Mixing
taking the anti-derivative(integration)
Clash Royale CLAN TAG#URR8PPP
up vote
-4
down vote
favorite
https://ww3.math.msu.edu/images/webwork/132/Examples/4.7.4a-eg.pdf
I'm very confused how this integral trig formula works.
(Just using $ cos x $ in this case)
$$ frac1kcos(kx) $$
How come if the equation is for example $ frac12cos(2x) $ would come be how it is? My book never uses this formula in the book. However I was given this in a PDF randomly on a problem on the online homework.
integration trigonometry
asked Aug 6 at 23:09
Disgruntled Student
104
add a comment |Â
up vote
-4
down vote
favorite
https://ww3.math.msu.edu/images/webwork/132/Examples/4.7.4a-eg.pdf
I'm very confused how this integral trig formula works.
(Just using $ cos x $ in this case)
$$ frac1kcos(kx) $$
How come if the equation is for example $ frac12cos(2x) $ would come be how it is? My book never uses this formula in the book. However I was given this in a PDF randomly on a problem on the online homework.
integration trigonometry
asked Aug 6 at 23:09
Disgruntled Student
104
2
Try taking the derivative of the solution to see why.
– Shaun_the_Post
Aug 6 at 23:12
A line above, the paper states that $int sin(kx) dx= - frac1kcos(kx) + C$. Then $k$ is replaced by $2$.
– Zacky
Aug 6 at 23:13
Welcome to stackexchange. We want to help you, but you have to help us do that. Please ask just one question at a time. Put the actual question in the post - no images. Use mathjax: math.meta.stackexchange.com/questions/5020/… Show us what you have done. or tried and where you are stuck. Hint for this question: to understand whether an antiderivative is correct, just differentiate. Remember the chain rule.
– Ethan Bolker
Aug 7 at 0:22
add a comment |Â
up vote
-4
down vote
favorite
up vote
-4
down vote
favorite
https://ww3.math.msu.edu/images/webwork/132/Examples/4.7.4a-eg.pdf
I'm very confused how this integral trig formula works.
(Just using $ cos x $ in this case)
$$ frac1kcos(kx) $$
How come if the equation is for example $ frac12cos(2x) $ would come be how it is? My book never uses this formula in the book. However I was given this in a PDF randomly on a problem on the online homework.
integration trigonometry
asked Aug 6 at 23:09
Disgruntled Student
104
https://ww3.math.msu.edu/images/webwork/132/Examples/4.7.4a-eg.pdf
I'm very confused how this integral trig formula works.
(Just using $ cos x $ in this case)
$$ frac1kcos(kx) $$
How come if the equation is for example $ frac12cos(2x) $ would come be how it is? My book never uses this formula in the book. However I was given this in a PDF randomly on a problem on the online homework.
integration trigonometry
asked Aug 6 at 23:09
Disgruntled Student
104
edited Aug 11 at 20:06
asked Aug 6 at 23:09
Disgruntled Student
104
asked Aug 6 at 23:09
Disgruntled Student
104
asked Aug 6 at 23:09
Disgruntled Student
104
104
2
Try taking the derivative of the solution to see why.
– Shaun_the_Post
Aug 6 at 23:12
A line above, the paper states that $int sin(kx) dx= - frac1kcos(kx) + C$. Then $k$ is replaced by $2$.
– Zacky
Aug 6 at 23:13
Welcome to stackexchange. We want to help you, but you have to help us do that. Please ask just one question at a time. Put the actual question in the post - no images. Use mathjax: math.meta.stackexchange.com/questions/5020/… Show us what you have done. or tried and where you are stuck. Hint for this question: to understand whether an antiderivative is correct, just differentiate. Remember the chain rule.
– Ethan Bolker
Aug 7 at 0:22
add a comment |Â
2
Try taking the derivative of the solution to see why.
– Shaun_the_Post
Aug 6 at 23:12
A line above, the paper states that $int sin(kx) dx= - frac1kcos(kx) + C$. Then $k$ is replaced by $2$.
– Zacky
Aug 6 at 23:13
Welcome to stackexchange. We want to help you, but you have to help us do that. Please ask just one question at a time. Put the actual question in the post - no images. Use mathjax: math.meta.stackexchange.com/questions/5020/… Show us what you have done. or tried and where you are stuck. Hint for this question: to understand whether an antiderivative is correct, just differentiate. Remember the chain rule.
– Ethan Bolker
Aug 7 at 0:22
2
2
Try taking the derivative of the solution to see why.
