Mixing
By creating a reduction formula for $I_n=int_0 ^pi x^n sin x ,dx$ Show that $I_4=pi^4 -12pi^2 + 48$
$begingroup$
By creating a reduction formula for
$$I_n=int_0 ^pi x^n sin x ,dx$$ Show that $I_4=pi^4 -12pi^2 + 48$
So by using integration by parts I wrote $I_n$ as $$I_n=int_0 ^pi x^n sin x ,dx = left[-x^ncos x right]_0 ^pi -int_0 ^pi nx^{n-1}times -cos x, dx$$ and after messing about with this I got $$I_n = pi^n +nI_{n-1}$$ However, the answer shows that this is wrong and I don't get the correct answer for $I_4$. The correct answer is $I_n = pi^n -n(n-1)I_{n-2}$. Can someone help explain how you get this reduction formula as I can't seem to get it and why my one doesn't work.
integration definite-integrals reduction-formula
asked Feb 1 at 21:05
H.LinkhornH.Linkhorn
495213
$endgroup$
add a comment |
$begingroup$
By creating a reduction formula for
$$I_n=int_0 ^pi x^n sin x ,dx$$ Show that $I_4=pi^4 -12pi^2 + 48$
So by using integration by parts I wrote $I_n$ as $$I_n=int_0 ^pi x^n sin x ,dx = left[-x^ncos x right]_0 ^pi -int_0 ^pi nx^{n-1}times -cos x, dx$$ and after messing about with this I got $$I_n = pi^n +nI_{n-1}$$ However, the answer shows that this is wrong and I don't get the correct answer for $I_4$. The correct answer is $I_n = pi^n -n(n-1)I_{n-2}$. Can someone help explain how you get this reduction formula as I can't seem to get it and why my one doesn't work.
integration definite-integrals reduction-formula
asked Feb 1 at 21:05
H.LinkhornH.Linkhorn
495213
$endgroup$
1
$begingroup$
You need to do by-parts again in order to compare the new integral with $I_n.$
$endgroup$
– Adrian Keister
Feb 1 at 21:08
$begingroup$
Can you show me what you mean?
$endgroup$
– H.Linkhorn
Feb 1 at 21:26
$begingroup$
Do you know an antiderivative of the function $cos$ is $sin$? Perform again an integration by parts with the integral in your formula. (in the right side of you formula)
$endgroup$
– FDP
Feb 1 at 21:47
add a comment |
$begingroup$
By creating a reduction formula for
$$I_n=int_0 ^pi x^n sin x ,dx$$ Show that $I_4=pi^4 -12pi^2 + 48$
So by using integration by parts I wrote $I_n$ as $$I_n=int_0 ^pi x^n sin x ,dx = left[-x^ncos x right]_0 ^pi -int_0 ^pi nx^{n-1}times -cos x, dx$$ and after messing about with this I got $$I_n = pi^n +nI_{n-1}$$ However, the answer shows that this is wrong and I don't get the correct answer for $I_4$. The correct answer is $I_n = pi^n -n(n-1)I_{n-2}$. Can someone help explain how you get this reduction formula as I can't seem to get it and why my one doesn't work.
integration definite-integrals reduction-formula
asked Feb 1 at 21:05
H.LinkhornH.Linkhorn
495213
$endgroup$
By creating a reduction formula for
$$I_n=int_0 ^pi x^n sin x ,dx$$ Show that $I_4=pi^4 -12pi^2 + 48$
So by using integration by parts I wrote $I_n$ as $$I_n=int_0 ^pi x^n sin x ,dx = left[-x^ncos x right]_0 ^pi -int_0 ^pi nx^{n-1}times -cos x, dx$$ and after messing about with this I got $$I_n = pi^n +nI_{n-1}$$ However, the answer shows that this is wrong and I don't get the correct answer for $I_4$. The correct answer is $I_n = pi^n -n(n-1)I_{n-2}$. Can someone help explain how you get this reduction formula as I can't seem to get it and why my one doesn't work.
