Integrate $frac{cos(2x)}{sin(x)}$ [closed]
Integrate $$intfrac{cos(2x)}{sin(x)}dx$$
I've tried all the trigonometric identities (known to me) but it just keeps getting more and more complicated.
I was also unable to solve another integral of similar form:
$$intfrac{cos(3x)}{1-sin(x)}dx$$ Solving the double chain is very confusing...
calculus integration substitution
closed as off-topic by amWhy, RRL, Cesareo, Zacky, Leucippus Jan 5 at 5:20
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Integrate $$intfrac{cos(2x)}{sin(x)}dx$$
I've tried all the trigonometric identities (known to me) but it just keeps getting more and more complicated.
I was also unable to solve another integral of similar form:
$$intfrac{cos(3x)}{1-sin(x)}dx$$ Solving the double chain is very confusing...
calculus integration substitution
closed as off-topic by amWhy, RRL, Cesareo, Zacky, Leucippus Jan 5 at 5:20
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." – amWhy, RRL, Cesareo, Zacky, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
Integrate $$intfrac{cos(2x)}{sin(x)}dx$$
I've tried all the trigonometric identities (known to me) but it just keeps getting more and more complicated.
I was also unable to solve another integral of similar form:
$$intfrac{cos(3x)}{1-sin(x)}dx$$ Solving the double chain is very confusing...
calculus integration substitution
Integrate $$intfrac{cos(2x)}{sin(x)}dx$$
I've tried all the trigonometric identities (known to me) but it just keeps getting more and more complicated.
I was also unable to solve another integral of similar form:
$$intfrac{cos(3x)}{1-sin(x)}dx$$ Solving the double chain is very confusing...
calculus integration substitution
calculus integration substitution
asked Jan 4 at 20:16
Conny DagoConny Dago
255
255
closed as off-topic by amWhy, RRL, Cesareo, Zacky, Leucippus Jan 5 at 5:20
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." – amWhy, RRL, Cesareo, Zacky, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by amWhy, RRL, Cesareo, Zacky, Leucippus Jan 5 at 5:20
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." – amWhy, RRL, Cesareo, Zacky, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
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4 Answers
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For the first one we can write $cos(2x) =1-2sin^2 x$.
$$Rightarrow int frac{cos(2x)}{sin x}dx= int frac{1}{sin x}dx-2int sin xdx$$
For the second one we have:
$$I=int frac{cos(3x)}{1-sin x}frac{1+sin x}{1+sin x}dx=int frac{cos(3x)}{cos^2 x}dx+intfrac{cos(3x)sin x}{cos^2 x}dx$$
Now we can use the following identity: $$cos(3x)=4cos^3 x-3cos x$$
$$Rightarrow I=4intcos x dx-3intfrac{1}{cos x}dx+2int sin(2x) dx -3int tan x dx$$
I think you can finish now, but as a hint for $displaystyle{int frac{dx}{sin x}}$ and $displaystyle{int frac{dx}{cos x}}$, write them as: $displaystyle{int frac{sin x}{1-cos^2 x}dx}$ respectively $displaystyle{int frac{cos x}{1-sin^2 x}dx}$.
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$$frac{cos2x}{sin x}=frac{1-2sin^2x}{sin x}==frac1{sin x}-2sin x$$
You may also want to observe that
$$intfrac1{sin x}dx=intfrac{1+t^2}{2t}cdotfrac2{1+t^2}dt=intfrac{dt}tldots$$
with the Weierstrass substitution $;t=tanfrac x2;$ ...which can also be used for the second integral.
add a comment |
Hint:
Bioche's rules suggest for the first integral to make the substitution
$$t=cos x,quadmathrm dt=-sin x ,mathrm dx,$$
and for the second integral:
$$t=sin x,quadmathrm dt=cos x ,mathrm dx.$$
For the latter, you'll also need to know the formula
$$cos3x=4cos^3x-3cos x.$$
You'll obtain ultimately the integral of a rational function in $t$, which you integrate with the standard method of partial fractions decomposition.
add a comment |
Hint:
For the second one, $$dfrac{cos x(4cos^2x-3)}{1-sin x},$$ set $1-sin x=u,implies cos^2x=1-(1-u)^2=?$
For the first, $$dfrac{2cos^2x-1}{1-cos^2x}sin x,$$ choose $cos x=v$
$$dfrac{2v^2-1}{v^2-1}=2+dfrac{v+1-(v-1)}{2(v^2-1)}$$
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
For the first one we can write $cos(2x) =1-2sin^2 x$.