– Shaun_the_Post
Aug 6 at 23:12
Try taking the derivative of the solution to see why.
– Shaun_the_Post
Aug 6 at 23:12
A line above, the paper states that $int sin(kx) dx= - frac1kcos(kx) + C$. Then $k$ is replaced by $2$.
– Zacky
Aug 6 at 23:13
A line above, the paper states that $int sin(kx) dx= - frac1kcos(kx) + C$. Then $k$ is replaced by $2$.
– Zacky
Aug 6 at 23:13
Welcome to stackexchange. We want to help you, but you have to help us do that. Please ask just one question at a time. Put the actual question in the post - no images. Use mathjax: math.meta.stackexchange.com/questions/5020/… Show us what you have done. or tried and where you are stuck. Hint for this question: to understand whether an antiderivative is correct, just differentiate. Remember the chain rule.
– Ethan Bolker
Aug 7 at 0:22
Welcome to stackexchange. We want to help you, but you have to help us do that. Please ask just one question at a time. Put the actual question in the post - no images. Use mathjax: math.meta.stackexchange.com/questions/5020/… Show us what you have done. or tried and where you are stuck. Hint for this question: to understand whether an antiderivative is correct, just differentiate. Remember the chain rule.
– Ethan Bolker
Aug 7 at 0:22
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
0
down vote
accepted
By the fundamental theorem of calculus
$frac ddx int f(x) dx = f(x)$
So, we can just take the derivative of $-frac 12 cos 2x + C$ and see what we get back. Which is exactly what is going on the the attached document.
But that is a little bit unsatisfying.
Nonetheless, from first principles you must differentiate $sin x$ and $cos x$ and show that $frac ddx sin x = cos x$ and $frac ddx cos x = -sin x.$ Which I have to assume you have already done in class and in your book and I am not going to repeat the derivations here.
Then when confronted with: $int sin 2x dx $
You are going to need to do a substitution to make it take on the form of a function whose anti-derivative you know.
$int sin 2x dx = int frac 12 sin u du = -frac 12 int sin u du = -frac 12 cos u = -frac 12 cos 2x$
answered Aug 6 at 23:21
Doug M
39.3k31749
I think what perplexes me the most is that my book does not go into detail about this at all. How does this relate to x^(n+1)/n+1
– Disgruntled Student
Aug 6 at 23:34
$int x^n dx = frac x^n+1n+1 + C$ is an expression of the "power rule." en.wikipedia.org/wiki/Power_rule You really should know this.
– Doug M
Aug 7 at 2:01
But why the k in cos(k)x be treated as if k was a power. I thought $cos x^k $ is when a k is a power of x and $cos^k x$ was the same as $(cos x)^k $
– Disgruntled Student
Aug 7 at 4:39
$cos kx, cos^kx,$ and $cos x^k$ should be treated as completely different objects with different rules. $int cos 2x dx = frac 12 sin 2x+C, int cos^2 x dx = frac 12 (x+cos xsin x) + C$ and $int cos x^2 dx$ cannot be integrated into elementary functions. Anyway, what you want to be doing is looking for substitutions such that you can make your integrals look like simpler functions that you know how to integrate.
– Doug M
Aug 7 at 4:57
Actually, you could say $sin 2x = 2sin xcos x, u = sin x, du = cos x dx implies int sin 2x dx = int u du = frac 12 u^2 + C = frac 12 sin^2 x + C = - frac 12 cos 2x + D\$ And by trigonometry show that since the two answers differ only by a constant, both are acceptable answers.
– Doug M
Aug 7 at 5:11
 |Â
show 1 more comment
up vote
-2
down vote
You have to U-sub to get $1/2$, that's the only way you can find the anti derivative of $sin 2x$
edited Aug 7 at 0:16
Suzet
2,216527
answered Aug 6 at 23:13
theoretical
1
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
By the fundamental theorem of calculus
$frac ddx int f(x) dx = f(x)$
So, we can just take the derivative of $-frac 12 cos 2x + C$ and see what we get back. Which is exactly what is going on the the attached document.
But that is a little bit unsatisfying.
Nonetheless, from first principles you must differentiate $sin x$ and $cos x$ and show that $frac ddx sin x = cos x$ and $frac ddx cos x = -sin x.$ Which I have to assume you have already done in class and in your book and I am not going to repeat the derivations here.
Then when confronted with: $int sin 2x dx $
You are going to need to do a substitution to make it take on the form of a function whose anti-derivative you know.