integration definite-integrals reduction-formula
integration definite-integrals reduction-formula
asked Feb 1 at 21:05
H.LinkhornH.Linkhorn
495213
asked Feb 1 at 21:05
H.LinkhornH.Linkhorn
495213
asked Feb 1 at 21:05
H.LinkhornH.Linkhorn
495213
asked Feb 1 at 21:05
H.LinkhornH.Linkhorn
495213
asked Feb 1 at 21:05
H.LinkhornH.Linkhorn
495213
495213
1
$begingroup$
You need to do by-parts again in order to compare the new integral with $I_n.$
$endgroup$
– Adrian Keister
Feb 1 at 21:08
$begingroup$
Can you show me what you mean?
$endgroup$
– H.Linkhorn
Feb 1 at 21:26
$begingroup$
Do you know an antiderivative of the function $cos$ is $sin$? Perform again an integration by parts with the integral in your formula. (in the right side of you formula)
$endgroup$
– FDP
Feb 1 at 21:47
add a comment |
1
$begingroup$
You need to do by-parts again in order to compare the new integral with $I_n.$
$endgroup$
– Adrian Keister
Feb 1 at 21:08
$begingroup$
Can you show me what you mean?
$endgroup$
– H.Linkhorn
Feb 1 at 21:26
$begingroup$
Do you know an antiderivative of the function $cos$ is $sin$? Perform again an integration by parts with the integral in your formula. (in the right side of you formula)
$endgroup$
– FDP
Feb 1 at 21:47
1
1
$begingroup$
You need to do by-parts again in order to compare the new integral with $I_n.$
$endgroup$
– Adrian Keister
Feb 1 at 21:08
$begingroup$
You need to do by-parts again in order to compare the new integral with $I_n.$
$endgroup$
– Adrian Keister
Feb 1 at 21:08
$begingroup$
Can you show me what you mean?
$endgroup$
– H.Linkhorn
Feb 1 at 21:26
$begingroup$
Can you show me what you mean?
$endgroup$
– H.Linkhorn
Feb 1 at 21:26
$begingroup$
Do you know an antiderivative of the function $cos$ is $sin$? Perform again an integration by parts with the integral in your formula. (in the right side of you formula)
$endgroup$
– FDP
Feb 1 at 21:47
$begingroup$
Do you know an antiderivative of the function $cos$ is $sin$? Perform again an integration by parts with the integral in your formula. (in the right side of you formula)
$endgroup$
– FDP
Feb 1 at 21:47
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Not really different, but all fully formatted and pretty :)
$$S(n)=int_0^pi x^nsin x,mathrm dx$$
IBP: $$mathrm dv=sin x,mathrm dxRightarrow v=-cos x\ u=x^nRightarrow mathrm du=nx^{n-1}mathrm dx$$
Thus
$$S(n)=-x^ncos xbig|_0^pi+nint_0^pi x^{n-1}cos x,mathrm dx$$
$$S(n)=pi^n+nint_0^pi x^{n-1}cos x,mathrm dx$$
IBP: $$mathrm dv=cos x,mathrm dxRightarrow v=sin x\ u=x^{n-1}Rightarrow mathrm du=(n-1)x^{n-2}mathrm dx$$
Thus
$$S(n)=pi^n+nleft[x^{n-1}sin xbig|_{0}^{pi}-(n-1)int_0^pi x^{n-2}sin x,mathrm dxright]$$
$$S(n)=pi^n-n(n-1)S(n-2)$$
QED
answered Feb 2 at 5:28
clathratusclathratus
5,1141439
$endgroup$
add a comment |
$begingroup$
$I_n=pi^n +int_0 ^pi nx^{n-1}cos x, dx ne pi^n + nI_{n-1}$
As $I$ is an integral with a sine factor and not a cosine factor.
