trigonometric integration problem
Hello guys can someone please help me find the answer :
$$int sin x tan x~dx$$
indefinite-integrals
add a comment |
Hello guys can someone please help me find the answer :
$$int sin x tan x~dx$$
indefinite-integrals
add a comment |
Hello guys can someone please help me find the answer :
$$int sin x tan x~dx$$
indefinite-integrals
Hello guys can someone please help me find the answer :
$$int sin x tan x~dx$$
indefinite-integrals
indefinite-integrals
edited 41 mins ago
mrtaurho
3,1051930
3,1051930
asked 42 mins ago
RS 2838484
61
61
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2 Answers
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Note that $$sin(x)tan(x)=frac{sin^2(x)}{cos(x)}=frac{1-cos^2(x)}{cos(x)}$$
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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.
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Note that $$sin(x)tan(x)=frac{sin^2(x)}{cos(x)}=frac{1-cos^2(x)}{cos(x)}$$
add a comment |
Note that $$sin(x)tan(x)=frac{sin^2(x)}{cos(x)}=frac{1-cos^2(x)}{cos(x)}$$
add a comment |
Note that $$sin(x)tan(x)=frac{sin^2(x)}{cos(x)}=frac{1-cos^2(x)}{cos(x)}$$
Note that $$sin(x)tan(x)=frac{sin^2(x)}{cos(x)}=frac{1-cos^2(x)}{cos(x)}$$
edited 22 mins ago
Bernard
117k637111
117k637111
answered 38 mins ago
Dr. Sonnhard Graubner
72.6k32865
72.6k32865
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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.
add a comment |
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.
add a comment |
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.
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 13 mins ago
Bernard
117k637111
117k637111
add a comment |
add a comment |
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