Integral $int frac{sin^n(x)}{cos(x)}dx$
In one of my exercises about integration we had to solve the following integral:
begin{equation}
int frac{sin^n(x)}{cos^m(x)}dx
end{equation}
We had to do this via a recursive integral. I found:
begin{equation}
mathcal{K_{m,n}} = frac{sin^{n-1}(x)}{(m-1)cdotcos^{m-1}(x)}-frac{n-1}{m-1}cdotmathcal{K}_{m-2,n-2}, qquad n,mgeq2
end{equation}
I know for a fact that this solution is correct because we solved this in class, but the other cases where m and/or n are not $geq$ 2, were left as an exercise for us at home. I've found a solution for every case except for the case where $m = 1$, which makes the following integral.
begin{equation}
int frac{sin^n(x)}{cos(x)}dx
end{equation}
I've tried different things, I tried integration by parts with many different u's and v's but none of them seem to work out. I tried for example:
$u = frac{sin^{n-1}(x)}{cos(x)}$, $v = sin(x)$
$u = tan(x)$, $v = sin^{n-1}(x)$
$u = sin^{n-1}(x)$, $v = tan(x)$
Would anyone know how to solve this integral? I'm not looking for a complete solution but rather for a method that should work so i can find the solution by myself from that point on.
integration indefinite-integrals recursion reduction-formula
edited Dec 27 '18 at 23:48
clathratus
3,243331
asked Dec 27 '18 at 14:30
Viktor
416
add a comment |
In one of my exercises about integration we had to solve the following integral:
begin{equation}
int frac{sin^n(x)}{cos^m(x)}dx
end{equation}
We had to do this via a recursive integral. I found:
begin{equation}
mathcal{K_{m,n}} = frac{sin^{n-1}(x)}{(m-1)cdotcos^{m-1}(x)}-frac{n-1}{m-1}cdotmathcal{K}_{m-2,n-2}, qquad n,mgeq2
end{equation}
I know for a fact that this solution is correct because we solved this in class, but the other cases where m and/or n are not $geq$ 2, were left as an exercise for us at home. I've found a solution for every case except for the case where $m = 1$, which makes the following integral.
begin{equation}
int frac{sin^n(x)}{cos(x)}dx
end{equation}
I've tried different things, I tried integration by parts with many different u's and v's but none of them seem to work out. I tried for example:
$u = frac{sin^{n-1}(x)}{cos(x)}$, $v = sin(x)$
$u = tan(x)$, $v = sin^{n-1}(x)$
$u = sin^{n-1}(x)$, $v = tan(x)$
Would anyone know how to solve this integral? I'm not looking for a complete solution but rather for a method that should work so i can find the solution by myself from that point on.
integration indefinite-integrals recursion reduction-formula
edited Dec 27 '18 at 23:48
clathratus
3,243331
asked Dec 27 '18 at 14:30
Viktor
416
add a comment |
In one of my exercises about integration we had to solve the following integral:
begin{equation}
int frac{sin^n(x)}{cos^m(x)}dx
end{equation}
We had to do this via a recursive integral. I found:
begin{equation}
mathcal{K_{m,n}} = frac{sin^{n-1}(x)}{(m-1)cdotcos^{m-1}(x)}-frac{n-1}{m-1}cdotmathcal{K}_{m-2,n-2}, qquad n,mgeq2
end{equation}
I know for a fact that this solution is correct because we solved this in class, but the other cases where m and/or n are not $geq$ 2, were left as an exercise for us at home. I've found a solution for every case except for the case where $m = 1$, which makes the following integral.
begin{equation}
int frac{sin^n(x)}{cos(x)}dx
end{equation}
I've tried different things, I tried integration by parts with many different u's and v's but none of them seem to work out. I tried for example:
$u = frac{sin^{n-1}(x)}{cos(x)}$, $v = sin(x)$
$u = tan(x)$, $v = sin^{n-1}(x)$
$u = sin^{n-1}(x)$, $v = tan(x)$
Would anyone know how to solve this integral? I'm not looking for a complete solution but rather for a method that should work so i can find the solution by myself from that point on.
