Finding $frac{1}{2pi}int_{0}^{2pi}phi^prime(x) dx$, where $phi(x)=arctanfrac{3cos x}{4(cos x+sin x)}$. Why...
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I'm tasked with the following problem:
Evaluate
$$I_C=frac{1}{2pi}int_{0}^{2pi}left(frac{d}{dtheta}phi(theta)right) dtheta,quadtext{where}; phi(theta)=arctanleft[frac{3cos(theta)}{4(cos(theta)+sin(theta))}right]$$
Am I correct in assuming that this is simply $phi(2pi)-phi(0)$? When I do that I get zero, but when I take the derivative, then evaluate the integral I get -1. How do I use the fundamental theorem of calculus to get the correct answer?
calculus
edited Dec 11 '18 at 9:10
Blue
48.3k870153
asked Dec 10 '18 at 6:49
PeetriusPeetrius
409111
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add a comment |
$begingroup$
I'm tasked with the following problem:
Evaluate
$$I_C=frac{1}{2pi}int_{0}^{2pi}left(frac{d}{dtheta}phi(theta)right) dtheta,quadtext{where}; phi(theta)=arctanleft[frac{3cos(theta)}{4(cos(theta)+sin(theta))}right]$$
Am I correct in assuming that this is simply $phi(2pi)-phi(0)$? When I do that I get zero, but when I take the derivative, then evaluate the integral I get -1. How do I use the fundamental theorem of calculus to get the correct answer?
calculus
edited Dec 11 '18 at 9:10
Blue
48.3k870153
asked Dec 10 '18 at 6:49
PeetriusPeetrius
409111
$endgroup$
add a comment |
$begingroup$
I'm tasked with the following problem:
Evaluate
$$I_C=frac{1}{2pi}int_{0}^{2pi}left(frac{d}{dtheta}phi(theta)right) dtheta,quadtext{where}; phi(theta)=arctanleft[frac{3cos(theta)}{4(cos(theta)+sin(theta))}right]$$
Am I correct in assuming that this is simply $phi(2pi)-phi(0)$? When I do that I get zero, but when I take the derivative, then evaluate the integral I get -1. How do I use the fundamental theorem of calculus to get the correct answer?
calculus
edited Dec 11 '18 at 9:10
Blue
48.3k870153
asked Dec 10 '18 at 6:49
PeetriusPeetrius
409111
$endgroup$
I'm tasked with the following problem:
Evaluate
$$I_C=frac{1}{2pi}int_{0}^{2pi}left(frac{d}{dtheta}phi(theta)right) dtheta,quadtext{where}; phi(theta)=arctanleft[frac{3cos(theta)}{4(cos(theta)+sin(theta))}right]$$
Am I correct in assuming that this is simply $phi(2pi)-phi(0)$? When I do that I get zero, but when I take the derivative, then evaluate the integral I get -1. How do I use the fundamental theorem of calculus to get the correct answer?
calculus
calculus
edited Dec 11 '18 at 9:10
Blue
48.3k870153
asked Dec 10 '18 at 6:49
PeetriusPeetrius
409111
edited Dec 11 '18 at 9:10
Blue
48.3k870153
asked Dec 10 '18 at 6:49
PeetriusPeetrius
409111
edited Dec 11 '18 at 9:10
Blue
48.3k870153
edited Dec 11 '18 at 9:10
Blue
48.3k870153
edited Dec 11 '18 at 9:10
Blue
48.3k870153
48.3k870153
asked Dec 10 '18 at 6:49
PeetriusPeetrius
409111
asked Dec 10 '18 at 6:49
PeetriusPeetrius
409111
asked Dec 10 '18 at 6:49
PeetriusPeetrius
409111
409111
add a comment |
add a comment |
1 Answer
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Hint: what happens at $theta =3pi/4, 7pi/4$? See the graph if required.
answered Dec 10 '18 at 8:28
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.3k42871
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add a comment |
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1 Answer
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$begingroup$
Hint: what happens at $theta =3pi/4, 7pi/4$? See the graph if required.
answered Dec 10 '18 at 8:28
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.3k42871
$endgroup$
add a comment |
$begingroup$
Hint: what happens at $theta =3pi/4, 7pi/4$? See the graph if required.
answered Dec 10 '18 at 8:28
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.3k42871
$endgroup$
add a comment |
$begingroup$
Hint: what happens at $theta =3pi/4, 7pi/4$? See the graph if required.
answered Dec 10 '18 at 8:28
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.3k42871
$endgroup$
Hint: what happens at $theta =3pi/4, 7pi/4$? See the graph if required.
answered Dec 10 '18 at 8:28
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.3k42871
edited Dec 11 '18 at 8:40
answered Dec 10 '18 at 8:28
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.3k42871
answered Dec 10 '18 at 8:28
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.3k42871
answered Dec 10 '18 at 8:28
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.3k42871
34.3k42871
add a comment |
add a comment |
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