Trigonometric and Exponential Integrationpositive integer $n$ in definite integrationFinding $sum^infty_n=1int^2(n+1)pi_2npifracxsin x+cos xx^2$Trigonometric exponential IntegrationValue of product of cosines definite integrationArrangement of definite integral in increasing orderFinding $ int^1_-1fracx^3sqrt1-x^2lnbigg(frac1+x1-xbigg)dx$Finding $int^pi_0sin(8x+8sin 3x)dx$If $int^infty_0fracln^2(x)(1-x)^2dx+kint^1_0fracln(1-x)xdx=0.$ Find $k$.Evaluation of inverse of $tan$ integrationWhat is $PQ^-1$ if $P=int^pi_0fracsin(994 x)sin xsin(1332x),dx$ and $Q=int^1_0fracx^338(x^1988-1)x^2-1,dx$?
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Trigonometric and Exponential Integration
positive integer $n$ in definite integrationFinding $sum^infty_n=1int^2(n+1)pi_2npifracxsin x+cos xx^2$Trigonometric exponential IntegrationValue of product of cosines definite integrationArrangement of definite integral in increasing orderFinding $ int^1_-1fracx^3sqrt1-x^2lnbigg(frac1+x1-xbigg)dx$Finding $int^pi_0sin(8x+8sin 3x)dx$If $int^infty_0fracln^2(x)(1-x)^2dx+kint^1_0fracln(1-x)xdx=0.$ Find $k$.Evaluation of inverse of $tan$ integrationWhat is $PQ^-1$ if $P=int^pi_0fracsin(994 x)sin xsin(1332x),dx$ and $Q=int^1_0fracx^338(x^1988-1)x^2-1,dx$?
$begingroup$
If $displaystyle I_n =int^infty_fracpi2e^-xcos^n(x)dx.$ Then $displaystyle fracI_2018I_2016$ is
Try: using by parts
$$I_n=int^infty_fracpi2e^-xcos^n(x)dx$$
$$I_n=-cos^n(x)cdot e^-xbigg|^infty_fracpi2-nint^infty_fracpi2cos^n-1(x)sin(x)cdot e^-xdx$$
$$I_n=nint^infty_fracpi2cos^n-1(x)sin(x)cdot e^-xdx$$
Could some help me to solve it, Thanks
definite-integrals
asked Apr 24 at 6:51
DXTDXT
5,9792733
$endgroup$
add a comment |
$begingroup$
If $displaystyle I_n =int^infty_fracpi2e^-xcos^n(x)dx.$ Then $displaystyle fracI_2018I_2016$ is
Try: using by parts
$$I_n=int^infty_fracpi2e^-xcos^n(x)dx$$
$$I_n=-cos^n(x)cdot e^-xbigg|^infty_fracpi2-nint^infty_fracpi2cos^n-1(x)sin(x)cdot e^-xdx$$
$$I_n=nint^infty_fracpi2cos^n-1(x)sin(x)cdot e^-xdx$$
Could some help me to solve it, Thanks
definite-integrals
asked Apr 24 at 6:51
DXTDXT
5,9792733
$endgroup$
$begingroup$
Integrate both sides of $$dfracd(e^-xcos^mxsin x)dx$$
$endgroup$
– lab bhattacharjee
Apr 24 at 6:59
add a comment |
$begingroup$
If $displaystyle I_n =int^infty_fracpi2e^-xcos^n(x)dx.$ Then $displaystyle fracI_2018I_2016$ is
Try: using by parts
$$I_n=int^infty_fracpi2e^-xcos^n(x)dx$$
$$I_n=-cos^n(x)cdot e^-xbigg|^infty_fracpi2-nint^infty_fracpi2cos^n-1(x)sin(x)cdot e^-xdx$$
