Mixing
How to integrate $tan(x)^n$?
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First, did use wolfram alpha and it does give me the answer in terms of a hypergeometric sequence.
I am a grade 12 student, and I am trying to investigate into this integral for my assessment.
Could someone please explain how I can go from the LHS to the RHS:
$$int tan^n(x), dx = frac {tan^{n + 1}(x); _2F_1(1, frac {n + 1}2, frac {n + 3}2, -tan^2(x))}{n + 1}, + ,text {constant}$$
Where, as usual, $_2F_1$ denotes the HyperGeometric Function
calculus
asked Jan 10 at 14:29
Redlion11Redlion11
133
$endgroup$
|
show 2 more comments
$begingroup$
First, did use wolfram alpha and it does give me the answer in terms of a hypergeometric sequence.
I am a grade 12 student, and I am trying to investigate into this integral for my assessment.
Could someone please explain how I can go from the LHS to the RHS:
$$int tan^n(x), dx = frac {tan^{n + 1}(x); _2F_1(1, frac {n + 1}2, frac {n + 3}2, -tan^2(x))}{n + 1}, + ,text {constant}$$
Where, as usual, $_2F_1$ denotes the HyperGeometric Function
calculus
asked Jan 10 at 14:29
Redlion11Redlion11
133
$endgroup$
1
$begingroup$
The image is unreadable
$endgroup$
– Don Thousand
Jan 10 at 14:34
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I'm not sure in the image if they are meant to be $()$ brackets or if there is a reason for the different shape
$endgroup$
– Henry Lee
Jan 10 at 14:36
2
$begingroup$
Note: I added what I believe to be the correct integral to your post, the link you included could not be read. Please advise if what I wrote was not what you intended. I left your link in place though, if what I wrote is what you wanted, then you might want to delete the broken link.
$endgroup$
– lulu
Jan 10 at 14:40
$begingroup$
This seems like quite a high level integral compared to what you will have learnt. If you can avoid it that would probably be better
$endgroup$
– Henry Lee
Jan 10 at 14:41
$begingroup$
Worth remarking; if $nin mathbb N$ then, generally, one just uses a simple reduction to integrate this.
$endgroup$
– lulu
Jan 10 at 14:44
|
show 2 more comments
$begingroup$
First, did use wolfram alpha and it does give me the answer in terms of a hypergeometric sequence.
I am a grade 12 student, and I am trying to investigate into this integral for my assessment.
Could someone please explain how I can go from the LHS to the RHS:
$$int tan^n(x), dx = frac {tan^{n + 1}(x); _2F_1(1, frac {n + 1}2, frac {n + 3}2, -tan^2(x))}{n + 1}, + ,text {constant}$$
Where, as usual, $_2F_1$ denotes the HyperGeometric Function
calculus
asked Jan 10 at 14:29
Redlion11Redlion11
133
$endgroup$
First, did use wolfram alpha and it does give me the answer in terms of a hypergeometric sequence.
I am a grade 12 student, and I am trying to investigate into this integral for my assessment.
Could someone please explain how I can go from the LHS to the RHS:
$$int tan^n(x), dx = frac {tan^{n + 1}(x); _2F_1(1, frac {n + 1}2, frac {n + 3}2, -tan^2(x))}{n + 1}, + ,text {constant}$$
Where, as usual, $_2F_1$ denotes the HyperGeometric Function
calculus
calculus
asked Jan 10 at 14:29
Redlion11Redlion11
133
asked Jan 10 at 14:29
Redlion11Redlion11
133
edited Jan 10 at 15:19
Redlion11
asked Jan 10 at 14:29
Redlion11Redlion11
133
asked Jan 10 at 14:29
Redlion11Redlion11
133
asked Jan 10 at 14:29
Redlion11Redlion11
133
133
1
$begingroup$
The image is unreadable
$endgroup$
– Don Thousand
Jan 10 at 14:34
$begingroup$
I'm not sure in the image if they are meant to be $()$ brackets or if there is a reason for the different shape
$endgroup$
– Henry Lee
Jan 10 at 14:36
2
$begingroup$
Note: I added what I believe to be the correct integral to your post, the link you included could not be read. Please advise if what I wrote was not what you intended. I left your link in place though, if what I wrote is what you wanted, then you might want to delete the broken link.
