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Compute $int_0^{frac {pi}2} ln(cos^2x+y^2sin^2x) dx$ [duplicate]
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This question already has an answer here:
Integrals depending on a parameter: $int_{0}^{pi/2} ln(a^2sin^2{x} + cos^2{x})dx $
2 answers
Compute: $$I(y)=int_0^{frac {pi}2} ln(cos^2x+y^2sin^2x) dx$$
So$$I'(y)=int_0^{frac {pi}2}frac {2ysin^2x}{cos^2x+y^2sin^2x}dx$$
Now what?
integration definite-integrals
edited Jan 24 at 18:52
Zacky
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asked Jan 24 at 18:38
C. CristiC. Cristi
1,634218
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marked as duplicate by xpaul, Zacky, Community♦ Jan 24 at 19:22
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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$begingroup$
This question already has an answer here:
Integrals depending on a parameter: $int_{0}^{pi/2} ln(a^2sin^2{x} + cos^2{x})dx $
2 answers
Compute: $$I(y)=int_0^{frac {pi}2} ln(cos^2x+y^2sin^2x) dx$$
So$$I'(y)=int_0^{frac {pi}2}frac {2ysin^2x}{cos^2x+y^2sin^2x}dx$$
Now what?
integration definite-integrals
edited Jan 24 at 18:52
Zacky
7,89011061
asked Jan 24 at 18:38
C. CristiC. Cristi
1,634218
$endgroup$
marked as duplicate by xpaul, Zacky, Community♦ Jan 24 at 19:22
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Integrals depending on a parameter: $int_{0}^{pi/2} ln(a^2sin^2{x} + cos^2{x})dx $
2 answers
Compute: $$I(y)=int_0^{frac {pi}2} ln(cos^2x+y^2sin^2x) dx$$
So$$I'(y)=int_0^{frac {pi}2}frac {2ysin^2x}{cos^2x+y^2sin^2x}dx$$
Now what?
integration definite-integrals
edited Jan 24 at 18:52
Zacky
7,89011061
asked Jan 24 at 18:38
C. CristiC. Cristi
1,634218
$endgroup$
This question already has an answer here:
Integrals depending on a parameter: $int_{0}^{pi/2} ln(a^2sin^2{x} + cos^2{x})dx $
2 answers
Compute: $$I(y)=int_0^{frac {pi}2} ln(cos^2x+y^2sin^2x) dx$$
So$$I'(y)=int_0^{frac {pi}2}frac {2ysin^2x}{cos^2x+y^2sin^2x}dx$$
Now what?
This question already has an answer here:
Integrals depending on a parameter: $int_{0}^{pi/2} ln(a^2sin^2{x} + cos^2{x})dx $
2 answers
integration definite-integrals
integration definite-integrals
edited Jan 24 at 18:52
Zacky
7,89011061
asked Jan 24 at 18:38
C. CristiC. Cristi
1,634218
edited Jan 24 at 18:52
Zacky
7,89011061
asked Jan 24 at 18:38
C. CristiC. Cristi
1,634218
edited Jan 24 at 18:52
Zacky
7,89011061
edited Jan 24 at 18:52
Zacky
7,89011061
edited Jan 24 at 18:52
Zacky
7,89011061
7,89011061
asked Jan 24 at 18:38
C. CristiC. Cristi
1,634218
asked Jan 24 at 18:38
C. CristiC. Cristi
1,634218
asked Jan 24 at 18:38
C. CristiC. Cristi
1,634218
1,634218
marked as duplicate by xpaul, Zacky, Community♦ Jan 24 at 19:22
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by xpaul, Zacky, Community♦ Jan 24 at 19:22
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
add a comment |
1 Answer
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$$I(y)=int_0^{frac {pi}2} ln(cos^2x+y^2sin^2x) dxRightarrow I'(y) =2yint_0^frac{pi}{2}frac{sin^2 x}{cos^2 x+y^2sin^2 x}dx$$
$$=2yint_0^frac{pi}{2} frac{1}{cot^2 x +y^2}dxoverset{cot x=t}=2yint_0^infty frac{1}{t^2+y^2}frac{dt}{t^2+1}$$
$$=frac{2y}{1-y^2}int_0^infty frac{dt}{t^2+y^2}-frac{2y}{1-y^2}int_0^infty frac{dt}{t^2+1}$$
$$=frac{2y}{1-y^2}frac{1}{y} frac{pi}{2} -frac{2y}{1-y^2}frac{pi}{2}=pifrac{1-y}{1-y^2}=frac{pi}{1+y}$$
This gives:$$I(y)=int frac{pi}{1+y} dy =pi ln(1+y)+C$$
In order to find the constant we can simply set $y=1$ in the original integral, that is because $sin^2 x+cos^2 x$ has a nice value.