$int sin 2x dx = int frac 12 sin u du = -frac 12 int sin u du = -frac 12 cos u = -frac 12 cos 2x$
answered Aug 6 at 23:21
Doug M
39.3k31749
I think what perplexes me the most is that my book does not go into detail about this at all. How does this relate to x^(n+1)/n+1
– Disgruntled Student
Aug 6 at 23:34
$int x^n dx = frac x^n+1n+1 + C$ is an expression of the "power rule." en.wikipedia.org/wiki/Power_rule You really should know this.
– Doug M
Aug 7 at 2:01
But why the k in cos(k)x be treated as if k was a power. I thought $cos x^k $ is when a k is a power of x and $cos^k x$ was the same as $(cos x)^k $
– Disgruntled Student
Aug 7 at 4:39
$cos kx, cos^kx,$ and $cos x^k$ should be treated as completely different objects with different rules. $int cos 2x dx = frac 12 sin 2x+C, int cos^2 x dx = frac 12 (x+cos xsin x) + C$ and $int cos x^2 dx$ cannot be integrated into elementary functions. Anyway, what you want to be doing is looking for substitutions such that you can make your integrals look like simpler functions that you know how to integrate.
– Doug M
Aug 7 at 4:57
Actually, you could say $sin 2x = 2sin xcos x, u = sin x, du = cos x dx implies int sin 2x dx = int u du = frac 12 u^2 + C = frac 12 sin^2 x + C = - frac 12 cos 2x + D\$ And by trigonometry show that since the two answers differ only by a constant, both are acceptable answers.
– Doug M
Aug 7 at 5:11
 |Â
show 1 more comment
up vote
0
down vote
accepted
By the fundamental theorem of calculus
$frac ddx int f(x) dx = f(x)$
So, we can just take the derivative of $-frac 12 cos 2x + C$ and see what we get back. Which is exactly what is going on the the attached document.
But that is a little bit unsatisfying.
Nonetheless, from first principles you must differentiate $sin x$ and $cos x$ and show that $frac ddx sin x = cos x$ and $frac ddx cos x = -sin x.$ Which I have to assume you have already done in class and in your book and I am not going to repeat the derivations here.
Then when confronted with: $int sin 2x dx $
You are going to need to do a substitution to make it take on the form of a function whose anti-derivative you know.
$int sin 2x dx = int frac 12 sin u du = -frac 12 int sin u du = -frac 12 cos u = -frac 12 cos 2x$
answered Aug 6 at 23:21
Doug M
39.3k31749
I think what perplexes me the most is that my book does not go into detail about this at all. How does this relate to x^(n+1)/n+1
– Disgruntled Student
Aug 6 at 23:34
$int x^n dx = frac x^n+1n+1 + C$ is an expression of the "power rule." en.wikipedia.org/wiki/Power_rule You really should know this.
– Doug M
Aug 7 at 2:01
But why the k in cos(k)x be treated as if k was a power. I thought $cos x^k $ is when a k is a power of x and $cos^k x$ was the same as $(cos x)^k $
– Disgruntled Student
Aug 7 at 4:39
$cos kx, cos^kx,$ and $cos x^k$ should be treated as completely different objects with different rules. $int cos 2x dx = frac 12 sin 2x+C, int cos^2 x dx = frac 12 (x+cos xsin x) + C$ and $int cos x^2 dx$ cannot be integrated into elementary functions. Anyway, what you want to be doing is looking for substitutions such that you can make your integrals look like simpler functions that you know how to integrate.
– Doug M
Aug 7 at 4:57
Actually, you could say $sin 2x = 2sin xcos x, u = sin x, du = cos x dx implies int sin 2x dx = int u du = frac 12 u^2 + C = frac 12 sin^2 x + C = - frac 12 cos 2x + D\$ And by trigonometry show that since the two answers differ only by a constant, both are acceptable answers.
– Doug M
Aug 7 at 5:11
 |Â
show 1 more comment
up vote
0
down vote
accepted
up vote
0
down vote
accepted
By the fundamental theorem of calculus
$frac ddx int f(x) dx = f(x)$
So, we can just take the derivative of $-frac 12 cos 2x + C$ and see what we get back. Which is exactly what is going on the the attached document.
But that is a little bit unsatisfying.
Nonetheless, from first principles you must differentiate $sin x$ and $cos x$ and show that $frac ddx sin x = cos x$ and $frac ddx cos x = -sin x.$ Which I have to assume you have already done in class and in your book and I am not going to repeat the derivations here.
Then when confronted with: $int sin 2x dx $
You are going to need to do a substitution to make it take on the form of a function whose anti-derivative you know.