$$Rightarrow int frac{cos(2x)}{sin x}dx= int frac{1}{sin x}dx-2int sin xdx$$
For the second one we have:
$$I=int frac{cos(3x)}{1-sin x}frac{1+sin x}{1+sin x}dx=int frac{cos(3x)}{cos^2 x}dx+intfrac{cos(3x)sin x}{cos^2 x}dx$$
Now we can use the following identity: $$cos(3x)=4cos^3 x-3cos x$$
$$Rightarrow I=4intcos x dx-3intfrac{1}{cos x}dx+2int sin(2x) dx -3int tan x dx$$
I think you can finish now, but as a hint for $displaystyle{int frac{dx}{sin x}}$ and $displaystyle{int frac{dx}{cos x}}$, write them as: $displaystyle{int frac{sin x}{1-cos^2 x}dx}$ respectively $displaystyle{int frac{cos x}{1-sin^2 x}dx}$.
add a comment |
For the first one we can write $cos(2x) =1-2sin^2 x$.
$$Rightarrow int frac{cos(2x)}{sin x}dx= int frac{1}{sin x}dx-2int sin xdx$$
For the second one we have:
$$I=int frac{cos(3x)}{1-sin x}frac{1+sin x}{1+sin x}dx=int frac{cos(3x)}{cos^2 x}dx+intfrac{cos(3x)sin x}{cos^2 x}dx$$
Now we can use the following identity: $$cos(3x)=4cos^3 x-3cos x$$
$$Rightarrow I=4intcos x dx-3intfrac{1}{cos x}dx+2int sin(2x) dx -3int tan x dx$$
I think you can finish now, but as a hint for $displaystyle{int frac{dx}{sin x}}$ and $displaystyle{int frac{dx}{cos x}}$, write them as: $displaystyle{int frac{sin x}{1-cos^2 x}dx}$ respectively $displaystyle{int frac{cos x}{1-sin^2 x}dx}$.
add a comment |
For the first one we can write $cos(2x) =1-2sin^2 x$.
$$Rightarrow int frac{cos(2x)}{sin x}dx= int frac{1}{sin x}dx-2int sin xdx$$
For the second one we have:
$$I=int frac{cos(3x)}{1-sin x}frac{1+sin x}{1+sin x}dx=int frac{cos(3x)}{cos^2 x}dx+intfrac{cos(3x)sin x}{cos^2 x}dx$$
Now we can use the following identity: $$cos(3x)=4cos^3 x-3cos x$$
$$Rightarrow I=4intcos x dx-3intfrac{1}{cos x}dx+2int sin(2x) dx -3int tan x dx$$
I think you can finish now, but as a hint for $displaystyle{int frac{dx}{sin x}}$ and $displaystyle{int frac{dx}{cos x}}$, write them as: $displaystyle{int frac{sin x}{1-cos^2 x}dx}$ respectively $displaystyle{int frac{cos x}{1-sin^2 x}dx}$.
For the first one we can write $cos(2x) =1-2sin^2 x$.
$$Rightarrow int frac{cos(2x)}{sin x}dx= int frac{1}{sin x}dx-2int sin xdx$$
For the second one we have:
$$I=int frac{cos(3x)}{1-sin x}frac{1+sin x}{1+sin x}dx=int frac{cos(3x)}{cos^2 x}dx+intfrac{cos(3x)sin x}{cos^2 x}dx$$
Now we can use the following identity: $$cos(3x)=4cos^3 x-3cos x$$
$$Rightarrow I=4intcos x dx-3intfrac{1}{cos x}dx+2int sin(2x) dx -3int tan x dx$$
I think you can finish now, but as a hint for $displaystyle{int frac{dx}{sin x}}$ and $displaystyle{int frac{dx}{cos x}}$, write them as: $displaystyle{int frac{sin x}{1-cos^2 x}dx}$ respectively $displaystyle{int frac{cos x}{1-sin^2 x}dx}$.
edited Jan 5 at 0:26
answered Jan 4 at 20:31
ZackyZacky
5,1531752
5,1531752
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$$frac{cos2x}{sin x}=frac{1-2sin^2x}{sin x}==frac1{sin x}-2sin x$$
You may also want to observe that
$$intfrac1{sin x}dx=intfrac{1+t^2}{2t}cdotfrac2{1+t^2}dt=intfrac{dt}tldots$$
with the Weierstrass substitution $;t=tanfrac x2;$ ...which can also be used for the second integral.
add a comment |
$$frac{cos2x}{sin x}=frac{1-2sin^2x}{sin x}==frac1{sin x}-2sin x$$
You may also want to observe that
$$intfrac1{sin x}dx=intfrac{1+t^2}{2t}cdotfrac2{1+t^2}dt=intfrac{dt}tldots$$
with the Weierstrass substitution $;t=tanfrac x2;$ ...which can also be used for the second integral.
