Trigonometric and Exponential Integration Unicorn Meta Zoo #1: Why another podcast? ...
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Trigonometric and Exponential Integration
Unicorn Meta Zoo #1: Why another podcast?
Announcing the arrival of Valued Associate #679: Cesar Manarapositive integer $n$ in definite integrationFinding $sum^{infty}_{n=1}int^{2(n+1)pi}_{2npi}frac{xsin x+cos x}{x^2}$Trigonometric exponential IntegrationValue of product of cosines definite integrationArrangement of definite integral in increasing orderFinding $ int^{1}_{-1}frac{x^3}{sqrt{1-x^2}}lnbigg(frac{1+x}{1-x}bigg)dx$Finding $int^{pi}_{0}sin(8x+8sin 3x)dx$If $int^{infty}_{0}frac{ln^2(x)}{(1-x)^2}dx+kint^{1}_{0}frac{ln(1-x)}{x}dx=0.$ Find $k$.Evaluation of inverse of $tan$ integrationWhat is $PQ^{-1}$ if $P=int^{pi}_{0}frac{sin(994 x)}{sin x}sin(1332x),dx$ and $Q=int^{1}_{0}frac{x^{338}(x^{1988}-1)}{x^2-1},dx$?
$begingroup$
If $displaystyle I_{n} =int^{infty}_{frac{pi}{2}}e^{-x}cos^{n}(x)dx.$ Then $displaystyle frac{I_{2018}}{I_{2016}}$ is
Try: using by parts
$$I_{n}=int^{infty}_{frac{pi}{2}}e^{-x}cos^{n}(x)dx$$
$$I_{n}=-cos^{n}(x)cdot e^{-x}bigg|^{infty}_{frac{pi}{2}}-nint^{infty}_{frac{pi}{2}}cos^{n-1}(x)sin(x)cdot e^{-x}dx$$
$$I_{n}=nint^{infty}_{frac{pi}{2}}cos^{n-1}(x)sin(x)cdot e^{-x}dx$$
Could some help me to solve it, Thanks
definite-integrals
asked 17 hours ago
DXTDXT
5,8872733
$endgroup$
add a comment |
$begingroup$
If $displaystyle I_{n} =int^{infty}_{frac{pi}{2}}e^{-x}cos^{n}(x)dx.$ Then $displaystyle frac{I_{2018}}{I_{2016}}$ is
Try: using by parts
$$I_{n}=int^{infty}_{frac{pi}{2}}e^{-x}cos^{n}(x)dx$$
$$I_{n}=-cos^{n}(x)cdot e^{-x}bigg|^{infty}_{frac{pi}{2}}-nint^{infty}_{frac{pi}{2}}cos^{n-1}(x)sin(x)cdot e^{-x}dx$$
$$I_{n}=nint^{infty}_{frac{pi}{2}}cos^{n-1}(x)sin(x)cdot e^{-x}dx$$
Could some help me to solve it, Thanks
definite-integrals
asked 17 hours ago
DXTDXT
5,8872733
$endgroup$
$begingroup$
Integrate both sides of $$dfrac{d(e^{-x}cos^mxsin x)}{dx}$$
$endgroup$
– lab bhattacharjee
17 hours ago
add a comment |
$begingroup$
If $displaystyle I_{n} =int^{infty}_{frac{pi}{2}}e^{-x}cos^{n}(x)dx.$ Then $displaystyle frac{I_{2018}}{I_{2016}}$ is
Try: using by parts
$$I_{n}=int^{infty}_{frac{pi}{2}}e^{-x}cos^{n}(x)dx$$
$$I_{n}=-cos^{n}(x)cdot e^{-x}bigg|^{infty}_{frac{pi}{2}}-nint^{infty}_{frac{pi}{2}}cos^{n-1}(x)sin(x)cdot e^{-x}dx$$
$$I_{n}=nint^{infty}_{frac{pi}{2}}cos^{n-1}(x)sin(x)cdot e^{-x}dx$$
Could some help me to solve it, Thanks
definite-integrals
asked 17 hours ago
DXTDXT
5,8872733
$endgroup$
If $displaystyle I_{n} =int^{infty}_{frac{pi}{2}}e^{-x}cos^{n}(x)dx.$ Then $displaystyle frac{I_{2018}}{I_{2016}}$ is
Try: using by parts
$$I_{n}=int^{infty}_{frac{pi}{2}}e^{-x}cos^{n}(x)dx$$
$$I_{n}=-cos^{n}(x)cdot e^{-x}bigg|^{infty}_{frac{pi}{2}}-nint^{infty}_{frac{pi}{2}}cos^{n-1}(x)sin(x)cdot e^{-x}dx$$
$$I_{n}=nint^{infty}_{frac{pi}{2}}cos^{n-1}(x)sin(x)cdot e^{-x}dx$$
Could some help me to solve it, Thanks
definite-integrals
definite-integrals
asked 17 hours ago
DXTDXT
5,8872733
asked 17 hours ago
DXTDXT
5,8872733
asked 17 hours ago
DXTDXT
5,8872733
asked 17 hours ago
DXTDXT
5,8872733
asked 17 hours ago
DXTDXT
5,8872733
5,8872733
$begingroup$
Integrate both sides of $$dfrac{d(e^{-x}cos^mxsin x)}{dx}$$
$endgroup$
– lab bhattacharjee
17 hours ago
add a comment |
$begingroup$
