Integral of $intfrac{sin x}{sin x+cos x}dx$
$begingroup$
The questions defines $$I=intfrac{sin x}{sin x +cos x}dx;;J=intfrac{cos x}{sin x +cos x}dx$$
It asked me to find $I+J$ and $J-I$ which I have done and I will show below but now I need to find the integral shown below and I'm unsure on what to do.
$$intfrac{sin x}{sin x+cos x}dx$$
I have found that:
$$I+J = x+c$$
$$J-I=ln{|cos x +sin x|} +c$$
But now i'm unsure on how to find just $I$
integration trigonometric-integrals
asked Dec 12 '18 at 17:35
H.LinkhornH.Linkhorn
456113
$endgroup$
add a comment |
$begingroup$
The questions defines $$I=intfrac{sin x}{sin x +cos x}dx;;J=intfrac{cos x}{sin x +cos x}dx$$
It asked me to find $I+J$ and $J-I$ which I have done and I will show below but now I need to find the integral shown below and I'm unsure on what to do.
$$intfrac{sin x}{sin x+cos x}dx$$
I have found that:
$$I+J = x+c$$
$$J-I=ln{|cos x +sin x|} +c$$
But now i'm unsure on how to find just $I$
integration trigonometric-integrals
asked Dec 12 '18 at 17:35
H.LinkhornH.Linkhorn
456113
$endgroup$
9
$begingroup$
If you know what $J-I$ is, you know what $I-J$ is. Try adding $I-J$ and $I+J$ together.
$endgroup$
– welshman500
Dec 12 '18 at 17:36
add a comment |
$begingroup$
The questions defines $$I=intfrac{sin x}{sin x +cos x}dx;;J=intfrac{cos x}{sin x +cos x}dx$$
It asked me to find $I+J$ and $J-I$ which I have done and I will show below but now I need to find the integral shown below and I'm unsure on what to do.
$$intfrac{sin x}{sin x+cos x}dx$$
I have found that:
$$I+J = x+c$$
$$J-I=ln{|cos x +sin x|} +c$$
But now i'm unsure on how to find just $I$
integration trigonometric-integrals
asked Dec 12 '18 at 17:35
H.LinkhornH.Linkhorn
456113
$endgroup$
The questions defines $$I=intfrac{sin x}{sin x +cos x}dx;;J=intfrac{cos x}{sin x +cos x}dx$$
It asked me to find $I+J$ and $J-I$ which I have done and I will show below but now I need to find the integral shown below and I'm unsure on what to do.
$$intfrac{sin x}{sin x+cos x}dx$$
I have found that:
$$I+J = x+c$$
$$J-I=ln{|cos x +sin x|} +c$$
But now i'm unsure on how to find just $I$
integration trigonometric-integrals
integration trigonometric-integrals
asked Dec 12 '18 at 17:35
H.LinkhornH.Linkhorn
456113
asked Dec 12 '18 at 17:35
H.LinkhornH.Linkhorn
456113
asked Dec 12 '18 at 17:35
H.LinkhornH.Linkhorn
456113
asked Dec 12 '18 at 17:35
H.LinkhornH.Linkhorn
456113
asked Dec 12 '18 at 17:35
H.LinkhornH.Linkhorn
456113
456113
9
$begingroup$
If you know what $J-I$ is, you know what $I-J$ is. Try adding $I-J$ and $I+J$ together.
$endgroup$
– welshman500
Dec 12 '18 at 17:36
add a comment |
9
$begingroup$
If you know what $J-I$ is, you know what $I-J$ is. Try adding $I-J$ and $I+J$ together.
$endgroup$
– welshman500
Dec 12 '18 at 17:36
9
9
$begingroup$
If you know what $J-I$ is, you know what $I-J$ is. Try adding $I-J$ and $I+J$ together.
$endgroup$
– welshman500
Dec 12 '18 at 17:36
$begingroup$
If you know what $J-I$ is, you know what $I-J$ is. Try adding $I-J$ and $I+J$ together.