add a comment |
$$frac{cos2x}{sin x}=frac{1-2sin^2x}{sin x}==frac1{sin x}-2sin x$$
You may also want to observe that
$$intfrac1{sin x}dx=intfrac{1+t^2}{2t}cdotfrac2{1+t^2}dt=intfrac{dt}tldots$$
with the Weierstrass substitution $;t=tanfrac x2;$ ...which can also be used for the second integral.
$$frac{cos2x}{sin x}=frac{1-2sin^2x}{sin x}==frac1{sin x}-2sin x$$
You may also want to observe that
$$intfrac1{sin x}dx=intfrac{1+t^2}{2t}cdotfrac2{1+t^2}dt=intfrac{dt}tldots$$
with the Weierstrass substitution $;t=tanfrac x2;$ ...which can also be used for the second integral.
answered Jan 4 at 20:31
DonAntonioDonAntonio
177k1492225
177k1492225
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Hint:
Bioche's rules suggest for the first integral to make the substitution
$$t=cos x,quadmathrm dt=-sin x ,mathrm dx,$$
and for the second integral:
$$t=sin x,quadmathrm dt=cos x ,mathrm dx.$$
For the latter, you'll also need to know the formula
$$cos3x=4cos^3x-3cos x.$$
You'll obtain ultimately the integral of a rational function in $t$, which you integrate with the standard method of partial fractions decomposition.
add a comment |
Hint:
Bioche's rules suggest for the first integral to make the substitution
$$t=cos x,quadmathrm dt=-sin x ,mathrm dx,$$
and for the second integral:
$$t=sin x,quadmathrm dt=cos x ,mathrm dx.$$
For the latter, you'll also need to know the formula
$$cos3x=4cos^3x-3cos x.$$
You'll obtain ultimately the integral of a rational function in $t$, which you integrate with the standard method of partial fractions decomposition.
add a comment |
Hint:
Bioche's rules suggest for the first integral to make the substitution
$$t=cos x,quadmathrm dt=-sin x ,mathrm dx,$$
and for the second integral:
$$t=sin x,quadmathrm dt=cos x ,mathrm dx.$$
For the latter, you'll also need to know the formula
$$cos3x=4cos^3x-3cos x.$$
You'll obtain ultimately the integral of a rational function in $t$, which you integrate with the standard method of partial fractions decomposition.
Hint:
Bioche's rules suggest for the first integral to make the substitution
$$t=cos x,quadmathrm dt=-sin x ,mathrm dx,$$
and for the second integral:
$$t=sin x,quadmathrm dt=cos x ,mathrm dx.$$
For the latter, you'll also need to know the formula
$$cos3x=4cos^3x-3cos x.$$
You'll obtain ultimately the integral of a rational function in $t$, which you integrate with the standard method of partial fractions decomposition.
edited Jan 4 at 22:02
answered Jan 4 at 20:31
BernardBernard
118k639112
118k639112
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Hint:
For the second one, $$dfrac{cos x(4cos^2x-3)}{1-sin x},$$ set $1-sin x=u,implies cos^2x=1-(1-u)^2=?$
For the first, $$dfrac{2cos^2x-1}{1-cos^2x}sin x,$$ choose $cos x=v$
$$dfrac{2v^2-1}{v^2-1}=2+dfrac{v+1-(v-1)}{2(v^2-1)}$$
add a comment |
Hint:
For the second one, $$dfrac{cos x(4cos^2x-3)}{1-sin x},$$ set $1-sin x=u,implies cos^2x=1-(1-u)^2=?$
For the first, $$dfrac{2cos^2x-1}{1-cos^2x}sin x,$$ choose $cos x=v$
$$dfrac{2v^2-1}{v^2-1}=2+dfrac{v+1-(v-1)}{2(v^2-1)}$$
add a comment |
Hint:
For the second one, $$dfrac{cos x(4cos^2x-3)}{1-sin x},$$ set $1-sin x=u,implies cos^2x=1-(1-u)^2=?$
For the first, $$dfrac{2cos^2x-1}{1-cos^2x}sin x,$$ choose $cos x=v$
$$dfrac{2v^2-1}{v^2-1}=2+dfrac{v+1-(v-1)}{2(v^2-1)}$$
Hint:
For the second one, $$dfrac{cos x(4cos^2x-3)}{1-sin x},$$ set $1-sin x=u,implies cos^2x=1-(1-u)^2=?$
For the first, $$dfrac{2cos^2x-1}{1-cos^2x}sin x,$$ choose $cos x=v$
$$dfrac{2v^2-1}{v^2-1}=2+dfrac{v+1-(v-1)}{2(v^2-1)}$$
answered Jan 5 at 2:54
lab bhattacharjeelab bhattacharjee
224k15156274
224k15156274
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