integration indefinite-integrals recursion reduction-formula
edited Dec 27 '18 at 23:48
clathratus
3,243331
asked Dec 27 '18 at 14:30
Viktor
416
In one of my exercises about integration we had to solve the following integral:
begin{equation}
int frac{sin^n(x)}{cos^m(x)}dx
end{equation}
We had to do this via a recursive integral. I found:
begin{equation}
mathcal{K_{m,n}} = frac{sin^{n-1}(x)}{(m-1)cdotcos^{m-1}(x)}-frac{n-1}{m-1}cdotmathcal{K}_{m-2,n-2}, qquad n,mgeq2
end{equation}
I know for a fact that this solution is correct because we solved this in class, but the other cases where m and/or n are not $geq$ 2, were left as an exercise for us at home. I've found a solution for every case except for the case where $m = 1$, which makes the following integral.
begin{equation}
int frac{sin^n(x)}{cos(x)}dx
end{equation}
I've tried different things, I tried integration by parts with many different u's and v's but none of them seem to work out. I tried for example:
$u = frac{sin^{n-1}(x)}{cos(x)}$, $v = sin(x)$
$u = tan(x)$, $v = sin^{n-1}(x)$
$u = sin^{n-1}(x)$, $v = tan(x)$
Would anyone know how to solve this integral? I'm not looking for a complete solution but rather for a method that should work so i can find the solution by myself from that point on.
integration indefinite-integrals recursion reduction-formula
integration indefinite-integrals recursion reduction-formula
edited Dec 27 '18 at 23:48
clathratus
3,243331
asked Dec 27 '18 at 14:30
Viktor
416
edited Dec 27 '18 at 23:48
clathratus
3,243331
asked Dec 27 '18 at 14:30
Viktor
416
edited Dec 27 '18 at 23:48
clathratus
3,243331
edited Dec 27 '18 at 23:48
clathratus
3,243331
edited Dec 27 '18 at 23:48
clathratus
3,243331
3,243331
asked Dec 27 '18 at 14:30
Viktor
416
asked Dec 27 '18 at 14:30
Viktor
416
asked Dec 27 '18 at 14:30
Viktor
416
416
add a comment |
add a comment |
1 Answer
1
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Denote $I_n=displaystyleint frac{sin^n(x)}{cos(x)}~mathrm dx$. We have
begin{align*}
I_{n}-I_{n+2}&=int frac{sin^n(x)(1-sin^2(x))}{cos(x)}~mathrm dx\
&=int sin^n(x)cos(x)~mathrm dx,
end{align*}
which is the reduction formula for $I_n$.
answered Dec 27 '18 at 14:41
Tianlalu
3,09621038
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Denote $I_n=displaystyleint frac{sin^n(x)}{cos(x)}~mathrm dx$. We have
begin{align*}
I_{n}-I_{n+2}&=int frac{sin^n(x)(1-sin^2(x))}{cos(x)}~mathrm dx\
&=int sin^n(x)cos(x)~mathrm dx,
end{align*}
which is the reduction formula for $I_n$.
answered Dec 27 '18 at 14:41
Tianlalu
3,09621038
add a comment |
Denote $I_n=displaystyleint frac{sin^n(x)}{cos(x)}~mathrm dx$. We have
begin{align*}
I_{n}-I_{n+2}&=int frac{sin^n(x)(1-sin^2(x))}{cos(x)}~mathrm dx\
&=int sin^n(x)cos(x)~mathrm dx,
end{align*}
which is the reduction formula for $I_n$.
answered Dec 27 '18 at 14:41
Tianlalu
3,09621038
add a comment |
Denote $I_n=displaystyleint frac{sin^n(x)}{cos(x)}~mathrm dx$. We have
begin{align*}
I_{n}-I_{n+2}&=int frac{sin^n(x)(1-sin^2(x))}{cos(x)}~mathrm dx\
&=int sin^n(x)cos(x)~mathrm dx,
end{align*}
which is the reduction formula for $I_n$.
answered Dec 27 '18 at 14:41
Tianlalu
3,09621038
Denote $I_n=displaystyleint frac{sin^n(x)}{cos(x)}~mathrm dx$. We have
begin{align*}
I_{n}-I_{n+2}&=int frac{sin^n(x)(1-sin^2(x))}{cos(x)}~mathrm dx\
&=int sin^n(x)cos(x)~mathrm dx,
end{align*}
which is the reduction formula for $I_n$.
answered Dec 27 '18 at 14:41
Tianlalu
3,09621038
answered Dec 27 '18 at 14:41
Tianlalu
3,09621038
answered Dec 27 '18 at 14:41
Tianlalu
3,09621038
answered Dec 27 '18 at 14:41
Tianlalu
3,09621038
3,09621038
add a comment |
add a comment |
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