$$I_n=nint^infty_fracpi2cos^n-1(x)sin(x)cdot e^-xdx$$
Could some help me to solve it, Thanks
definite-integrals
asked Apr 24 at 6:51
DXTDXT
5,9792733
$endgroup$
If $displaystyle I_n =int^infty_fracpi2e^-xcos^n(x)dx.$ Then $displaystyle fracI_2018I_2016$ is
Try: using by parts
$$I_n=int^infty_fracpi2e^-xcos^n(x)dx$$
$$I_n=-cos^n(x)cdot e^-xbigg|^infty_fracpi2-nint^infty_fracpi2cos^n-1(x)sin(x)cdot e^-xdx$$
$$I_n=nint^infty_fracpi2cos^n-1(x)sin(x)cdot e^-xdx$$
Could some help me to solve it, Thanks
definite-integrals
definite-integrals
asked Apr 24 at 6:51
DXTDXT
5,9792733
asked Apr 24 at 6:51
DXTDXT
5,9792733
asked Apr 24 at 6:51
DXTDXT
5,9792733
asked Apr 24 at 6:51
DXTDXT
5,9792733
asked Apr 24 at 6:51
DXTDXT
5,9792733
5,9792733
$begingroup$
Integrate both sides of $$dfracd(e^-xcos^mxsin x)dx$$
$endgroup$
– lab bhattacharjee
Apr 24 at 6:59
add a comment |
$begingroup$
Integrate both sides of $$dfracd(e^-xcos^mxsin x)dx$$
$endgroup$
– lab bhattacharjee
Apr 24 at 6:59
$begingroup$
Integrate both sides of $$dfracd(e^-xcos^mxsin x)dx$$
$endgroup$
– lab bhattacharjee
Apr 24 at 6:59
$begingroup$
Integrate both sides of $$dfracd(e^-xcos^mxsin x)dx$$
$endgroup$
– lab bhattacharjee
Apr 24 at 6:59
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
As you say,
$$
fracI_nn = - int_pi/2^infty mathrme^-x cos^n-1(x)sin(x) mathrmdx
$$
I believe the trick is to use integration by parts again, by differentiating the term multiplying the exponential, that is
beginalign*
fracmathrmd mathrmd x , cos^n-1(x)sin(x) &= cos^n(x) - (n-1)cos^n-2(x)colorbluesin^2(x) \
& = cos^n(x) - (n-1)cos^n-2(x)left[colorblue1-cos^2(x)right] \
& = colorredcos^n(x) - (n-1) cos^n-2(x) + (ncolorred-1)cos^n(x) \
& = n cos^n(x) - (n-1) cos^n-2(x).
endalign*
Then the whole integral becomes
$$
fracI_nn = - displaystyle colorredleft[ -mathrme^-x cos^n-1(x)sin(x)right]^infty_pi/2 - int_pi/2^infty mathrme^-x left[ n cos^n(x) - (n-1) cos^n-2(x)right] mathrmdx
$$
with the red term being zero, we obtain
beginalign*
fracI_nn &= -nI_n + (n-1)I_n-2, \
I_n & = - n^2 I_n + n(n-1)I_n-2, \
(n^2 + 1)I_n &= n(n-1)I_n-2, \
fracI_nI_n-2 &= fracn(n-1)n^2+1, \
endalign*
In particular,
$$
fracI_2018I_2016 = frac2018 times 20172018^2+1 = frac40703064072325.
$$
answered Apr 24 at 9:06
Bennett GardinerBennett Gardiner
3,16211540
$endgroup$
3
$begingroup$
No, the result is $dfrac40703064072325$. (Checked with Alpha.)
$endgroup$
– Yves Daoust
Apr 24 at 10:50
1
$begingroup$
Thanks, I took a typo from OP as my starting point (oops).
