trigonometric integration problem [closed]
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Hello guys can someone please help me find the answer :
$$int sin x tan x~dx$$
indefinite-integrals
edited Dec 22 '18 at 18:05
mrtaurho
4,16121234
asked Dec 22 '18 at 18:04
RS 2838484RS 2838484
51
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closed as off-topic by mrtaurho, amWhy, user10354138, José Carlos Santos, Eric Wofsey Dec 22 '18 at 23:12
This question appears to be off-topic. The users who voted to close gave this specific reason:
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$begingroup$
Hello guys can someone please help me find the answer :
$$int sin x tan x~dx$$
indefinite-integrals
edited Dec 22 '18 at 18:05
mrtaurho
4,16121234
asked Dec 22 '18 at 18:04
RS 2838484RS 2838484
51
$endgroup$
closed as off-topic by mrtaurho, amWhy, user10354138, José Carlos Santos, Eric Wofsey Dec 22 '18 at 23:12
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, amWhy, user10354138, José Carlos Santos, Eric Wofsey
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Hello guys can someone please help me find the answer :
$$int sin x tan x~dx$$
indefinite-integrals
edited Dec 22 '18 at 18:05
mrtaurho
4,16121234
asked Dec 22 '18 at 18:04
RS 2838484RS 2838484
51
$endgroup$
Hello guys can someone please help me find the answer :
$$int sin x tan x~dx$$
indefinite-integrals
indefinite-integrals
edited Dec 22 '18 at 18:05
mrtaurho
4,16121234
asked Dec 22 '18 at 18:04
RS 2838484RS 2838484
51
edited Dec 22 '18 at 18:05
mrtaurho
4,16121234
asked Dec 22 '18 at 18:04
RS 2838484RS 2838484
51
edited Dec 22 '18 at 18:05
mrtaurho
4,16121234
edited Dec 22 '18 at 18:05
mrtaurho
4,16121234
edited Dec 22 '18 at 18:05
mrtaurho
4,16121234
4,16121234
asked Dec 22 '18 at 18:04
RS 2838484RS 2838484
51
asked Dec 22 '18 at 18:04
RS 2838484RS 2838484
51
asked Dec 22 '18 at 18:04
RS 2838484RS 2838484
51
51
closed as off-topic by mrtaurho, amWhy, user10354138, José Carlos Santos, Eric Wofsey Dec 22 '18 at 23:12
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, amWhy, user10354138, José Carlos Santos, Eric Wofsey
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by mrtaurho, amWhy, user10354138, José Carlos Santos, Eric Wofsey Dec 22 '18 at 23:12
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, amWhy, user10354138, José Carlos Santos, Eric Wofsey
If this question can be reworded to fit the rules in the help center, please edit the question.
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2 Answers
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Note that $$sin(x)tan(x)=frac{sin^2(x)}{cos(x)}=frac{1-cos^2(x)}{cos(x)}$$
edited Dec 22 '18 at 18:24
Bernard
119k740113
answered Dec 22 '18 at 18:08
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74k42865
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add a comment |
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Bioche's rules say you should make the substitution $;u=sin x$, $mathrm d u=cos x,mathrm dx$, so that
$$int sin x tan x,mathrm dx=intfrac{sin^2x}{cos x},frac{mathrm du}{cos x}=intfrac{u^2}{1-u^2},mathrm du,$$
the integral of a rational function. Thus it comes down to a decomposition into partial fractions.
answered Dec 22 '18 at 18:32
BernardBernard
119k740113
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add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that $$sin(x)tan(x)=frac{sin^2(x)}{cos(x)}=frac{1-cos^2(x)}{cos(x)}$$
edited Dec 22 '18 at 18:24
Bernard
119k740113
answered Dec 22 '18 at 18:08
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74k42865
$endgroup$
add a comment |
$begingroup$
Note that $$sin(x)tan(x)=frac{sin^2(x)}{cos(x)}=frac{1-cos^2(x)}{cos(x)}$$
edited Dec 22 '18 at 18:24
Bernard
119k740113
answered Dec 22 '18 at 18:08
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74k42865
$endgroup$
add a comment |
$begingroup$
Note that $$sin(x)tan(x)=frac{sin^2(x)}{cos(x)}=frac{1-cos^2(x)}{cos(x)}$$
edited Dec 22 '18 at 18:24
Bernard
119k740113
answered Dec 22 '18 at 18:08
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74k42865
$endgroup$
Note that $$sin(x)tan(x)=frac{sin^2(x)}{cos(x)}=frac{1-cos^2(x)}{cos(x)}$$
edited Dec 22 '18 at 18:24
Bernard
119k740113
answered Dec 22 '18 at 18:08
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74k42865
edited Dec 22 '18 at 18:24
Bernard
119k740113
edited Dec 22 '18 at 18:24
Bernard
119k740113
edited Dec 22 '18 at 18:24
Bernard
119k740113
119k740113
answered Dec 22 '18 at 18:08
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74k42865
answered Dec 22 '18 at 18:08
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74k42865
answered Dec 22 '18 at 18:08
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74k42865
74k42865
add a comment |
add a comment |
$begingroup$
Bioche's rules say you should make the substitution $;u=sin x$, $mathrm d u=cos x,mathrm dx$, so that
$$int sin x tan x,mathrm dx=intfrac{sin^2x}{cos x},frac{mathrm du}{cos x}=intfrac{u^2}{1-u^2},mathrm du,$$
the integral of a rational function. Thus it comes down to a decomposition into partial fractions.
answered Dec 22 '18 at 18:32
BernardBernard
119k740113
$endgroup$
add a comment |
$begingroup$
Bioche's rules say you should make the substitution $;u=sin x$, $mathrm d u=cos x,mathrm dx$, so that
$$int sin x tan x,mathrm dx=intfrac{sin^2x}{cos x},frac{mathrm du}{cos x}=intfrac{u^2}{1-u^2},mathrm du,$$
the integral of a rational function. Thus it comes down to a decomposition into partial fractions.
answered Dec 22 '18 at 18:32
BernardBernard
119k740113
$endgroup$
add a comment |
$begingroup$
Bioche's rules say you should make the substitution $;u=sin x$, $mathrm d u=cos x,mathrm dx$, so that
$$int sin x tan x,mathrm dx=intfrac{sin^2x}{cos x},frac{mathrm du}{cos x}=intfrac{u^2}{1-u^2},mathrm du,$$
the integral of a rational function. Thus it comes down to a decomposition into partial fractions.
answered Dec 22 '18 at 18:32
BernardBernard
119k740113
$endgroup$
Bioche's rules say you should make the substitution $;u=sin x$, $mathrm d u=cos x,mathrm dx$, so that
$$int sin x tan x,mathrm dx=intfrac{sin^2x}{cos x},frac{mathrm du}{cos x}=intfrac{u^2}{1-u^2},mathrm du,$$
the integral of a rational function. Thus it comes down to a decomposition into partial fractions.
answered Dec 22 '18 at 18:32
BernardBernard
119k740113
answered Dec 22 '18 at 18:32
BernardBernard
119k740113
answered Dec 22 '18 at 18:32
BernardBernard
119k740113
answered Dec 22 '18 at 18:32
BernardBernard
119k740113
119k740113
add a comment |
add a comment |
