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Integral of $fraccos(f_1(x))cos(f_2(x))$
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When I analyze something in my work.
I need to calculate following form related to cosine function.
Even though I try to find indefinite form or integration by parts, I can not calculate following integral. Please let me know if you have any good idea.
$$int_b_1^b_2fraccos(f_1(x))cos(f_2(x))(f_3(x))' dx=? $$
where $cos(f_2(b_1))neq0$ and $cos(f_2(b_2))neq0$
Have a nice day.
analysis
asked Jul 24 at 11:24
user145377
306
add a comment |Â
up vote
-2
down vote
favorite
When I analyze something in my work.
I need to calculate following form related to cosine function.
Even though I try to find indefinite form or integration by parts, I can not calculate following integral. Please let me know if you have any good idea.
$$int_b_1^b_2fraccos(f_1(x))cos(f_2(x))(f_3(x))' dx=? $$
where $cos(f_2(b_1))neq0$ and $cos(f_2(b_2))neq0$
Have a nice day.
analysis
asked Jul 24 at 11:24
user145377
306
4
It looks hideous. For one thing the denominator might have zeroes, and there is not enough information to either cancel these with zeroes in the numerator or to control the rate at which the denominator tends to zero. Perhaps the trigonometric identities allow those to be simplified, but in the general form stated I have no suggestions.
– hardmath
Jul 24 at 11:28
add a comment |Â
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
When I analyze something in my work.
I need to calculate following form related to cosine function.
Even though I try to find indefinite form or integration by parts, I can not calculate following integral. Please let me know if you have any good idea.
$$int_b_1^b_2fraccos(f_1(x))cos(f_2(x))(f_3(x))' dx=? $$
where $cos(f_2(b_1))neq0$ and $cos(f_2(b_2))neq0$
Have a nice day.
analysis
asked Jul 24 at 11:24
user145377
306
When I analyze something in my work.
I need to calculate following form related to cosine function.
Even though I try to find indefinite form or integration by parts, I can not calculate following integral. Please let me know if you have any good idea.
$$int_b_1^b_2fraccos(f_1(x))cos(f_2(x))(f_3(x))' dx=? $$
where $cos(f_2(b_1))neq0$ and $cos(f_2(b_2))neq0$
Have a nice day.
analysis
asked Jul 24 at 11:24
user145377
306
edited Jul 24 at 11:40
asked Jul 24 at 11:24
user145377
306
asked Jul 24 at 11:24
user145377
306
asked Jul 24 at 11:24
user145377
306
306
4
It looks hideous. For one thing the denominator might have zeroes, and there is not enough information to either cancel these with zeroes in the numerator or to control the rate at which the denominator tends to zero. Perhaps the trigonometric identities allow those to be simplified, but in the general form stated I have no suggestions.
– hardmath
Jul 24 at 11:28
add a comment |Â
4
It looks hideous. For one thing the denominator might have zeroes, and there is not enough information to either cancel these with zeroes in the numerator or to control the rate at which the denominator tends to zero. Perhaps the trigonometric identities allow those to be simplified, but in the general form stated I have no suggestions.
– hardmath
Jul 24 at 11:28
4
4
It looks hideous. For one thing the denominator might have zeroes, and there is not enough information to either cancel these with zeroes in the numerator or to control the rate at which the denominator tends to zero. Perhaps the trigonometric identities allow those to be simplified, but in the general form stated I have no suggestions.
– hardmath
Jul 24 at 11:28
It looks hideous. For one thing the denominator might have zeroes, and there is not enough information to either cancel these with zeroes in the numerator or to control the rate at which the denominator tends to zero. Perhaps the trigonometric identities allow those to be simplified, but in the general form stated I have no suggestions.
– hardmath
Jul 24 at 11:28
add a comment |Â
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4
It looks hideous. For one thing the denominator might have zeroes, and there is not enough information to either cancel these with zeroes in the numerator or to control the rate at which the denominator tends to zero. Perhaps the trigonometric identities allow those to be simplified, but in the general form stated I have no suggestions.
– hardmath
Jul 24 at 11:28