$endgroup$
– Bennett Gardiner
Apr 24 at 11:00
add a comment |
$begingroup$
For brevity we set
$$f_c,s(x):= e^-xcos^c(x)sin^s(x),$$
$$F_c,s(x):=int f_c,s(x),dx.$$
Then by parts, integrating on $e^-x$,
$$F_c,0=-f_c,0-cF_c-1,1$$
and
$$F_c-1,1=-f_c-1,1-(c-1)F_c-2,2+F_c,0.$$
Using $sin^2x=1-cos^2x$ (i.e. $F_c-2,2=F_c-2,0-F_c,0$) and the fact that the $f$ terms vanish, we have after simplification,
$$(c^2+1)F_c,0=c(c-1)F_c-2,0.$$
answered Apr 24 at 9:57
Yves DaoustYves Daoust
135k676233
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
As you say,
$$
fracI_nn = - int_pi/2^infty mathrme^-x cos^n-1(x)sin(x) mathrmdx
$$
I believe the trick is to use integration by parts again, by differentiating the term multiplying the exponential, that is
beginalign*
fracmathrmd mathrmd x , cos^n-1(x)sin(x) &= cos^n(x) - (n-1)cos^n-2(x)colorbluesin^2(x) \
& = cos^n(x) - (n-1)cos^n-2(x)left[colorblue1-cos^2(x)right] \
& = colorredcos^n(x) - (n-1) cos^n-2(x) + (ncolorred-1)cos^n(x) \
& = n cos^n(x) - (n-1) cos^n-2(x).
endalign*
Then the whole integral becomes
$$
fracI_nn = - displaystyle colorredleft[ -mathrme^-x cos^n-1(x)sin(x)right]^infty_pi/2 - int_pi/2^infty mathrme^-x left[ n cos^n(x) - (n-1) cos^n-2(x)right] mathrmdx
$$
with the red term being zero, we obtain
beginalign*
fracI_nn &= -nI_n + (n-1)I_n-2, \
I_n & = - n^2 I_n + n(n-1)I_n-2, \
(n^2 + 1)I_n &= n(n-1)I_n-2, \
fracI_nI_n-2 &= fracn(n-1)n^2+1, \
endalign*
In particular,
$$
fracI_2018I_2016 = frac2018 times 20172018^2+1 = frac40703064072325.
$$
answered Apr 24 at 9:06
Bennett GardinerBennett Gardiner
3,16211540
$endgroup$
3
$begingroup$
No, the result is $dfrac40703064072325$. (Checked with Alpha.)
$endgroup$
– Yves Daoust
Apr 24 at 10:50
1
$begingroup$
Thanks, I took a typo from OP as my starting point (oops).
$endgroup$
– Bennett Gardiner
Apr 24 at 11:00
add a comment |
$begingroup$
As you say,
$$
fracI_nn = - int_pi/2^infty mathrme^-x cos^n-1(x)sin(x) mathrmdx
$$
I believe the trick is to use integration by parts again, by differentiating the term multiplying the exponential, that is
beginalign*
fracmathrmd mathrmd x , cos^n-1(x)sin(x) &= cos^n(x) - (n-1)cos^n-2(x)colorbluesin^2(x) \
& = cos^n(x) - (n-1)cos^n-2(x)left[colorblue1-cos^2(x)right] \
& = colorredcos^n(x) - (n-1) cos^n-2(x) + (ncolorred-1)cos^n(x) \
& = n cos^n(x) - (n-1) cos^n-2(x).
endalign*
Then the whole integral becomes
$$
fracI_nn = - displaystyle colorredleft[ -mathrme^-x cos^n-1(x)sin(x)right]^infty_pi/2 - int_pi/2^infty mathrme^-x left[ n cos^n(x) - (n-1) cos^n-2(x)right] mathrmdx
$$
with the red term being zero, we obtain
beginalign*
fracI_nn &= -nI_n + (n-1)I_n-2, \
I_n & = - n^2 I_n + n(n-1)I_n-2, \
(n^2 + 1)I_n &= n(n-1)I_n-2, \
fracI_nI_n-2 &= fracn(n-1)n^2+1, \
endalign*
In particular,
$$
fracI_2018I_2016 = frac2018 times 20172018^2+1 = frac40703064072325.
$$
answered Apr 24 at 9:06
Bennett GardinerBennett Gardiner
3,16211540
$endgroup$
3
$begingroup$
No, the result is $dfrac40703064072325$. (Checked with Alpha.)
$endgroup$
– Yves Daoust
Apr 24 at 10:50
1
$begingroup$
Thanks, I took a typo from OP as my starting point (oops).