Integrate both sides of $$dfrac{d(e^{-x}cos^mxsin x)}{dx}$$
$endgroup$
– lab bhattacharjee
17 hours ago
$begingroup$
Integrate both sides of $$dfrac{d(e^{-x}cos^mxsin x)}{dx}$$
$endgroup$
– lab bhattacharjee
17 hours ago
$begingroup$
Integrate both sides of $$dfrac{d(e^{-x}cos^mxsin x)}{dx}$$
$endgroup$
– lab bhattacharjee
17 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
As you say,
$$
frac{I_n}{n} = - int_{pi/2}^{infty} mathrm{e}^{-x} cos^{n-1}(x)sin(x) mathrm{d}x
$$
I believe the trick is to use integration by parts again, by differentiating the term multiplying the exponential, that is
begin{align*}
frac{mathrm{d} }{mathrm{d} x} , cos^{n-1}(x)sin(x) &= cos^n(x) - (n-1)cos^{n-2}(x)color{blue}{sin^2(x)} \
& = cos^n(x) - (n-1)cos^{n-2}(x)left[color{blue}{1-cos^2(x)}right] \
& = color{red}{cos^n(x)} - (n-1) cos^{n-2}(x) + (ncolor{red}{-1})cos^n(x) \
& = n cos^n(x) - (n-1) cos^{n-2}(x).
end{align*}
Then the whole integral becomes
$$
frac{I_n}{n} = - displaystyle color{red}{left[ -mathrm{e}^{-x} cos^{n-1}(x)sin(x)right]^{infty}_{pi/2}} - int_{pi/2}^{infty} mathrm{e}^{-x} left[ n cos^n(x) - (n-1) cos^{n-2}(x)right] mathrm{d}x
$$
with the red term being zero, we obtain
begin{align*}
frac{I_n}{n} &= -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}, \
frac{I_n}{I_{n-2}} &= frac{n(n-1)}{n^2+1}, \
end{align*}
In particular,
$$
frac{I_{2018}}{I_{2016}} = frac{2018 times 2017}{2018^2+1} = frac{4070306}{4072325}.
$$
answered 15 hours ago
Bennett GardinerBennett Gardiner
3,16211540
$endgroup$
2
$begingroup$
No, the result is $dfrac{4070306}{4072325}$. (Checked with Alpha.)
$endgroup$
– Yves Daoust
13 hours ago
1
$begingroup$
Thanks, I took a typo from OP as my starting point (oops).
$endgroup$
– Bennett Gardiner
13 hours ago
add a comment |
$begingroup$
For brevity we set
$$f_{c,s}(x):= e^{-x}cos^c(x)sin^s(x),$$
$$F_{c,s}(x):=int f_{c,s}(x),dx.$$
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 14 hours ago
Yves DaoustYves Daoust
134k676232
$endgroup$
add a comment |
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2 Answers
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$begingroup$
As you say,
$$
frac{I_n}{n} = - int_{pi/2}^{infty} mathrm{e}^{-x} cos^{n-1}(x)sin(x) mathrm{d}x
$$
I believe the trick is to use integration by parts again, by differentiating the term multiplying the exponential, that is
begin{align*}
frac{mathrm{d} }{mathrm{d} x} , cos^{n-1}(x)sin(x) &= cos^n(x) - (n-1)cos^{n-2}(x)color{blue}{sin^2(x)} \
& = cos^n(x) - (n-1)cos^{n-2}(x)left[color{blue}{1-cos^2(x)}right] \
& = color{red}{cos^n(x)} - (n-1) cos^{n-2}(x) + (ncolor{red}{-1})cos^n(x) \
& = n cos^n(x) - (n-1) cos^{n-2}(x).
end{align*}
Then the whole integral becomes
$$
frac{I_n}{n} = - displaystyle color{red}{left[ -mathrm{e}^{-x} cos^{n-1}(x)sin(x)right]^{infty}_{pi/2}} - int_{pi/2}^{infty} mathrm{e}^{-x} left[ n cos^n(x) - (n-1) cos^{n-2}(x)right] mathrm{d}x
$$
with the red term being zero, we obtain
begin{align*}
frac{I_n}{n} &= -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}, \
frac{I_n}{I_{n-2}} &= frac{n(n-1)}{n^2+1}, \
end{align*}
In particular,
$$
frac{I_{2018}}{I_{2016}} = frac{2018 times 2017}{2018^2+1} = frac{4070306}{4072325}.
$$
answered 15 hours ago
Bennett GardinerBennett Gardiner
3,16211540
$endgroup$
2
$begingroup$
No, the result is $dfrac{4070306}{4072325}$. (Checked with Alpha.)