You need to do integration by parts a second time.
$nint_0 ^pi nx^{n-1}cos x, dx = n(n-1)sin x|_0^{pi} - n(n-1)int_0^pi x^{n-2}sin x dx$
$I_n=pi^n - n(n-1)I_{n-2}$
And $I_0 = 2$
answered Feb 1 at 22:07
Doug MDoug M
45.4k31954
$endgroup$
add a comment |
$begingroup$
Let $$I_n(a)=int_0^{pi}x^nsin(ax)dx$$
Then $$begin{align*}I_{n-2}^{''}(a)
&=frac{d^2}{da^2}int_0^{pi}x^{n-2}sin(ax)dx\
&=int_0^{pi}x^{n-2}left(frac{d^2}{da^2}sin(ax)right)dx\
&=-int_0^{pi}x^nsin(ax)dx\
&=-I_n(a)end{align*}$$
and $$begin{align*}I_n(1)
&=-lim_{ato 1}frac{d^2}{da^2}int_0^{pi}x^{n-2}sin(ax)dx\
&=-lim_{ato 1}frac{d^2}{da^2}left(frac1{a^{n-1}}int_0^{api}x^{n-2}sin(x)dxright)\
&=-lim_{ato 1}left(frac{n(n-1)}{a^{n+1}}int_0^{api}x^{n-2}sin(x)dx-2frac{n-1}{a^{n}}pi (api)^{n-2}sin(api)right.\&qquad+left.frac{pi^{n-1}}{a^{n-1}}left((n-2)a^{n-3}sin(api)+a^{n-2}picos(api)right)right)\
&=-n(n-1)I_{n-2}(1)+pi^nend{align*}$$
which is what we wanted.
answered Feb 1 at 22:35
Edward H.Edward H.
1689
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3096730%2fby-creating-a-reduction-formula-for-i-n-int-0-pi-xn-sin-x-dx-show-that%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Not really different, but all fully formatted and pretty :)
$$S(n)=int_0^pi x^nsin x,mathrm dx$$
IBP: $$mathrm dv=sin x,mathrm dxRightarrow v=-cos x\ u=x^nRightarrow mathrm du=nx^{n-1}mathrm dx$$
Thus
$$S(n)=-x^ncos xbig|_0^pi+nint_0^pi x^{n-1}cos x,mathrm dx$$
$$S(n)=pi^n+nint_0^pi x^{n-1}cos x,mathrm dx$$
IBP: $$mathrm dv=cos x,mathrm dxRightarrow v=sin x\ u=x^{n-1}Rightarrow mathrm du=(n-1)x^{n-2}mathrm dx$$
Thus
$$S(n)=pi^n+nleft[x^{n-1}sin xbig|_{0}^{pi}-(n-1)int_0^pi x^{n-2}sin x,mathrm dxright]$$
$$S(n)=pi^n-n(n-1)S(n-2)$$
QED
answered Feb 2 at 5:28
clathratusclathratus
5,1141439
$endgroup$
add a comment |
$begingroup$
Not really different, but all fully formatted and pretty :)
$$S(n)=int_0^pi x^nsin x,mathrm dx$$
IBP: $$mathrm dv=sin x,mathrm dxRightarrow v=-cos x\ u=x^nRightarrow mathrm du=nx^{n-1}mathrm dx$$
Thus
$$S(n)=-x^ncos xbig|_0^pi+nint_0^pi x^{n-1}cos x,mathrm dx$$
$$S(n)=pi^n+nint_0^pi x^{n-1}cos x,mathrm dx$$
IBP: $$mathrm dv=cos x,mathrm dxRightarrow v=sin x\ u=x^{n-1}Rightarrow mathrm du=(n-1)x^{n-2}mathrm dx$$
Thus
$$S(n)=pi^n+nleft[x^{n-1}sin xbig|_{0}^{pi}-(n-1)int_0^pi x^{n-2}sin x,mathrm dxright]$$
$$S(n)=pi^n-n(n-1)S(n-2)$$
QED
answered Feb 2 at 5:28
clathratusclathratus
