Integration by Substitution of $cos(log(x))$
$begingroup$
I need to find the indefinite integral:
$int cos(log(x)) dx$.
I've tried using u-substitution, setting $u=log(x)$. Then I get:
$int cos(u)du$, where $du$ is $frac{1}{x}dx$.
The worked solutions I can see put an $e^u$ in there as well, and I'm not really sure why. How should I go about solving this problem?
Thanks for your time.
integration substitution
edited May 13 '18 at 7:56
farruhota
20.4k2739
asked May 13 '18 at 7:27
Three OneFourThree OneFour
1099
$endgroup$
add a comment |
$begingroup$
I need to find the indefinite integral:
$int cos(log(x)) dx$.
I've tried using u-substitution, setting $u=log(x)$. Then I get:
$int cos(u)du$, where $du$ is $frac{1}{x}dx$.
The worked solutions I can see put an $e^u$ in there as well, and I'm not really sure why. How should I go about solving this problem?
Thanks for your time.
integration substitution
edited May 13 '18 at 7:56
farruhota
20.4k2739
asked May 13 '18 at 7:27
Three OneFourThree OneFour
1099
$endgroup$
add a comment |
$begingroup$
I need to find the indefinite integral:
$int cos(log(x)) dx$.
I've tried using u-substitution, setting $u=log(x)$. Then I get:
$int cos(u)du$, where $du$ is $frac{1}{x}dx$.
The worked solutions I can see put an $e^u$ in there as well, and I'm not really sure why. How should I go about solving this problem?
Thanks for your time.
integration substitution
edited May 13 '18 at 7:56
farruhota
20.4k2739
asked May 13 '18 at 7:27
Three OneFourThree OneFour
1099
$endgroup$
I need to find the indefinite integral:
$int cos(log(x)) dx$.
I've tried using u-substitution, setting $u=log(x)$. Then I get:
$int cos(u)du$, where $du$ is $frac{1}{x}dx$.
The worked solutions I can see put an $e^u$ in there as well, and I'm not really sure why. How should I go about solving this problem?
Thanks for your time.
integration substitution
integration substitution
edited May 13 '18 at 7:56
farruhota
20.4k2739
asked May 13 '18 at 7:27
Three OneFourThree OneFour
1099
edited May 13 '18 at 7:56
farruhota
20.4k2739
asked May 13 '18 at 7:27
Three OneFourThree OneFour
1099
edited May 13 '18 at 7:56
farruhota
20.4k2739
edited May 13 '18 at 7:56
farruhota
20.4k2739
edited May 13 '18 at 7:56
farruhota
20.4k2739
20.4k2739
asked May 13 '18 at 7:27
Three OneFourThree OneFour
1099
asked May 13 '18 at 7:27
Three OneFourThree OneFour
1099
asked May 13 '18 at 7:27
Three OneFourThree OneFour
1099
1099
add a comment |
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
Let $ln x=uimplies x=e^u,dx=e^u du$
$$intcos(ln x)dx=int e^ucos u du $$
Now use Integration of $e^{ax}cos bx$ and $e^{ax}sin bx$
See also : Set $n=-1$ in $100$-th derivative of the function $f(x)=e^{x}cos(x)$
answered May 13 '18 at 7:32
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
$begingroup$
How do you know that $dx=e^u du$?
$endgroup$
– Three OneFour
May 13 '18 at 7:34
$begingroup$
@ThreeOneFour, $$du=dfrac{dx}x=dfrac{dx}{e^u}$$ right?
$endgroup$
– lab bhattacharjee
May 13 '18 at 7:35
add a comment |
$begingroup$
Better way:
Recall that $$cos x=frac{e^{ix}+e^{-ix}}2$$
So $$cos log x=frac{e^{ilog x}+e^{-ilog x}}2=frac{x^i+x^{-i}}2$$
And hence
$$intcoslog x,mathrm dx=frac12left[frac1{1+i}x^{1+i}+frac{1}{1-i}x^{1-i}right]+C$$
Then using $$sin x=frac{e^{ix}-e^{-ix}}{2i}$$
We have that $$intcoslog x,mathrm dx=frac{x}2[coslog x+sinlog x]+C$$
answered Jan 21 at 1:42
clathratusclathratus
4,615337
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
Let $ln x=uimplies x=e^u,dx=e^u du$
$$intcos(ln x)dx=int e^ucos u du $$
Now use Integration of $e^{ax}cos bx$ and $e^{ax}sin bx$
See also : Set $n=-1$ in $100$-th derivative of the function $f(x)=e^{x}cos(x)$
answered May 13 '18 at 7:32
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
$begingroup$
How do you know that $dx=e^u du$?
$endgroup$
– Three OneFour
May 13 '18 at 7:34
$begingroup$
@ThreeOneFour, $$du=dfrac{dx}x=dfrac{dx}{e^u}$$ right?
$endgroup$
– lab bhattacharjee
May 13 '18 at 7:35
add a comment |
$begingroup$
Let $ln x=uimplies x=e^u,dx=e^u du$
$$intcos(ln x)dx=int e^ucos u du $$
Now use Integration of $e^{ax}cos bx$ and $e^{ax}sin bx$
See also : Set $n=-1$ in $100$-th derivative of the function $f(x)=e^{x}cos(x)$
answered May 13 '18 at 7:32
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
$begingroup$
How do you know that $dx=e^u du$?
$endgroup$
– Three OneFour
May 13 '18 at 7:34
$begingroup$
@ThreeOneFour, $$du=dfrac{dx}x=dfrac{dx}{e^u}$$ right?