$int sin 2x dx = int frac 12 sin u du = -frac 12 int sin u du = -frac 12 cos u = -frac 12 cos 2x$
answered Aug 6 at 23:21
Doug M
39.3k31749
By the fundamental theorem of calculus
$frac ddx int f(x) dx = f(x)$
So, we can just take the derivative of $-frac 12 cos 2x + C$ and see what we get back. Which is exactly what is going on the the attached document.
But that is a little bit unsatisfying.
Nonetheless, from first principles you must differentiate $sin x$ and $cos x$ and show that $frac ddx sin x = cos x$ and $frac ddx cos x = -sin x.$ Which I have to assume you have already done in class and in your book and I am not going to repeat the derivations here.
Then when confronted with: $int sin 2x dx $
You are going to need to do a substitution to make it take on the form of a function whose anti-derivative you know.
$int sin 2x dx = int frac 12 sin u du = -frac 12 int sin u du = -frac 12 cos u = -frac 12 cos 2x$
answered Aug 6 at 23:21
Doug M
39.3k31749
answered Aug 6 at 23:21
Doug M
39.3k31749
answered Aug 6 at 23:21
Doug M
39.3k31749
answered Aug 6 at 23:21
Doug M
39.3k31749
39.3k31749
I think what perplexes me the most is that my book does not go into detail about this at all. How does this relate to x^(n+1)/n+1
– Disgruntled Student
Aug 6 at 23:34
$int x^n dx = frac x^n+1n+1 + C$ is an expression of the "power rule." en.wikipedia.org/wiki/Power_rule You really should know this.
– Doug M
Aug 7 at 2:01
But why the k in cos(k)x be treated as if k was a power. I thought $cos x^k $ is when a k is a power of x and $cos^k x$ was the same as $(cos x)^k $
– Disgruntled Student
Aug 7 at 4:39
$cos kx, cos^kx,$ and $cos x^k$ should be treated as completely different objects with different rules. $int cos 2x dx = frac 12 sin 2x+C, int cos^2 x dx = frac 12 (x+cos xsin x) + C$ and $int cos x^2 dx$ cannot be integrated into elementary functions. Anyway, what you want to be doing is looking for substitutions such that you can make your integrals look like simpler functions that you know how to integrate.
– Doug M
Aug 7 at 4:57
Actually, you could say $sin 2x = 2sin xcos x, u = sin x, du = cos x dx implies int sin 2x dx = int u du = frac 12 u^2 + C = frac 12 sin^2 x + C = - frac 12 cos 2x + D\$ And by trigonometry show that since the two answers differ only by a constant, both are acceptable answers.
– Doug M
Aug 7 at 5:11
 |Â
show 1 more comment
I think what perplexes me the most is that my book does not go into detail about this at all. How does this relate to x^(n+1)/n+1
– Disgruntled Student
Aug 6 at 23:34
$int x^n dx = frac x^n+1n+1 + C$ is an expression of the "power rule." en.wikipedia.org/wiki/Power_rule You really should know this.
– Doug M
Aug 7 at 2:01
But why the k in cos(k)x be treated as if k was a power. I thought $cos x^k $ is when a k is a power of x and $cos^k x$ was the same as $(cos x)^k $
– Disgruntled Student
Aug 7 at 4:39
$cos kx, cos^kx,$ and $cos x^k$ should be treated as completely different objects with different rules. $int cos 2x dx = frac 12 sin 2x+C, int cos^2 x dx = frac 12 (x+cos xsin x) + C$ and $int cos x^2 dx$ cannot be integrated into elementary functions. Anyway, what you want to be doing is looking for substitutions such that you can make your integrals look like simpler functions that you know how to integrate.
– Doug M
Aug 7 at 4:57
Actually, you could say $sin 2x = 2sin xcos x, u = sin x, du = cos x dx implies int sin 2x dx = int u du = frac 12 u^2 + C = frac 12 sin^2 x + C = - frac 12 cos 2x + D\$ And by trigonometry show that since the two answers differ only by a constant, both are acceptable answers.
– Doug M
Aug 7 at 5:11
I think what perplexes me the most is that my book does not go into detail about this at all. How does this relate to x^(n+1)/n+1
– Disgruntled Student
Aug 6 at 23:34
I think what perplexes me the most is that my book does not go into detail about this at all. How does this relate to x^(n+1)/n+1
– Disgruntled Student
Aug 6 at 23:34
$int x^n dx = frac x^n+1n+1 + C$ is an expression of the "power rule." en.wikipedia.org/wiki/Power_rule You really should know this.