5,1141439
$endgroup$
add a comment |
$begingroup$
Not really different, but all fully formatted and pretty :)
$$S(n)=int_0^pi x^nsin x,mathrm dx$$
IBP: $$mathrm dv=sin x,mathrm dxRightarrow v=-cos x\ u=x^nRightarrow mathrm du=nx^{n-1}mathrm dx$$
Thus
$$S(n)=-x^ncos xbig|_0^pi+nint_0^pi x^{n-1}cos x,mathrm dx$$
$$S(n)=pi^n+nint_0^pi x^{n-1}cos x,mathrm dx$$
IBP: $$mathrm dv=cos x,mathrm dxRightarrow v=sin x\ u=x^{n-1}Rightarrow mathrm du=(n-1)x^{n-2}mathrm dx$$
Thus
$$S(n)=pi^n+nleft[x^{n-1}sin xbig|_{0}^{pi}-(n-1)int_0^pi x^{n-2}sin x,mathrm dxright]$$
$$S(n)=pi^n-n(n-1)S(n-2)$$
QED
answered Feb 2 at 5:28
clathratusclathratus
5,1141439
$endgroup$
Not really different, but all fully formatted and pretty :)
$$S(n)=int_0^pi x^nsin x,mathrm dx$$
IBP: $$mathrm dv=sin x,mathrm dxRightarrow v=-cos x\ u=x^nRightarrow mathrm du=nx^{n-1}mathrm dx$$
Thus
$$S(n)=-x^ncos xbig|_0^pi+nint_0^pi x^{n-1}cos x,mathrm dx$$
$$S(n)=pi^n+nint_0^pi x^{n-1}cos x,mathrm dx$$
IBP: $$mathrm dv=cos x,mathrm dxRightarrow v=sin x\ u=x^{n-1}Rightarrow mathrm du=(n-1)x^{n-2}mathrm dx$$
Thus
$$S(n)=pi^n+nleft[x^{n-1}sin xbig|_{0}^{pi}-(n-1)int_0^pi x^{n-2}sin x,mathrm dxright]$$
$$S(n)=pi^n-n(n-1)S(n-2)$$
QED
answered Feb 2 at 5:28
clathratusclathratus
5,1141439
answered Feb 2 at 5:28
clathratusclathratus
5,1141439
answered Feb 2 at 5:28
clathratusclathratus
5,1141439
answered Feb 2 at 5:28
clathratusclathratus
5,1141439
5,1141439
add a comment |
add a comment |
$begingroup$
$I_n=pi^n +int_0 ^pi nx^{n-1}cos x, dx ne pi^n + nI_{n-1}$
As $I$ is an integral with a sine factor and not a cosine factor.
You need to do integration by parts a second time.
$nint_0 ^pi nx^{n-1}cos x, dx = n(n-1)sin x|_0^{pi} - n(n-1)int_0^pi x^{n-2}sin x dx$
$I_n=pi^n - n(n-1)I_{n-2}$
And $I_0 = 2$
answered Feb 1 at 22:07
Doug MDoug M
45.4k31954
$endgroup$
add a comment |
$begingroup$
$I_n=pi^n +int_0 ^pi nx^{n-1}cos x, dx ne pi^n + nI_{n-1}$
As $I$ is an integral with a sine factor and not a cosine factor.
You need to do integration by parts a second time.
$nint_0 ^pi nx^{n-1}cos x, dx = n(n-1)sin x|_0^{pi} - n(n-1)int_0^pi x^{n-2}sin x dx$
$I_n=pi^n - n(n-1)I_{n-2}$
And $I_0 = 2$
answered Feb 1 at 22:07
Doug MDoug M
45.4k31954
$endgroup$
add a comment |
$begingroup$
$I_n=pi^n +int_0 ^pi nx^{n-1}cos x, dx ne pi^n + nI_{n-1}$
As $I$ is an integral with a sine factor and not a cosine factor.