$$Rightarrow I(1)=int_0^frac{pi}{2} ln(1)dx =piln(1+1)+C Rightarrow 0=piln 2 +C Rightarrow C=-pi ln 2$$
$$Rightarrow I(y)=piln(1+y) -pi ln 2$$
answered Jan 24 at 18:49
ZackyZacky
7,89011061
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$I = pi ln(1+y)-frac {pi}4 ln 2?$
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– C. Cristi
Jan 24 at 19:21
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Because we have to integrate with respect to y back to get the original right? And then you have the constant of integration
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– C. Cristi
Jan 26 at 8:28
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@C.Cristi I have updated the answer.
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– Zacky
Jan 26 at 9:58
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$I(y)=int_0^{frac {pi}2} ln(cos^2x+y^2sin^2x) dxRightarrow I'(y) =2yint_0^frac{pi}{2}frac{sin^2 x}{cos^2 x+y^2sin^2 x}dx$$
$$=2yint_0^frac{pi}{2} frac{1}{cot^2 x +y^2}dxoverset{cot x=t}=2yint_0^infty frac{1}{t^2+y^2}frac{dt}{t^2+1}$$
$$=frac{2y}{1-y^2}int_0^infty frac{dt}{t^2+y^2}-frac{2y}{1-y^2}int_0^infty frac{dt}{t^2+1}$$
$$=frac{2y}{1-y^2}frac{1}{y} frac{pi}{2} -frac{2y}{1-y^2}frac{pi}{2}=pifrac{1-y}{1-y^2}=frac{pi}{1+y}$$
This gives:$$I(y)=int frac{pi}{1+y} dy =pi ln(1+y)+C$$
In order to find the constant we can simply set $y=1$ in the original integral, that is because $sin^2 x+cos^2 x$ has a nice value.
$$Rightarrow I(1)=int_0^frac{pi}{2} ln(1)dx =piln(1+1)+C Rightarrow 0=piln 2 +C Rightarrow C=-pi ln 2$$
$$Rightarrow I(y)=piln(1+y) -pi ln 2$$
answered Jan 24 at 18:49
ZackyZacky
7,89011061
$endgroup$
$begingroup$
$I = pi ln(1+y)-frac {pi}4 ln 2?$
$endgroup$
– C. Cristi
Jan 24 at 19:21
$begingroup$
Because we have to integrate with respect to y back to get the original right? And then you have the constant of integration
$endgroup$
– C. Cristi
Jan 26 at 8:28
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@C.Cristi I have updated the answer.
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– Zacky
Jan 26 at 9:58
add a comment |
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$$I(y)=int_0^{frac {pi}2} ln(cos^2x+y^2sin^2x) dxRightarrow I'(y) =2yint_0^frac{pi}{2}frac{sin^2 x}{cos^2 x+y^2sin^2 x}dx$$
$$=2yint_0^frac{pi}{2} frac{1}{cot^2 x +y^2}dxoverset{cot x=t}=2yint_0^infty frac{1}{t^2+y^2}frac{dt}{t^2+1}$$
$$=frac{2y}{1-y^2}int_0^infty frac{dt}{t^2+y^2}-frac{2y}{1-y^2}int_0^infty frac{dt}{t^2+1}$$
$$=frac{2y}{1-y^2}frac{1}{y} frac{pi}{2} -frac{2y}{1-y^2}frac{pi}{2}=pifrac{1-y}{1-y^2}=frac{pi}{1+y}$$
This gives:$$I(y)=int frac{pi}{1+y} dy =pi ln(1+y)+C$$
In order to find the constant we can simply set $y=1$ in the original integral, that is because $sin^2 x+cos^2 x$ has a nice value.
$$Rightarrow I(1)=int_0^frac{pi}{2} ln(1)dx =piln(1+1)+C Rightarrow 0=piln 2 +C Rightarrow C=-pi ln 2$$
$$Rightarrow I(y)=piln(1+y) -pi ln 2$$
answered Jan 24 at 18:49
ZackyZacky
7,89011061
$endgroup$
$begingroup$
$I = pi ln(1+y)-frac {pi}4 ln 2?$
$endgroup$
– C. Cristi
Jan 24 at 19:21
$begingroup$
Because we have to integrate with respect to y back to get the original right? And then you have the constant of integration
$endgroup$
– C. Cristi
Jan 26 at 8:28
$begingroup$
@C.Cristi I have updated the answer.