$endgroup$
– lulu
Jan 10 at 14:40
$begingroup$
This seems like quite a high level integral compared to what you will have learnt. If you can avoid it that would probably be better
$endgroup$
– Henry Lee
Jan 10 at 14:41
$begingroup$
Worth remarking; if $nin mathbb N$ then, generally, one just uses a simple reduction to integrate this.
$endgroup$
– lulu
Jan 10 at 14:44
|
show 2 more comments
1
$begingroup$
The image is unreadable
$endgroup$
– Don Thousand
Jan 10 at 14:34
$begingroup$
I'm not sure in the image if they are meant to be $()$ brackets or if there is a reason for the different shape
$endgroup$
– Henry Lee
Jan 10 at 14:36
2
$begingroup$
Note: I added what I believe to be the correct integral to your post, the link you included could not be read. Please advise if what I wrote was not what you intended. I left your link in place though, if what I wrote is what you wanted, then you might want to delete the broken link.
$endgroup$
– lulu
Jan 10 at 14:40
$begingroup$
This seems like quite a high level integral compared to what you will have learnt. If you can avoid it that would probably be better
$endgroup$
– Henry Lee
Jan 10 at 14:41
$begingroup$
Worth remarking; if $nin mathbb N$ then, generally, one just uses a simple reduction to integrate this.
$endgroup$
– lulu
Jan 10 at 14:44
1
1
$begingroup$
The image is unreadable
$endgroup$
– Don Thousand
Jan 10 at 14:34
$begingroup$
The image is unreadable
$endgroup$
– Don Thousand
Jan 10 at 14:34
$begingroup$
I'm not sure in the image if they are meant to be $()$ brackets or if there is a reason for the different shape
$endgroup$
– Henry Lee
Jan 10 at 14:36
$begingroup$
I'm not sure in the image if they are meant to be $()$ brackets or if there is a reason for the different shape
$endgroup$
– Henry Lee
Jan 10 at 14:36
2
2
$begingroup$
Note: I added what I believe to be the correct integral to your post, the link you included could not be read. Please advise if what I wrote was not what you intended. I left your link in place though, if what I wrote is what you wanted, then you might want to delete the broken link.
$endgroup$
– lulu
Jan 10 at 14:40
$begingroup$
Note: I added what I believe to be the correct integral to your post, the link you included could not be read. Please advise if what I wrote was not what you intended. I left your link in place though, if what I wrote is what you wanted, then you might want to delete the broken link.
$endgroup$
– lulu
Jan 10 at 14:40
$begingroup$
This seems like quite a high level integral compared to what you will have learnt. If you can avoid it that would probably be better
$endgroup$
– Henry Lee
Jan 10 at 14:41
$begingroup$
This seems like quite a high level integral compared to what you will have learnt. If you can avoid it that would probably be better
$endgroup$
– Henry Lee
Jan 10 at 14:41
$begingroup$
Worth remarking; if $nin mathbb N$ then, generally, one just uses a simple reduction to integrate this.
$endgroup$
– lulu
Jan 10 at 14:44
$begingroup$
Worth remarking; if $nin mathbb N$ then, generally, one just uses a simple reduction to integrate this.