$endgroup$
– Yves Daoust
13 hours ago
1
$begingroup$
Thanks, I took a typo from OP as my starting point (oops).
$endgroup$
– Bennett Gardiner
13 hours ago
add a comment |
$begingroup$
As you say,
$$
frac{I_n}{n} = - int_{pi/2}^{infty} mathrm{e}^{-x} cos^{n-1}(x)sin(x) mathrm{d}x
$$
I believe the trick is to use integration by parts again, by differentiating the term multiplying the exponential, that is
begin{align*}
frac{mathrm{d} }{mathrm{d} x} , cos^{n-1}(x)sin(x) &= cos^n(x) - (n-1)cos^{n-2}(x)color{blue}{sin^2(x)} \
& = cos^n(x) - (n-1)cos^{n-2}(x)left[color{blue}{1-cos^2(x)}right] \
& = color{red}{cos^n(x)} - (n-1) cos^{n-2}(x) + (ncolor{red}{-1})cos^n(x) \
& = n cos^n(x) - (n-1) cos^{n-2}(x).
end{align*}
Then the whole integral becomes
$$
frac{I_n}{n} = - displaystyle color{red}{left[ -mathrm{e}^{-x} cos^{n-1}(x)sin(x)right]^{infty}_{pi/2}} - int_{pi/2}^{infty} mathrm{e}^{-x} left[ n cos^n(x) - (n-1) cos^{n-2}(x)right] mathrm{d}x
$$
with the red term being zero, we obtain
begin{align*}
frac{I_n}{n} &= -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}, \
frac{I_n}{I_{n-2}} &= frac{n(n-1)}{n^2+1}, \
end{align*}
In particular,
$$
frac{I_{2018}}{I_{2016}} = frac{2018 times 2017}{2018^2+1} = frac{4070306}{4072325}.
$$
answered 15 hours ago
Bennett GardinerBennett Gardiner
3,16211540
$endgroup$
2
$begingroup$
No, the result is $dfrac{4070306}{4072325}$. (Checked with Alpha.)
$endgroup$
– Yves Daoust
13 hours ago
1
$begingroup$
Thanks, I took a typo from OP as my starting point (oops).
$endgroup$
– Bennett Gardiner
13 hours ago
add a comment |
$begingroup$
As you say,
$$
frac{I_n}{n} = - int_{pi/2}^{infty} mathrm{e}^{-x} cos^{n-1}(x)sin(x) mathrm{d}x
$$
I believe the trick is to use integration by parts again, by differentiating the term multiplying the exponential, that is
begin{align*}
frac{mathrm{d} }{mathrm{d} x} , cos^{n-1}(x)sin(x) &= cos^n(x) - (n-1)cos^{n-2}(x)color{blue}{sin^2(x)} \
& = cos^n(x) - (n-1)cos^{n-2}(x)left[color{blue}{1-cos^2(x)}right] \
& = color{red}{cos^n(x)} - (n-1) cos^{n-2}(x) + (ncolor{red}{-1})cos^n(x) \
& = n cos^n(x) - (n-1) cos^{n-2}(x).
end{align*}
Then the whole integral becomes
$$
frac{I_n}{n} = - displaystyle color{red}{left[ -mathrm{e}^{-x} cos^{n-1}(x)sin(x)right]^{infty}_{pi/2}} - int_{pi/2}^{infty} mathrm{e}^{-x} left[ n cos^n(x) - (n-1) cos^{n-2}(x)right] mathrm{d}x
$$
with the red term being zero, we obtain
begin{align*}
frac{I_n}{n} &= -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}, \
frac{I_n}{I_{n-2}} &= frac{n(n-1)}{n^2+1}, \
end{align*}
In particular,
$$
frac{I_{2018}}{I_{2016}} = frac{2018 times 2017}{2018^2+1} = frac{4070306}{4072325}.