$endgroup$
– lab bhattacharjee
May 13 '18 at 7:35
add a comment |
$begingroup$
Let $ln x=uimplies x=e^u,dx=e^u du$
$$intcos(ln x)dx=int e^ucos u du $$
Now use Integration of $e^{ax}cos bx$ and $e^{ax}sin bx$
See also : Set $n=-1$ in $100$-th derivative of the function $f(x)=e^{x}cos(x)$
answered May 13 '18 at 7:32
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
Let $ln x=uimplies x=e^u,dx=e^u du$
$$intcos(ln x)dx=int e^ucos u du $$
Now use Integration of $e^{ax}cos bx$ and $e^{ax}sin bx$
See also : Set $n=-1$ in $100$-th derivative of the function $f(x)=e^{x}cos(x)$
answered May 13 '18 at 7:32
lab bhattacharjeelab bhattacharjee
226k15157275
answered May 13 '18 at 7:32
lab bhattacharjeelab bhattacharjee
226k15157275
answered May 13 '18 at 7:32
lab bhattacharjeelab bhattacharjee
226k15157275
answered May 13 '18 at 7:32
lab bhattacharjeelab bhattacharjee
226k15157275
226k15157275
$begingroup$
How do you know that $dx=e^u du$?
$endgroup$
– Three OneFour
May 13 '18 at 7:34
$begingroup$
@ThreeOneFour, $$du=dfrac{dx}x=dfrac{dx}{e^u}$$ right?
$endgroup$
– lab bhattacharjee
May 13 '18 at 7:35
add a comment |
$begingroup$
How do you know that $dx=e^u du$?
$endgroup$
– Three OneFour
May 13 '18 at 7:34
$begingroup$
@ThreeOneFour, $$du=dfrac{dx}x=dfrac{dx}{e^u}$$ right?
$endgroup$
– lab bhattacharjee
May 13 '18 at 7:35
$begingroup$
How do you know that $dx=e^u du$?
$endgroup$
– Three OneFour
May 13 '18 at 7:34
$begingroup$
How do you know that $dx=e^u du$?
$endgroup$
– Three OneFour
May 13 '18 at 7:34
$begingroup$
@ThreeOneFour, $$du=dfrac{dx}x=dfrac{dx}{e^u}$$ right?
$endgroup$
– lab bhattacharjee
May 13 '18 at 7:35
$begingroup$
@ThreeOneFour, $$du=dfrac{dx}x=dfrac{dx}{e^u}$$ right?
$endgroup$
– lab bhattacharjee
May 13 '18 at 7:35
add a comment |
$begingroup$
Better way:
Recall that $$cos x=frac{e^{ix}+e^{-ix}}2$$
So $$cos log x=frac{e^{ilog x}+e^{-ilog x}}2=frac{x^i+x^{-i}}2$$
And hence
$$intcoslog x,mathrm dx=frac12left[frac1{1+i}x^{1+i}+frac{1}{1-i}x^{1-i}right]+C$$
Then using $$sin x=frac{e^{ix}-e^{-ix}}{2i}$$
We have that $$intcoslog x,mathrm dx=frac{x}2[coslog x+sinlog x]+C$$
answered Jan 21 at 1:42
clathratusclathratus
4,615337
$endgroup$
add a comment |
$begingroup$
Better way:
Recall that $$cos x=frac{e^{ix}+e^{-ix}}2$$
So $$cos log x=frac{e^{ilog x}+e^{-ilog x}}2=frac{x^i+x^{-i}}2$$
And hence
$$intcoslog x,mathrm dx=frac12left[frac1{1+i}x^{1+i}+frac{1}{1-i}x^{1-i}right]+C$$
Then using $$sin x=frac{e^{ix}-e^{-ix}}{2i}$$
We have that $$intcoslog x,mathrm dx=frac{x}2[coslog x+sinlog x]+C$$
answered Jan 21 at 1:42
clathratusclathratus
4,615337
$endgroup$
add a comment |
$begingroup$
Better way:
Recall that $$cos x=frac{e^{ix}+e^{-ix}}2$$
So $$cos log x=frac{e^{ilog x}+e^{-ilog x}}2=frac{x^i+x^{-i}}2$$
And hence
$$intcoslog x,mathrm dx=frac12left[frac1{1+i}x^{1+i}+frac{1}{1-i}x^{1-i}right]+C$$
Then using $$sin x=frac{e^{ix}-e^{-ix}}{2i}$$
We have that $$intcoslog x,mathrm dx=frac{x}2[coslog x+sinlog x]+C$$
answered Jan 21 at 1:42
clathratusclathratus
4,615337
$endgroup$
Better way:
Recall that $$cos x=frac{e^{ix}+e^{-ix}}2$$
So $$cos log x=frac{e^{ilog x}+e^{-ilog x}}2=frac{x^i+x^{-i}}2$$
And hence
$$intcoslog x,mathrm dx=frac12left[frac1{1+i}x^{1+i}+frac{1}{1-i}x^{1-i}right]+C$$
Then using $$sin x=frac{e^{ix}-e^{-ix}}{2i}$$
We have that $$intcoslog x,mathrm dx=frac{x}2[coslog x+sinlog x]+C$$
answered Jan 21 at 1:42
clathratusclathratus
4,615337
answered Jan 21 at 1:42
clathratusclathratus
4,615337
answered Jan 21 at 1:42
clathratusclathratus
4,615337
answered Jan 21 at 1:42
clathratusclathratus
4,615337
4,615337
add a comment |
add a comment |
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