$endgroup$
– lulu
Jan 10 at 14:44
|
show 2 more comments
3 Answers
3
active
oldest
votes
$begingroup$
Let's consider the case when $n=2k+1$ is odd: we can write
$$
inttan^nx,dx=inttan^{2k+1}x,dx=
intfrac{(sin^2x)^k}{cos^{2k+1}x}sin x,dx=
-intfrac{(1-cos^2x)^k}{cos^{2k+1}x},d(cos x)
$$
which is elementary. Example with $k=2$: we reduce to
begin{align}
-intfrac{(1-cos^2x)^2}{cos^5x},d(cos x)
&=-intleft(frac{1}{cos^5x}-frac{2}{cos^3x}+frac{1}{cos x}right),d(cos x)\
&=frac{1}{4cos^4x}-frac{1}{cos^2x}-loglvertcos xrvert+c
end{align}
For $n=2k$ even, it's a bit different:
$$
inttan^nx,dx=intfrac{sin^{2k}x}{cos^{2k}x},dx=
intfrac{(1-cos^2x)^k}{cos^{2k}x},dx
$$
and the problem is reduced to computing
$$
intfrac{1}{cos^{2m}x},dx
$$
which can be dealt with the substitution $t=tan x$. Example with $m=2$; since
$$
cos^2x=frac{1}{1+tan^2x}
$$
we have
$$
intfrac{1}{cos^4x}=Bigl[t=tan x,dt=frac{1}{cos^2x},dxBigr]=
int(1+t^2),dt=t+frac{t^3}{3}+c=tan x+frac{1}{3}tan^3x,dx
$$
In general, for $mge1$, with the same substitution,
$$
intfrac{1}{cos^{2m}x}=int(1+t^2)^{m-1},dt
$$
answered Jan 10 at 14:45
egregegreg
181k1485203
$endgroup$
$begingroup$
Could you show how to integrate 1/(cos(x))^(2m)? And is it possible to further breakdown the first case where you split tan(x)^n in terms of odd powers of sin and cos? sorry if it an obvious question, I just find it slightly hard to understand this.
$endgroup$
– Redlion11
Jan 10 at 15:55
$begingroup$
@Redlion11 I added a couple of examples
$endgroup$
– egreg
Jan 10 at 16:08
$begingroup$
Thank you, I get it now.
$endgroup$
– Redlion11
Jan 10 at 16:13
add a comment |
$begingroup$
Notice that $$(tan x)'=tan^2x+1$$ so that
$$tan^n x=tan^{n-2} x,(tan x)'-tan^{n-2}x.$$
This gives you a recurrence relation that ends with the integrand $tan x$ or $1$ depending on the parity of $n$.
$$I_n=frac{tan^{n-1}x}{n-1}-I_{n-2}.$$
This leads to better expressions than those given by Alpha, even with a numerical exponent.
E.g.
https://www.wolframalpha.com/input/?i=integrate+tan%5E8x
versus
$$frac{tan^7 x}7-frac{tan^5 x}5+frac{tan^3 x}3-tan x+x.$$
answered Jan 10 at 14:55
Yves DaoustYves Daoust
127k673226
$endgroup$
$begingroup$
Thanks for the response, but could you elaborate a bit more on how you arrived at tan^n x=tan^{n-2} x,(tan x)'-tan^{n-2}x
$endgroup$
– Redlion11
Jan 10 at 16:18
$begingroup$
@Redlion11: no, this is immediate. Expand.