– Doug M
Aug 7 at 2:01
$int x^n dx = frac x^n+1n+1 + C$ is an expression of the "power rule." en.wikipedia.org/wiki/Power_rule You really should know this.
– Doug M
Aug 7 at 2:01
But why the k in cos(k)x be treated as if k was a power. I thought $cos x^k $ is when a k is a power of x and $cos^k x$ was the same as $(cos x)^k $
– Disgruntled Student
Aug 7 at 4:39
But why the k in cos(k)x be treated as if k was a power. I thought $cos x^k $ is when a k is a power of x and $cos^k x$ was the same as $(cos x)^k $
– Disgruntled Student
Aug 7 at 4:39
$cos kx, cos^kx,$ and $cos x^k$ should be treated as completely different objects with different rules. $int cos 2x dx = frac 12 sin 2x+C, int cos^2 x dx = frac 12 (x+cos xsin x) + C$ and $int cos x^2 dx$ cannot be integrated into elementary functions. Anyway, what you want to be doing is looking for substitutions such that you can make your integrals look like simpler functions that you know how to integrate.
– Doug M
Aug 7 at 4:57
$cos kx, cos^kx,$ and $cos x^k$ should be treated as completely different objects with different rules. $int cos 2x dx = frac 12 sin 2x+C, int cos^2 x dx = frac 12 (x+cos xsin x) + C$ and $int cos x^2 dx$ cannot be integrated into elementary functions. Anyway, what you want to be doing is looking for substitutions such that you can make your integrals look like simpler functions that you know how to integrate.
– Doug M
Aug 7 at 4:57
Actually, you could say $sin 2x = 2sin xcos x, u = sin x, du = cos x dx implies int sin 2x dx = int u du = frac 12 u^2 + C = frac 12 sin^2 x + C = - frac 12 cos 2x + D\$ And by trigonometry show that since the two answers differ only by a constant, both are acceptable answers.
– Doug M
Aug 7 at 5:11
Actually, you could say $sin 2x = 2sin xcos x, u = sin x, du = cos x dx implies int sin 2x dx = int u du = frac 12 u^2 + C = frac 12 sin^2 x + C = - frac 12 cos 2x + D\$ And by trigonometry show that since the two answers differ only by a constant, both are acceptable answers.
– Doug M
Aug 7 at 5:11
 |Â
show 1 more comment
up vote
-2
down vote
You have to U-sub to get $1/2$, that's the only way you can find the anti derivative of $sin 2x$
edited Aug 7 at 0:16
Suzet
2,216527
answered Aug 6 at 23:13
theoretical
1
add a comment |Â
up vote
-2
down vote
You have to U-sub to get $1/2$, that's the only way you can find the anti derivative of $sin 2x$
edited Aug 7 at 0:16
Suzet
2,216527
answered Aug 6 at 23:13
theoretical
1
add a comment |Â
up vote
-2
down vote
up vote
-2
down vote
You have to U-sub to get $1/2$, that's the only way you can find the anti derivative of $sin 2x$
edited Aug 7 at 0:16
Suzet
2,216527
answered Aug 6 at 23:13
theoretical
1
You have to U-sub to get $1/2$, that's the only way you can find the anti derivative of $sin 2x$
edited Aug 7 at 0:16
Suzet
2,216527
answered Aug 6 at 23:13
theoretical
1
edited Aug 7 at 0:16
Suzet
2,216527
edited Aug 7 at 0:16
Suzet
2,216527
edited Aug 7 at 0:16
Suzet
2,216527
2,216527
answered Aug 6 at 23:13
theoretical
1
answered Aug 6 at 23:13
theoretical
1
answered Aug 6 at 23:13
theoretical
1
1
add a comment |Â
add a comment |Â
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2
Try taking the derivative of the solution to see why.
– Shaun_the_Post
Aug 6 at 23:12
A line above, the paper states that $int sin(kx) dx= - frac1kcos(kx) + C$. Then $k$ is replaced by $2$.
– Zacky
Aug 6 at 23:13
Welcome to stackexchange. We want to help you, but you have to help us do that. Please ask just one question at a time. Put the actual question in the post - no images. Use mathjax: math.meta.stackexchange.com/questions/5020/… Show us what you have done. or tried and where you are stuck. Hint for this question: to understand whether an antiderivative is correct, just differentiate. Remember the chain rule.
– Ethan Bolker
Aug 7 at 0:22