$endgroup$
– Bennett Gardiner
Apr 24 at 11:00
add a comment |
$begingroup$
As you say,
$$
fracI_nn = - int_pi/2^infty mathrme^-x cos^n-1(x)sin(x) mathrmdx
$$
I believe the trick is to use integration by parts again, by differentiating the term multiplying the exponential, that is
beginalign*
fracmathrmd mathrmd x , cos^n-1(x)sin(x) &= cos^n(x) - (n-1)cos^n-2(x)colorbluesin^2(x) \
& = cos^n(x) - (n-1)cos^n-2(x)left[colorblue1-cos^2(x)right] \
& = colorredcos^n(x) - (n-1) cos^n-2(x) + (ncolorred-1)cos^n(x) \
& = n cos^n(x) - (n-1) cos^n-2(x).
endalign*
Then the whole integral becomes
$$
fracI_nn = - displaystyle colorredleft[ -mathrme^-x cos^n-1(x)sin(x)right]^infty_pi/2 - int_pi/2^infty mathrme^-x left[ n cos^n(x) - (n-1) cos^n-2(x)right] mathrmdx
$$
with the red term being zero, we obtain
beginalign*
fracI_nn &= -nI_n + (n-1)I_n-2, \
I_n & = - n^2 I_n + n(n-1)I_n-2, \
(n^2 + 1)I_n &= n(n-1)I_n-2, \
fracI_nI_n-2 &= fracn(n-1)n^2+1, \
endalign*
In particular,
$$
fracI_2018I_2016 = frac2018 times 20172018^2+1 = frac40703064072325.
$$
answered Apr 24 at 9:06
Bennett GardinerBennett Gardiner
3,16211540
$endgroup$
As you say,
$$
fracI_nn = - int_pi/2^infty mathrme^-x cos^n-1(x)sin(x) mathrmdx
$$
I believe the trick is to use integration by parts again, by differentiating the term multiplying the exponential, that is
beginalign*
fracmathrmd mathrmd x , cos^n-1(x)sin(x) &= cos^n(x) - (n-1)cos^n-2(x)colorbluesin^2(x) \
& = cos^n(x) - (n-1)cos^n-2(x)left[colorblue1-cos^2(x)right] \
& = colorredcos^n(x) - (n-1) cos^n-2(x) + (ncolorred-1)cos^n(x) \
& = n cos^n(x) - (n-1) cos^n-2(x).
endalign*
Then the whole integral becomes
$$
fracI_nn = - displaystyle colorredleft[ -mathrme^-x cos^n-1(x)sin(x)right]^infty_pi/2 - int_pi/2^infty mathrme^-x left[ n cos^n(x) - (n-1) cos^n-2(x)right] mathrmdx
$$
with the red term being zero, we obtain
beginalign*
fracI_nn &= -nI_n + (n-1)I_n-2, \
I_n & = - n^2 I_n + n(n-1)I_n-2, \
(n^2 + 1)I_n &= n(n-1)I_n-2, \
fracI_nI_n-2 &= fracn(n-1)n^2+1, \
endalign*
In particular,
$$
fracI_2018I_2016 = frac2018 times 20172018^2+1 = frac40703064072325.
$$
answered Apr 24 at 9:06
Bennett GardinerBennett Gardiner
3,16211540
edited Apr 24 at 11:00
answered Apr 24 at 9:06
Bennett GardinerBennett Gardiner
3,16211540
answered Apr 24 at 9:06
Bennett GardinerBennett Gardiner
3,16211540
answered Apr 24 at 9:06
Bennett GardinerBennett Gardiner
3,16211540
3,16211540
3
$begingroup$
No, the result is $dfrac40703064072325$. (Checked with Alpha.)
$endgroup$
– Yves Daoust
Apr 24 at 10:50
1
$begingroup$
Thanks, I took a typo from OP as my starting point (oops).
$endgroup$
– Bennett Gardiner
Apr 24 at 11:00
add a comment |
3
$begingroup$
No, the result is $dfrac40703064072325$. (Checked with Alpha.)
$endgroup$
– Yves Daoust
Apr 24 at 10:50
1
$begingroup$
Thanks, I took a typo from OP as my starting point (oops).
$endgroup$
– Bennett Gardiner
Apr 24 at 11:00
3
3
$begingroup$
No, the result is $dfrac40703064072325$. (Checked with Alpha.)