$endgroup$
– Yves Daoust
Jan 10 at 17:23
add a comment |
$begingroup$
Depending on the limits, the Beta function can be used:
$$B(m+1,n+1)=2int_0^{pi/2}cos^{2m+1}(x)sin^{2n+1}(x)dx=frac{m!n!}{(m+n+1)!}$$
For outside of this range you can use that:
$$B(z;a,b)=int_0^zx^{a-1}(1-x)^{b-1}dx=z^asum_{n=0}^inftyfrac{(1-b)_n}{n!(a+n)}z^n$$
All this can be found here:
http://mathworld.wolfram.com/BetaFunction.html
http://mathworld.wolfram.com/IncompleteBetaFunction.html
answered Jan 10 at 14:35
Henry LeeHenry Lee
1,921219
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let's consider the case when $n=2k+1$ is odd: we can write
$$
inttan^nx,dx=inttan^{2k+1}x,dx=
intfrac{(sin^2x)^k}{cos^{2k+1}x}sin x,dx=
-intfrac{(1-cos^2x)^k}{cos^{2k+1}x},d(cos x)
$$
which is elementary. Example with $k=2$: we reduce to
begin{align}
-intfrac{(1-cos^2x)^2}{cos^5x},d(cos x)
&=-intleft(frac{1}{cos^5x}-frac{2}{cos^3x}+frac{1}{cos x}right),d(cos x)\
&=frac{1}{4cos^4x}-frac{1}{cos^2x}-loglvertcos xrvert+c
end{align}
For $n=2k$ even, it's a bit different:
$$
inttan^nx,dx=intfrac{sin^{2k}x}{cos^{2k}x},dx=
intfrac{(1-cos^2x)^k}{cos^{2k}x},dx
$$
and the problem is reduced to computing
$$
intfrac{1}{cos^{2m}x},dx
$$
which can be dealt with the substitution $t=tan x$. Example with $m=2$; since
$$
cos^2x=frac{1}{1+tan^2x}
$$
we have
$$
intfrac{1}{cos^4x}=Bigl[t=tan x,dt=frac{1}{cos^2x},dxBigr]=
int(1+t^2),dt=t+frac{t^3}{3}+c=tan x+frac{1}{3}tan^3x,dx
$$
In general, for $mge1$, with the same substitution,
$$
intfrac{1}{cos^{2m}x}=int(1+t^2)^{m-1},dt
$$
answered Jan 10 at 14:45
egregegreg
181k1485203
$endgroup$
$begingroup$
Could you show how to integrate 1/(cos(x))^(2m)? And is it possible to further breakdown the first case where you split tan(x)^n in terms of odd powers of sin and cos? sorry if it an obvious question, I just find it slightly hard to understand this.
$endgroup$
– Redlion11
Jan 10 at 15:55
$begingroup$
@Redlion11 I added a couple of examples
$endgroup$
– egreg
Jan 10 at 16:08
$begingroup$
Thank you, I get it now.
$endgroup$
– Redlion11
Jan 10 at 16:13
add a comment |
$begingroup$
Let's consider the case when $n=2k+1$ is odd: we can write
$$
inttan^nx,dx=inttan^{2k+1}x,dx=
intfrac{(sin^2x)^k}{cos^{2k+1}x}sin x,dx=
-intfrac{(1-cos^2x)^k}{cos^{2k+1}x},d(cos x)
$$
which is elementary. Example with $k=2$: we reduce to
begin{align}
-intfrac{(1-cos^2x)^2}{cos^5x},d(cos x)
&=-intleft(frac{1}{cos^5x}-frac{2}{cos^3x}+frac{1}{cos x}right),d(cos x)\
&=frac{1}{4cos^4x}-frac{1}{cos^2x}-loglvertcos xrvert+c
end{align}
For $n=2k$ even, it's a bit different:
$$
inttan^nx,dx=intfrac{sin^{2k}x}{cos^{2k}x},dx=
intfrac{(1-cos^2x)^k}{cos^{2k}x},dx
$$
and the problem is reduced to computing
$$
intfrac{1}{cos^{2m}x},dx
$$
which can be dealt with the substitution $t=tan x$. Example with $m=2$; since
$$
cos^2x=frac{1}{1+tan^2x}
$$
we have
$$
intfrac{1}{cos^4x}=Bigl[t=tan x,dt=frac{1}{cos^2x},dxBigr]=
int(1+t^2),dt=t+frac{t^3}{3}+c=tan x+frac{1}{3}tan^3x,dx
$$
In general, for $mge1$, with the same substitution,
$$
intfrac{1}{cos^{2m}x}=int(1+t^2)^{m-1},dt
$$
answered Jan 10 at 14:45
egregegreg
181k1485203
$endgroup$
$begingroup$
Could you show how to integrate 1/(cos(x))^(2m)? And is it possible to further breakdown the first case where you split tan(x)^n in terms of odd powers of sin and cos? sorry if it an obvious question, I just find it slightly hard to understand this.
