What's the the integral of $tan(4x)$?
How is $displaystyle int tan(4x) dx = cos^2 (4x)$? Shouldn't it be $ln(sec(4x))$? I don't understand... Please, help.
Thank you.
calculus integration indefinite-integrals
edited Nov 24 at 3:27
Rócherz
2,7612721
asked Oct 16 '13 at 11:35
Joshua Ree
1813617
add a comment |
How is $displaystyle int tan(4x) dx = cos^2 (4x)$? Shouldn't it be $ln(sec(4x))$? I don't understand... Please, help.
Thank you.
calculus integration indefinite-integrals
edited Nov 24 at 3:27
Rócherz
2,7612721
asked Oct 16 '13 at 11:35
Joshua Ree
1813617
1
It isn't. The $ln sec (4x)$ is also not right, but that misses only a constant factor ($frac14$) and the arbitrary integration constant.
– Daniel Fischer♦
Oct 16 '13 at 11:37
add a comment |
How is $displaystyle int tan(4x) dx = cos^2 (4x)$? Shouldn't it be $ln(sec(4x))$? I don't understand... Please, help.
Thank you.
calculus integration indefinite-integrals
edited Nov 24 at 3:27
Rócherz
2,7612721
asked Oct 16 '13 at 11:35
Joshua Ree
1813617
How is $displaystyle int tan(4x) dx = cos^2 (4x)$? Shouldn't it be $ln(sec(4x))$? I don't understand... Please, help.
Thank you.
calculus integration indefinite-integrals
calculus integration indefinite-integrals
edited Nov 24 at 3:27
Rócherz
2,7612721
asked Oct 16 '13 at 11:35
Joshua Ree
1813617
edited Nov 24 at 3:27
Rócherz
2,7612721
asked Oct 16 '13 at 11:35
Joshua Ree
1813617
edited Nov 24 at 3:27
Rócherz
2,7612721
edited Nov 24 at 3:27
Rócherz
2,7612721
edited Nov 24 at 3:27
Rócherz
2,7612721
2,7612721
asked Oct 16 '13 at 11:35
Joshua Ree
1813617
asked Oct 16 '13 at 11:35
Joshua Ree
1813617
asked Oct 16 '13 at 11:35
Joshua Ree
1813617
1813617
1
It isn't. The $ln sec (4x)$ is also not right, but that misses only a constant factor ($frac14$) and the arbitrary integration constant.
– Daniel Fischer♦
Oct 16 '13 at 11:37
add a comment |
1
It isn't. The $ln sec (4x)$ is also not right, but that misses only a constant factor ($frac14$) and the arbitrary integration constant.
– Daniel Fischer♦
Oct 16 '13 at 11:37
1
1
It isn't. The $ln sec (4x)$ is also not right, but that misses only a constant factor ($frac14$) and the arbitrary integration constant.
– Daniel Fischer♦
Oct 16 '13 at 11:37
It isn't. The $ln sec (4x)$ is also not right, but that misses only a constant factor ($frac14$) and the arbitrary integration constant.
– Daniel Fischer♦
Oct 16 '13 at 11:37
add a comment |
3 Answers
3
active
oldest
votes
$$inttan x,dx=-logcos x +C=logsec x+Cimplies$$
$$inttan 4x,dx=frac14int (4dx)tan 4x=frac14logsec 4x+C$$
answered Oct 16 '13 at 11:38
DonAntonio
176k1491225
add a comment |
This is a correct method for the function as stated above,
begin{eqnarray}
int tan(4x)dx &=& frac{1}{4}ln (sec(4x)) + C.
end{eqnarray}
answered Oct 16 '13 at 11:40
Autolatry
2,69111015
add a comment |
$$
int{tan(4x)dx}=intfrac{sin(4x)dx}{cos(4x)}=|u=cos(4x)Rightarrow -4sin(4x)dx=dtRightarrow sin(4x)dx=-frac{1}{4}dt|
$$
$$
=-frac{1}{4}intfrac{dt}{t}=-frac{1}{4}ln|t|=-frac{1}{4}ln|cos(4x)|=frac{1}{4}(-1)ln|cos(4x)|=frac{1}{4}ln|cos(4x)|^{-1}
$$
$$
=frac{1}{4}ln|frac{1}{cos(4x)}|=frac{1}{4}ln|sec(4x)|+C
$$
answered Oct 16 '13 at 12:15
Madrit Zhaku
4,79621019
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
$$inttan x,dx=-logcos x +C=logsec x+Cimplies$$
$$inttan 4x,dx=frac14int (4dx)tan 4x=frac14logsec 4x+C$$
answered Oct 16 '13 at 11:38
DonAntonio
176k1491225
add a comment |
$$inttan x,dx=-logcos x +C=logsec x+Cimplies$$
$$inttan 4x,dx=frac14int (4dx)tan 4x=frac14logsec 4x+C$$
answered Oct 16 '13 at 11:38
DonAntonio
176k1491225
add a comment |
$$inttan x,dx=-logcos x +C=logsec x+Cimplies$$
$$inttan 4x,dx=frac14int (4dx)tan 4x=frac14logsec 4x+C$$
answered Oct 16 '13 at 11:38
DonAntonio
176k1491225
$$inttan x,dx=-logcos x +C=logsec x+Cimplies$$
$$inttan 4x,dx=frac14int (4dx)tan 4x=frac14logsec 4x+C$$
answered Oct 16 '13 at 11:38
DonAntonio
176k1491225
answered Oct 16 '13 at 11:38
DonAntonio
176k1491225
answered Oct 16 '13 at 11:38
DonAntonio
176k1491225
answered Oct 16 '13 at 11:38
DonAntonio
176k1491225
176k1491225
add a comment |
add a comment |
This is a correct method for the function as stated above,
begin{eqnarray}
int tan(4x)dx &=& frac{1}{4}ln (sec(4x)) + C.