$endgroup$
– Yves Daoust
Apr 24 at 10:50
$begingroup$
No, the result is $dfrac40703064072325$. (Checked with Alpha.)
$endgroup$
– Yves Daoust
Apr 24 at 10:50
1
1
$begingroup$
Thanks, I took a typo from OP as my starting point (oops).
$endgroup$
– Bennett Gardiner
Apr 24 at 11:00
$begingroup$
Thanks, I took a typo from OP as my starting point (oops).
$endgroup$
– Bennett Gardiner
Apr 24 at 11:00
add a comment |
$begingroup$
For brevity we set
$$f_c,s(x):= e^-xcos^c(x)sin^s(x),$$
$$F_c,s(x):=int f_c,s(x),dx.$$
Then by parts, integrating on $e^-x$,
$$F_c,0=-f_c,0-cF_c-1,1$$
and
$$F_c-1,1=-f_c-1,1-(c-1)F_c-2,2+F_c,0.$$
Using $sin^2x=1-cos^2x$ (i.e. $F_c-2,2=F_c-2,0-F_c,0$) and the fact that the $f$ terms vanish, we have after simplification,
$$(c^2+1)F_c,0=c(c-1)F_c-2,0.$$
answered Apr 24 at 9:57
Yves DaoustYves Daoust
135k676233
$endgroup$
add a comment |
$begingroup$
For brevity we set
$$f_c,s(x):= e^-xcos^c(x)sin^s(x),$$
$$F_c,s(x):=int f_c,s(x),dx.$$
Then by parts, integrating on $e^-x$,
$$F_c,0=-f_c,0-cF_c-1,1$$
and
$$F_c-1,1=-f_c-1,1-(c-1)F_c-2,2+F_c,0.$$
Using $sin^2x=1-cos^2x$ (i.e. $F_c-2,2=F_c-2,0-F_c,0$) and the fact that the $f$ terms vanish, we have after simplification,
$$(c^2+1)F_c,0=c(c-1)F_c-2,0.$$
answered Apr 24 at 9:57
Yves DaoustYves Daoust
135k676233
$endgroup$
add a comment |
$begingroup$
For brevity we set
$$f_c,s(x):= e^-xcos^c(x)sin^s(x),$$
$$F_c,s(x):=int f_c,s(x),dx.$$
Then by parts, integrating on $e^-x$,
$$F_c,0=-f_c,0-cF_c-1,1$$
and
$$F_c-1,1=-f_c-1,1-(c-1)F_c-2,2+F_c,0.$$
Using $sin^2x=1-cos^2x$ (i.e. $F_c-2,2=F_c-2,0-F_c,0$) and the fact that the $f$ terms vanish, we have after simplification,
$$(c^2+1)F_c,0=c(c-1)F_c-2,0.$$
answered Apr 24 at 9:57
Yves DaoustYves Daoust
135k676233
$endgroup$
For brevity we set
$$f_c,s(x):= e^-xcos^c(x)sin^s(x),$$
$$F_c,s(x):=int f_c,s(x),dx.$$
Then by parts, integrating on $e^-x$,
$$F_c,0=-f_c,0-cF_c-1,1$$
and
$$F_c-1,1=-f_c-1,1-(c-1)F_c-2,2+F_c,0.$$
Using $sin^2x=1-cos^2x$ (i.e. $F_c-2,2=F_c-2,0-F_c,0$) and the fact that the $f$ terms vanish, we have after simplification,
$$(c^2+1)F_c,0=c(c-1)F_c-2,0.$$
answered Apr 24 at 9:57
Yves DaoustYves Daoust
135k676233
edited Apr 24 at 12:56
answered Apr 24 at 9:57
Yves DaoustYves Daoust
135k676233
answered Apr 24 at 9:57
Yves DaoustYves Daoust
135k676233
answered Apr 24 at 9:57
Yves DaoustYves Daoust
135k676233
135k676233
add a comment |
add a comment |
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Required, but never shown
Required, but never shown
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Integrate both sides of $$dfracd(e^-xcos^mxsin x)dx$$
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– lab bhattacharjee
Apr 24 at 6:59