$endgroup$
– welshman500
Dec 12 '18 at 17:36
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
what you have is:
$$I+J=x+C$$
$$I-J=-ln|cos(x)+sin(x)|+C$$
so:
$$2I=x-ln|cos(x)+sin(x)|+C$$
answered Dec 12 '18 at 19:44
Henry LeeHenry Lee
2,054219
$endgroup$
add a comment |
$begingroup$
Hint :
Consider the following system of equations :
$$begin{cases} J+I = x+c \ J - I = ln|cos x + sin x | + cend{cases}$$
See an easy way out for $I$ ?
answered Dec 12 '18 at 17:45
RebellosRebellos
14.8k31248
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
active
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votes
$begingroup$
what you have is:
$$I+J=x+C$$
$$I-J=-ln|cos(x)+sin(x)|+C$$
so:
$$2I=x-ln|cos(x)+sin(x)|+C$$
answered Dec 12 '18 at 19:44
Henry LeeHenry Lee
2,054219
$endgroup$
add a comment |
$begingroup$
what you have is:
$$I+J=x+C$$
$$I-J=-ln|cos(x)+sin(x)|+C$$
so:
$$2I=x-ln|cos(x)+sin(x)|+C$$
answered Dec 12 '18 at 19:44
Henry LeeHenry Lee
2,054219
$endgroup$
add a comment |
$begingroup$
what you have is:
$$I+J=x+C$$
$$I-J=-ln|cos(x)+sin(x)|+C$$
so:
$$2I=x-ln|cos(x)+sin(x)|+C$$
answered Dec 12 '18 at 19:44
Henry LeeHenry Lee
2,054219
$endgroup$
what you have is:
$$I+J=x+C$$
$$I-J=-ln|cos(x)+sin(x)|+C$$
so:
$$2I=x-ln|cos(x)+sin(x)|+C$$
answered Dec 12 '18 at 19:44
Henry LeeHenry Lee
2,054219
answered Dec 12 '18 at 19:44
Henry LeeHenry Lee
2,054219
answered Dec 12 '18 at 19:44
Henry LeeHenry Lee
2,054219
answered Dec 12 '18 at 19:44
Henry LeeHenry Lee
2,054219
2,054219
add a comment |
add a comment |
$begingroup$
Hint :
Consider the following system of equations :
$$begin{cases} J+I = x+c \ J - I = ln|cos x + sin x | + cend{cases}$$
See an easy way out for $I$ ?
answered Dec 12 '18 at 17:45
RebellosRebellos
14.8k31248
$endgroup$
add a comment |
$begingroup$
Hint :
Consider the following system of equations :
$$begin{cases} J+I = x+c \ J - I = ln|cos x + sin x | + cend{cases}$$
See an easy way out for $I$ ?
answered Dec 12 '18 at 17:45
RebellosRebellos
14.8k31248
$endgroup$
add a comment |
$begingroup$
Hint :
Consider the following system of equations :
$$begin{cases} J+I = x+c \ J - I = ln|cos x + sin x | + cend{cases}$$
See an easy way out for $I$ ?
answered Dec 12 '18 at 17:45
RebellosRebellos
14.8k31248
$endgroup$
Hint :
Consider the following system of equations :
$$begin{cases} J+I = x+c \ J - I = ln|cos x + sin x | + cend{cases}$$
See an easy way out for $I$ ?
answered Dec 12 '18 at 17:45
RebellosRebellos
14.8k31248
answered Dec 12 '18 at 17:45
RebellosRebellos
14.8k31248
answered Dec 12 '18 at 17:45
RebellosRebellos
14.8k31248
answered Dec 12 '18 at 17:45
RebellosRebellos
14.8k31248
14.8k31248
add a comment |
add a comment |
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9
$begingroup$
If you know what $J-I$ is, you know what $I-J$ is. Try adding $I-J$ and $I+J$ together.
$endgroup$
– welshman500
Dec 12 '18 at 17:36