$$
answered 15 hours ago
Bennett GardinerBennett Gardiner
3,16211540
$endgroup$
As you say,
$$
frac{I_n}{n} = - int_{pi/2}^{infty} mathrm{e}^{-x} cos^{n-1}(x)sin(x) mathrm{d}x
$$
I believe the trick is to use integration by parts again, by differentiating the term multiplying the exponential, that is
begin{align*}
frac{mathrm{d} }{mathrm{d} x} , cos^{n-1}(x)sin(x) &= cos^n(x) - (n-1)cos^{n-2}(x)color{blue}{sin^2(x)} \
& = cos^n(x) - (n-1)cos^{n-2}(x)left[color{blue}{1-cos^2(x)}right] \
& = color{red}{cos^n(x)} - (n-1) cos^{n-2}(x) + (ncolor{red}{-1})cos^n(x) \
& = n cos^n(x) - (n-1) cos^{n-2}(x).
end{align*}
Then the whole integral becomes
$$
frac{I_n}{n} = - displaystyle color{red}{left[ -mathrm{e}^{-x} cos^{n-1}(x)sin(x)right]^{infty}_{pi/2}} - int_{pi/2}^{infty} mathrm{e}^{-x} left[ n cos^n(x) - (n-1) cos^{n-2}(x)right] mathrm{d}x
$$
with the red term being zero, we obtain
begin{align*}
frac{I_n}{n} &= -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}, \
frac{I_n}{I_{n-2}} &= frac{n(n-1)}{n^2+1}, \
end{align*}
In particular,
$$
frac{I_{2018}}{I_{2016}} = frac{2018 times 2017}{2018^2+1} = frac{4070306}{4072325}.
$$
answered 15 hours ago
Bennett GardinerBennett Gardiner
3,16211540
edited 13 hours ago
answered 15 hours ago
Bennett GardinerBennett Gardiner
3,16211540
answered 15 hours ago
Bennett GardinerBennett Gardiner
3,16211540
answered 15 hours ago
Bennett GardinerBennett Gardiner
3,16211540
3,16211540
2
$begingroup$
No, the result is $dfrac{4070306}{4072325}$. (Checked with Alpha.)
$endgroup$
– Yves Daoust
13 hours ago
1
$begingroup$
Thanks, I took a typo from OP as my starting point (oops).
$endgroup$
– Bennett Gardiner
13 hours ago
add a comment |
2
$begingroup$
No, the result is $dfrac{4070306}{4072325}$. (Checked with Alpha.)
$endgroup$
– Yves Daoust
13 hours ago
1
$begingroup$
Thanks, I took a typo from OP as my starting point (oops).
$endgroup$
– Bennett Gardiner
13 hours ago
2
2
$begingroup$
No, the result is $dfrac{4070306}{4072325}$. (Checked with Alpha.)
$endgroup$
– Yves Daoust
13 hours ago
$begingroup$
No, the result is $dfrac{4070306}{4072325}$. (Checked with Alpha.)
$endgroup$
– Yves Daoust
13 hours ago
1
1
$begingroup$
Thanks, I took a typo from OP as my starting point (oops).
$endgroup$
– Bennett Gardiner
13 hours ago
$begingroup$
Thanks, I took a typo from OP as my starting point (oops).
$endgroup$
– Bennett Gardiner
13 hours ago
add a comment |
$begingroup$
For brevity we set
$$f_{c,s}(x):= e^{-x}cos^c(x)sin^s(x),$$
$$F_{c,s}(x):=int f_{c,s}(x),dx.$$
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 14 hours ago
Yves DaoustYves Daoust
134k676232
$endgroup$
add a comment |
$begingroup$
For brevity we set
$$f_{c,s}(x):= e^{-x}cos^c(x)sin^s(x),$$
$$F_{c,s}(x):=int f_{c,s}(x),dx.$$
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 14 hours ago
Yves DaoustYves Daoust
134k676232
$endgroup$
add a comment |
$begingroup$
For brevity we set
$$f_{c,s}(x):= e^{-x}cos^c(x)sin^s(x),$$
$$F_{c,s}(x):=int f_{c,s}(x),dx.$$
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 14 hours ago
Yves DaoustYves Daoust
134k676232
$endgroup$
For brevity we set
$$f_{c,s}(x):= e^{-x}cos^c(x)sin^s(x),$$
$$F_{c,s}(x):=int f_{c,s}(x),dx.$$
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 14 hours ago
Yves DaoustYves Daoust
134k676232
edited 11 hours ago
answered 14 hours ago
Yves DaoustYves Daoust
134k676232
answered 14 hours ago
Yves DaoustYves Daoust
134k676232
answered 14 hours ago
Yves DaoustYves Daoust
134k676232
134k676232
add a comment |
add a comment |
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$begingroup$
Integrate both sides of $$dfrac{d(e^{-x}cos^mxsin x)}{dx}$$
$endgroup$
– lab bhattacharjee
17 hours ago