$endgroup$
– Redlion11
Jan 10 at 15:55
$begingroup$
@Redlion11 I added a couple of examples
$endgroup$
– egreg
Jan 10 at 16:08
$begingroup$
Thank you, I get it now.
$endgroup$
– Redlion11
Jan 10 at 16:13
add a comment |
$begingroup$
Let's consider the case when $n=2k+1$ is odd: we can write
$$
inttan^nx,dx=inttan^{2k+1}x,dx=
intfrac{(sin^2x)^k}{cos^{2k+1}x}sin x,dx=
-intfrac{(1-cos^2x)^k}{cos^{2k+1}x},d(cos x)
$$
which is elementary. Example with $k=2$: we reduce to
begin{align}
-intfrac{(1-cos^2x)^2}{cos^5x},d(cos x)
&=-intleft(frac{1}{cos^5x}-frac{2}{cos^3x}+frac{1}{cos x}right),d(cos x)\
&=frac{1}{4cos^4x}-frac{1}{cos^2x}-loglvertcos xrvert+c
end{align}
For $n=2k$ even, it's a bit different:
$$
inttan^nx,dx=intfrac{sin^{2k}x}{cos^{2k}x},dx=
intfrac{(1-cos^2x)^k}{cos^{2k}x},dx
$$
and the problem is reduced to computing
$$
intfrac{1}{cos^{2m}x},dx
$$
which can be dealt with the substitution $t=tan x$. Example with $m=2$; since
$$
cos^2x=frac{1}{1+tan^2x}
$$
we have
$$
intfrac{1}{cos^4x}=Bigl[t=tan x,dt=frac{1}{cos^2x},dxBigr]=
int(1+t^2),dt=t+frac{t^3}{3}+c=tan x+frac{1}{3}tan^3x,dx
$$
In general, for $mge1$, with the same substitution,
$$
intfrac{1}{cos^{2m}x}=int(1+t^2)^{m-1},dt
$$
answered Jan 10 at 14:45
egregegreg
181k1485203
$endgroup$
Let's consider the case when $n=2k+1$ is odd: we can write
$$
inttan^nx,dx=inttan^{2k+1}x,dx=
intfrac{(sin^2x)^k}{cos^{2k+1}x}sin x,dx=
-intfrac{(1-cos^2x)^k}{cos^{2k+1}x},d(cos x)
$$
which is elementary. Example with $k=2$: we reduce to
begin{align}
-intfrac{(1-cos^2x)^2}{cos^5x},d(cos x)
&=-intleft(frac{1}{cos^5x}-frac{2}{cos^3x}+frac{1}{cos x}right),d(cos x)\
&=frac{1}{4cos^4x}-frac{1}{cos^2x}-loglvertcos xrvert+c
end{align}
For $n=2k$ even, it's a bit different:
$$
inttan^nx,dx=intfrac{sin^{2k}x}{cos^{2k}x},dx=
intfrac{(1-cos^2x)^k}{cos^{2k}x},dx
$$
and the problem is reduced to computing
$$
intfrac{1}{cos^{2m}x},dx
$$
which can be dealt with the substitution $t=tan x$. Example with $m=2$; since
$$
cos^2x=frac{1}{1+tan^2x}
$$
we have
$$
intfrac{1}{cos^4x}=Bigl[t=tan x,dt=frac{1}{cos^2x},dxBigr]=
int(1+t^2),dt=t+frac{t^3}{3}+c=tan x+frac{1}{3}tan^3x,dx
$$
In general, for $mge1$, with the same substitution,
$$
intfrac{1}{cos^{2m}x}=int(1+t^2)^{m-1},dt
$$
answered Jan 10 at 14:45
egregegreg
181k1485203
edited Jan 10 at 16:08
answered Jan 10 at 14:45
egregegreg
181k1485203
answered Jan 10 at 14:45
egregegreg
181k1485203
answered Jan 10 at 14:45
egregegreg
181k1485203
181k1485203
$begingroup$
Could you show how to integrate 1/(cos(x))^(2m)? And is it possible to further breakdown the first case where you split tan(x)^n in terms of odd powers of sin and cos? sorry if it an obvious question, I just find it slightly hard to understand this.
