 '18 at 9:38
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This double recurrence relationship looks a deadend. I have found the general formula $$int_0^{pi/2} (cos x)^{nu-2} sin (nu x) dx=dfrac{1}{nu-1} $$ in a book. It can help to buid a more sympathetic recurrence...
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– Jean Marie
Dec 15 '18 at 9:51