You need to do integration by parts a second time.
$nint_0 ^pi nx^{n-1}cos x, dx = n(n-1)sin x|_0^{pi} - n(n-1)int_0^pi x^{n-2}sin x dx$
$I_n=pi^n - n(n-1)I_{n-2}$
And $I_0 = 2$
answered Feb 1 at 22:07
Doug MDoug M
45.4k31954
$endgroup$
$I_n=pi^n +int_0 ^pi nx^{n-1}cos x, dx ne pi^n + nI_{n-1}$
As $I$ is an integral with a sine factor and not a cosine factor.
You need to do integration by parts a second time.
$nint_0 ^pi nx^{n-1}cos x, dx = n(n-1)sin x|_0^{pi} - n(n-1)int_0^pi x^{n-2}sin x dx$
$I_n=pi^n - n(n-1)I_{n-2}$
And $I_0 = 2$
answered Feb 1 at 22:07
Doug MDoug M
45.4k31954
answered Feb 1 at 22:07
Doug MDoug M
45.4k31954
answered Feb 1 at 22:07
Doug MDoug M
45.4k31954
answered Feb 1 at 22:07
Doug MDoug M
45.4k31954
45.4k31954
add a comment |
add a comment |
$begingroup$
Let $$I_n(a)=int_0^{pi}x^nsin(ax)dx$$
Then $$begin{align*}I_{n-2}^{''}(a)
&=frac{d^2}{da^2}int_0^{pi}x^{n-2}sin(ax)dx\
&=int_0^{pi}x^{n-2}left(frac{d^2}{da^2}sin(ax)right)dx\
&=-int_0^{pi}x^nsin(ax)dx\
&=-I_n(a)end{align*}$$
and $$begin{align*}I_n(1)
&=-lim_{ato 1}frac{d^2}{da^2}int_0^{pi}x^{n-2}sin(ax)dx\
&=-lim_{ato 1}frac{d^2}{da^2}left(frac1{a^{n-1}}int_0^{api}x^{n-2}sin(x)dxright)\
&=-lim_{ato 1}left(frac{n(n-1)}{a^{n+1}}int_0^{api}x^{n-2}sin(x)dx-2frac{n-1}{a^{n}}pi (api)^{n-2}sin(api)right.\&qquad+left.frac{pi^{n-1}}{a^{n-1}}left((n-2)a^{n-3}sin(api)+a^{n-2}picos(api)right)right)\
&=-n(n-1)I_{n-2}(1)+pi^nend{align*}$$
which is what we wanted.
answered Feb 1 at 22:35
Edward H.Edward H.
1689
$endgroup$
add a comment |
$begingroup$
Let $$I_n(a)=int_0^{pi}x^nsin(ax)dx$$
Then $$begin{align*}I_{n-2}^{''}(a)
&=frac{d^2}{da^2}int_0^{pi}x^{n-2}sin(ax)dx\
&=int_0^{pi}x^{n-2}left(frac{d^2}{da^2}sin(ax)right)dx\
&=-int_0^{pi}x^nsin(ax)dx\
&=-I_n(a)end{align*}$$
and $$begin{align*}I_n(1)
&=-lim_{ato 1}frac{d^2}{da^2}int_0^{pi}x^{n-2}sin(ax)dx\
&=-lim_{ato 1}frac{d^2}{da^2}left(frac1{a^{n-1}}int_0^{api}x^{n-2}sin(x)dxright)\
&=-lim_{ato 1}left(frac{n(n-1)}{a^{n+1}}int_0^{api}x^{n-2}sin(x)dx-2frac{n-1}{a^{n}}pi (api)^{n-2}sin(api)right.\&qquad+left.frac{pi^{n-1}}{a^{n-1}}left((n-2)a^{n-3}sin(api)+a^{n-2}picos(api)right)right)\
&=-n(n-1)I_{n-2}(1)+pi^nend{align*}$$
which is what we wanted.
answered Feb 1 at 22:35
Edward H.Edward H.