$endgroup$
– Redlion11
Jan 10 at 15:55
$begingroup$
@Redlion11 I added a couple of examples
$endgroup$
– egreg
Jan 10 at 16:08
$begingroup$
Thank you, I get it now.
$endgroup$
– Redlion11
Jan 10 at 16:13
add a comment |
$begingroup$
Could you show how to integrate 1/(cos(x))^(2m)? And is it possible to further breakdown the first case where you split tan(x)^n in terms of odd powers of sin and cos? sorry if it an obvious question, I just find it slightly hard to understand this.
$endgroup$
– Redlion11
Jan 10 at 15:55
$begingroup$
@Redlion11 I added a couple of examples
$endgroup$
– egreg
Jan 10 at 16:08
$begingroup$
Thank you, I get it now.
$endgroup$
– Redlion11
Jan 10 at 16:13
$begingroup$
Could you show how to integrate 1/(cos(x))^(2m)? And is it possible to further breakdown the first case where you split tan(x)^n in terms of odd powers of sin and cos? sorry if it an obvious question, I just find it slightly hard to understand this.
$endgroup$
– Redlion11
Jan 10 at 15:55
$begingroup$
Could you show how to integrate 1/(cos(x))^(2m)? And is it possible to further breakdown the first case where you split tan(x)^n in terms of odd powers of sin and cos? sorry if it an obvious question, I just find it slightly hard to understand this.
$endgroup$
– Redlion11
Jan 10 at 15:55
$begingroup$
@Redlion11 I added a couple of examples
$endgroup$
– egreg
Jan 10 at 16:08
$begingroup$
@Redlion11 I added a couple of examples
$endgroup$
– egreg
Jan 10 at 16:08
$begingroup$
Thank you, I get it now.
$endgroup$
– Redlion11
Jan 10 at 16:13
$begingroup$
Thank you, I get it now.
$endgroup$
– Redlion11
Jan 10 at 16:13
add a comment |
$begingroup$
Notice that $$(tan x)'=tan^2x+1$$ so that
$$tan^n x=tan^{n-2} x,(tan x)'-tan^{n-2}x.$$
This gives you a recurrence relation that ends with the integrand $tan x$ or $1$ depending on the parity of $n$.
$$I_n=frac{tan^{n-1}x}{n-1}-I_{n-2}.$$
This leads to better expressions than those given by Alpha, even with a numerical exponent.
E.g.
https://www.wolframalpha.com/input/?i=integrate+tan%5E8x
versus
$$frac{tan^7 x}7-frac{tan^5 x}5+frac{tan^3 x}3-tan x+x.$$
answered Jan 10 at 14:55
Yves DaoustYves Daoust
127k673226
$endgroup$
$begingroup$
Thanks for the response, but could you elaborate a bit more on how you arrived at tan^n x=tan^{n-2} x,(tan x)'-tan^{n-2}x
$endgroup$
– Redlion11
Jan 10 at 16:18
$begingroup$
@Redlion11: no, this is immediate. Expand.
$endgroup$
– Yves Daoust
Jan 10 at 17:23
add a comment |
$begingroup$
Notice that $$(tan x)'=tan^2x+1$$ so that
$$tan^n x=tan^{n-2} x,(tan x)'-tan^{n-2}x.$$
This gives you a recurrence relation that ends with the integrand $tan x$ or $1$ depending on the parity of $n$.
$$I_n=frac{tan^{n-1}x}{n-1}-I_{n-2}.$$
This leads to better expressions than those given by Alpha, even with a numerical exponent.