end{eqnarray}
answered Oct 16 '13 at 11:40
Autolatry
2,69111015
add a comment |
This is a correct method for the function as stated above,
begin{eqnarray}
int tan(4x)dx &=& frac{1}{4}ln (sec(4x)) + C.
end{eqnarray}
answered Oct 16 '13 at 11:40
Autolatry
2,69111015
add a comment |
This is a correct method for the function as stated above,
begin{eqnarray}
int tan(4x)dx &=& frac{1}{4}ln (sec(4x)) + C.
end{eqnarray}
answered Oct 16 '13 at 11:40
Autolatry
2,69111015
This is a correct method for the function as stated above,
begin{eqnarray}
int tan(4x)dx &=& frac{1}{4}ln (sec(4x)) + C.
end{eqnarray}
answered Oct 16 '13 at 11:40
Autolatry
2,69111015
answered Oct 16 '13 at 11:40
Autolatry
2,69111015
answered Oct 16 '13 at 11:40
Autolatry
2,69111015
answered Oct 16 '13 at 11:40
Autolatry
2,69111015
2,69111015
add a comment |
add a comment |
$$
int{tan(4x)dx}=intfrac{sin(4x)dx}{cos(4x)}=|u=cos(4x)Rightarrow -4sin(4x)dx=dtRightarrow sin(4x)dx=-frac{1}{4}dt|
$$
$$
=-frac{1}{4}intfrac{dt}{t}=-frac{1}{4}ln|t|=-frac{1}{4}ln|cos(4x)|=frac{1}{4}(-1)ln|cos(4x)|=frac{1}{4}ln|cos(4x)|^{-1}
$$
$$
=frac{1}{4}ln|frac{1}{cos(4x)}|=frac{1}{4}ln|sec(4x)|+C
$$
answered Oct 16 '13 at 12:15
Madrit Zhaku
4,79621019
add a comment |
$$
int{tan(4x)dx}=intfrac{sin(4x)dx}{cos(4x)}=|u=cos(4x)Rightarrow -4sin(4x)dx=dtRightarrow sin(4x)dx=-frac{1}{4}dt|
$$
$$
=-frac{1}{4}intfrac{dt}{t}=-frac{1}{4}ln|t|=-frac{1}{4}ln|cos(4x)|=frac{1}{4}(-1)ln|cos(4x)|=frac{1}{4}ln|cos(4x)|^{-1}
$$
$$
=frac{1}{4}ln|frac{1}{cos(4x)}|=frac{1}{4}ln|sec(4x)|+C
$$
answered Oct 16 '13 at 12:15
Madrit Zhaku
4,79621019
add a comment |
$$
int{tan(4x)dx}=intfrac{sin(4x)dx}{cos(4x)}=|u=cos(4x)Rightarrow -4sin(4x)dx=dtRightarrow sin(4x)dx=-frac{1}{4}dt|
$$
$$
=-frac{1}{4}intfrac{dt}{t}=-frac{1}{4}ln|t|=-frac{1}{4}ln|cos(4x)|=frac{1}{4}(-1)ln|cos(4x)|=frac{1}{4}ln|cos(4x)|^{-1}
$$
$$
=frac{1}{4}ln|frac{1}{cos(4x)}|=frac{1}{4}ln|sec(4x)|+C
$$
answered Oct 16 '13 at 12:15
Madrit Zhaku
4,79621019
$$
int{tan(4x)dx}=intfrac{sin(4x)dx}{cos(4x)}=|u=cos(4x)Rightarrow -4sin(4x)dx=dtRightarrow sin(4x)dx=-frac{1}{4}dt|
$$
$$
=-frac{1}{4}intfrac{dt}{t}=-frac{1}{4}ln|t|=-frac{1}{4}ln|cos(4x)|=frac{1}{4}(-1)ln|cos(4x)|=frac{1}{4}ln|cos(4x)|^{-1}
$$
$$
=frac{1}{4}ln|frac{1}{cos(4x)}|=frac{1}{4}ln|sec(4x)|+C
$$
answered Oct 16 '13 at 12:15
Madrit Zhaku
4,79621019
answered Oct 16 '13 at 12:15
Madrit Zhaku
4,79621019
answered Oct 16 '13 at 12:15
Madrit Zhaku
4,79621019
answered Oct 16 '13 at 12:15
Madrit Zhaku
4,79621019
4,79621019
add a comment |
add a comment |
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1
It isn't. The $ln sec (4x)$ is also not right, but that misses only a constant factor ($frac14$) and the arbitrary integration constant.
– Daniel Fischer♦
Oct 16 '13 at 11:37