1689
$endgroup$
add a comment |
$begingroup$
Let $$I_n(a)=int_0^{pi}x^nsin(ax)dx$$
Then $$begin{align*}I_{n-2}^{''}(a)
&=frac{d^2}{da^2}int_0^{pi}x^{n-2}sin(ax)dx\
&=int_0^{pi}x^{n-2}left(frac{d^2}{da^2}sin(ax)right)dx\
&=-int_0^{pi}x^nsin(ax)dx\
&=-I_n(a)end{align*}$$
and $$begin{align*}I_n(1)
&=-lim_{ato 1}frac{d^2}{da^2}int_0^{pi}x^{n-2}sin(ax)dx\
&=-lim_{ato 1}frac{d^2}{da^2}left(frac1{a^{n-1}}int_0^{api}x^{n-2}sin(x)dxright)\
&=-lim_{ato 1}left(frac{n(n-1)}{a^{n+1}}int_0^{api}x^{n-2}sin(x)dx-2frac{n-1}{a^{n}}pi (api)^{n-2}sin(api)right.\&qquad+left.frac{pi^{n-1}}{a^{n-1}}left((n-2)a^{n-3}sin(api)+a^{n-2}picos(api)right)right)\
&=-n(n-1)I_{n-2}(1)+pi^nend{align*}$$
which is what we wanted.
answered Feb 1 at 22:35
Edward H.Edward H.
1689
$endgroup$
Let $$I_n(a)=int_0^{pi}x^nsin(ax)dx$$
Then $$begin{align*}I_{n-2}^{''}(a)
&=frac{d^2}{da^2}int_0^{pi}x^{n-2}sin(ax)dx\
&=int_0^{pi}x^{n-2}left(frac{d^2}{da^2}sin(ax)right)dx\
&=-int_0^{pi}x^nsin(ax)dx\
&=-I_n(a)end{align*}$$
and $$begin{align*}I_n(1)
&=-lim_{ato 1}frac{d^2}{da^2}int_0^{pi}x^{n-2}sin(ax)dx\
&=-lim_{ato 1}frac{d^2}{da^2}left(frac1{a^{n-1}}int_0^{api}x^{n-2}sin(x)dxright)\
&=-lim_{ato 1}left(frac{n(n-1)}{a^{n+1}}int_0^{api}x^{n-2}sin(x)dx-2frac{n-1}{a^{n}}pi (api)^{n-2}sin(api)right.\&qquad+left.frac{pi^{n-1}}{a^{n-1}}left((n-2)a^{n-3}sin(api)+a^{n-2}picos(api)right)right)\
&=-n(n-1)I_{n-2}(1)+pi^nend{align*}$$
which is what we wanted.
answered Feb 1 at 22:35
Edward H.Edward H.
1689
answered Feb 1 at 22:35
Edward H.Edward H.
1689
answered Feb 1 at 22:35
Edward H.Edward H.
1689
answered Feb 1 at 22:35
Edward H.Edward H.
1689
1689
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3096730%2fby-creating-a-reduction-formula-for-i-n-int-0-pi-xn-sin-x-dx-show-that%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
You need to do by-parts again in order to compare the new integral with $I_n.$
$endgroup$
– Adrian Keister
Feb 1 at 21:08
$begingroup$
Can you show me what you mean?
$endgroup$
– H.Linkhorn
Feb 1 at 21:26
$begingroup$
Do you know an antiderivative of the function $cos$ is $sin$? Perform again an integration by parts with the integral in your formula. (in the right side of you formula)
$endgroup$
– FDP
Feb 1 at 21:47