E.g.
https://www.wolframalpha.com/input/?i=integrate+tan%5E8x
versus
$$frac{tan^7 x}7-frac{tan^5 x}5+frac{tan^3 x}3-tan x+x.$$
answered Jan 10 at 14:55
Yves DaoustYves Daoust
127k673226
$endgroup$
$begingroup$
Thanks for the response, but could you elaborate a bit more on how you arrived at tan^n x=tan^{n-2} x,(tan x)'-tan^{n-2}x
$endgroup$
– Redlion11
Jan 10 at 16:18
$begingroup$
@Redlion11: no, this is immediate. Expand.
$endgroup$
– Yves Daoust
Jan 10 at 17:23
add a comment |
$begingroup$
Notice that $$(tan x)'=tan^2x+1$$ so that
$$tan^n x=tan^{n-2} x,(tan x)'-tan^{n-2}x.$$
This gives you a recurrence relation that ends with the integrand $tan x$ or $1$ depending on the parity of $n$.
$$I_n=frac{tan^{n-1}x}{n-1}-I_{n-2}.$$
This leads to better expressions than those given by Alpha, even with a numerical exponent.
E.g.
https://www.wolframalpha.com/input/?i=integrate+tan%5E8x
versus
$$frac{tan^7 x}7-frac{tan^5 x}5+frac{tan^3 x}3-tan x+x.$$
answered Jan 10 at 14:55
Yves DaoustYves Daoust
127k673226
$endgroup$
Notice that $$(tan x)'=tan^2x+1$$ so that
$$tan^n x=tan^{n-2} x,(tan x)'-tan^{n-2}x.$$
This gives you a recurrence relation that ends with the integrand $tan x$ or $1$ depending on the parity of $n$.
$$I_n=frac{tan^{n-1}x}{n-1}-I_{n-2}.$$
This leads to better expressions than those given by Alpha, even with a numerical exponent.
E.g.
https://www.wolframalpha.com/input/?i=integrate+tan%5E8x
versus
$$frac{tan^7 x}7-frac{tan^5 x}5+frac{tan^3 x}3-tan x+x.$$
answered Jan 10 at 14:55
Yves DaoustYves Daoust
127k673226
edited Jan 10 at 15:10
answered Jan 10 at 14:55
Yves DaoustYves Daoust
127k673226
answered Jan 10 at 14:55
Yves DaoustYves Daoust
127k673226
answered Jan 10 at 14:55
Yves DaoustYves Daoust
127k673226
127k673226
$begingroup$
Thanks for the response, but could you elaborate a bit more on how you arrived at tan^n x=tan^{n-2} x,(tan x)'-tan^{n-2}x
$endgroup$
– Redlion11
Jan 10 at 16:18
$begingroup$
@Redlion11: no, this is immediate. Expand.
$endgroup$
– Yves Daoust
Jan 10 at 17:23
add a comment |
$begingroup$
Thanks for the response, but could you elaborate a bit more on how you arrived at tan^n x=tan^{n-2} x,(tan x)'-tan^{n-2}x
$endgroup$
– Redlion11
Jan 10 at 16:18
$begingroup$
@Redlion11: no, this is immediate. Expand.
$endgroup$
– Yves Daoust
Jan 10 at 17:23
$begingroup$
Thanks for the response, but could you elaborate a bit more on how you arrived at tan^n x=tan^{n-2} x,(tan x)'-tan^{n-2}x
$endgroup$
– Redlion11
Jan 10 at 16:18
$begingroup$
Thanks for the response, but could you elaborate a bit more on how you arrived at tan^n x=tan^{n-2} x,(tan x)'-tan^{n-2}x
$endgroup$
– Redlion11
Jan 10 at 16:18
$begingroup$
@Redlion11: no, this is immediate. Expand.
$endgroup$
– Yves Daoust
Jan 10 at 17:23
$begingroup$
@Redlion11: no, this is immediate. Expand.
$endgroup$
– Yves Daoust
Jan 10 at 17:23
add a comment |
$begingroup$
Depending on the limits, the Beta function can be used:
$$B(m+1,n+1)=2int_0^{pi/2}cos^{2m+1}(x)sin^{2n+1}(x)dx=frac{m!n!}{(m+n+1)!}$$
For outside of this range you can use that:
$$B(z;a,b)=int_0^zx^{a-1}(1-x)^{b-1}dx=z^asum_{n=0}^inftyfrac{(1-b)_n}{n!(a+n)}z^n$$
All this can be found here:
http://mathworld.wolfram.com/BetaFunction.html
http://mathworld.wolfram.com/IncompleteBetaFunction.html
answered Jan 10 at 14:35
Henry LeeHenry Lee
1,921219
$endgroup$
add a comment |
$begingroup$
Depending on the limits, the Beta function can be used:
$$B(m+1,n+1)=2int_0^{pi/2}cos^{2m+1}(x)sin^{2n+1}(x)dx=frac{m!n!}{(m+n+1)!}$$
For outside of this range you can use that:
$$B(z;a,b)=int_0^zx^{a-1}(1-x)^{b-1}dx=z^asum_{n=0}^inftyfrac{(1-b)_n}{n!(a+n)}z^n$$
All this can be found here:
http://mathworld.wolfram.com/BetaFunction.html
http://mathworld.wolfram.com/IncompleteBetaFunction.html
answered Jan 10 at 14:35
Henry LeeHenry Lee
1,921219
$endgroup$
add a comment |
$begingroup$
Depending on the limits, the Beta function can be used:
$$B(m+1,n+1)=2int_0^{pi/2}cos^{2m+1}(x)sin^{2n+1}(x)dx=frac{m!n!}{(m+n+1)!}$$
For outside of this range you can use that:
$$B(z;a,b)=int_0^zx^{a-1}(1-x)^{b-1}dx=z^asum_{n=0}^inftyfrac{(1-b)_n}{n!(a+n)}z^n$$
All this can be found here:
http://mathworld.wolfram.com/BetaFunction.html
http://mathworld.wolfram.com/IncompleteBetaFunction.html
answered Jan 10 at 14:35
Henry LeeHenry Lee
1,921219
$endgroup$
Depending on the limits, the Beta function can be used:
$$B(m+1,n+1)=2int_0^{pi/2}cos^{2m+1}(x)sin^{2n+1}(x)dx=frac{m!n!}{(m+n+1)!}$$
For outside of this range you can use that:
$$B(z;a,b)=int_0^zx^{a-1}(1-x)^{b-1}dx=z^asum_{n=0}^inftyfrac{(1-b)_n}{n!(a+n)}z^n$$
All this can be found here:
http://mathworld.wolfram.com/BetaFunction.html
http://mathworld.wolfram.com/IncompleteBetaFunction.html
answered Jan 10 at 14:35
Henry LeeHenry Lee
1,921219
answered Jan 10 at 14:35
Henry LeeHenry Lee
1,921219
answered Jan 10 at 14:35
Henry LeeHenry Lee
1,921219
answered Jan 10 at 14:35
Henry LeeHenry Lee
1,921219
1,921219
add a comment |
add a comment |
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1
$begingroup$
The image is unreadable
$endgroup$
– Don Thousand
Jan 10 at 14:34
$begingroup$
I'm not sure in the image if they are meant to be $()$ brackets or if there is a reason for the different shape
$endgroup$
– Henry Lee
Jan 10 at 14:36
2
$begingroup$
Note: I added what I believe to be the correct integral to your post, the link you included could not be read. Please advise if what I wrote was not what you intended. I left your link in place though, if what I wrote is what you wanted, then you might want to delete the broken link.
$endgroup$
– lulu
Jan 10 at 14:40
$begingroup$
This seems like quite a high level integral compared to what you will have learnt. If you can avoid it that would probably be better
$endgroup$
– Henry Lee
Jan 10 at 14:41
$begingroup$
Worth remarking; if $nin mathbb N$ then, generally, one just uses a simple reduction to integrate this.
$endgroup$
– lulu
Jan 10 at 14:44