$endgroup$
– Zacky
Jan 26 at 9:58
add a comment |
$begingroup$
$$I(y)=int_0^{frac {pi}2} ln(cos^2x+y^2sin^2x) dxRightarrow I'(y) =2yint_0^frac{pi}{2}frac{sin^2 x}{cos^2 x+y^2sin^2 x}dx$$
$$=2yint_0^frac{pi}{2} frac{1}{cot^2 x +y^2}dxoverset{cot x=t}=2yint_0^infty frac{1}{t^2+y^2}frac{dt}{t^2+1}$$
$$=frac{2y}{1-y^2}int_0^infty frac{dt}{t^2+y^2}-frac{2y}{1-y^2}int_0^infty frac{dt}{t^2+1}$$
$$=frac{2y}{1-y^2}frac{1}{y} frac{pi}{2} -frac{2y}{1-y^2}frac{pi}{2}=pifrac{1-y}{1-y^2}=frac{pi}{1+y}$$
This gives:$$I(y)=int frac{pi}{1+y} dy =pi ln(1+y)+C$$
In order to find the constant we can simply set $y=1$ in the original integral, that is because $sin^2 x+cos^2 x$ has a nice value.
$$Rightarrow I(1)=int_0^frac{pi}{2} ln(1)dx =piln(1+1)+C Rightarrow 0=piln 2 +C Rightarrow C=-pi ln 2$$
$$Rightarrow I(y)=piln(1+y) -pi ln 2$$
answered Jan 24 at 18:49
ZackyZacky
7,89011061
$endgroup$
$$I(y)=int_0^{frac {pi}2} ln(cos^2x+y^2sin^2x) dxRightarrow I'(y) =2yint_0^frac{pi}{2}frac{sin^2 x}{cos^2 x+y^2sin^2 x}dx$$
$$=2yint_0^frac{pi}{2} frac{1}{cot^2 x +y^2}dxoverset{cot x=t}=2yint_0^infty frac{1}{t^2+y^2}frac{dt}{t^2+1}$$
$$=frac{2y}{1-y^2}int_0^infty frac{dt}{t^2+y^2}-frac{2y}{1-y^2}int_0^infty frac{dt}{t^2+1}$$
$$=frac{2y}{1-y^2}frac{1}{y} frac{pi}{2} -frac{2y}{1-y^2}frac{pi}{2}=pifrac{1-y}{1-y^2}=frac{pi}{1+y}$$
This gives:$$I(y)=int frac{pi}{1+y} dy =pi ln(1+y)+C$$
In order to find the constant we can simply set $y=1$ in the original integral, that is because $sin^2 x+cos^2 x$ has a nice value.
$$Rightarrow I(1)=int_0^frac{pi}{2} ln(1)dx =piln(1+1)+C Rightarrow 0=piln 2 +C Rightarrow C=-pi ln 2$$
$$Rightarrow I(y)=piln(1+y) -pi ln 2$$
answered Jan 24 at 18:49
ZackyZacky
7,89011061
edited Jan 26 at 9:57
answered Jan 24 at 18:49
ZackyZacky
7,89011061
answered Jan 24 at 18:49
ZackyZacky
7,89011061
answered Jan 24 at 18:49
ZackyZacky
7,89011061
7,89011061
$begingroup$
$I = pi ln(1+y)-frac {pi}4 ln 2?$
$endgroup$
– C. Cristi
Jan 24 at 19:21
$begingroup$
Because we have to integrate with respect to y back to get the original right? And then you have the constant of integration
$endgroup$
– C. Cristi
Jan 26 at 8:28
$begingroup$
@C.Cristi I have updated the answer.
$endgroup$
– Zacky
Jan 26 at 9:58
add a comment |
$begingroup$
$I = pi ln(1+y)-frac {pi}4 ln 2?$
$endgroup$
– C. Cristi
Jan 24 at 19:21
$begingroup$
Because we have to integrate with respect to y back to get the original right? And then you have the constant of integration
$endgroup$
– C. Cristi
Jan 26 at 8:28
$begingroup$
@C.Cristi I have updated the answer.
$endgroup$
– Zacky
Jan 26 at 9:58
$begingroup$
$I = pi ln(1+y)-frac {pi}4 ln 2?$
$endgroup$
– C. Cristi
Jan 24 at 19:21
$begingroup$
$I = pi ln(1+y)-frac {pi}4 ln 2?$
$endgroup$
– C. Cristi
Jan 24 at 19:21
$begingroup$
Because we have to integrate with respect to y back to get the original right? And then you have the constant of integration
$endgroup$
– C. Cristi
Jan 26 at 8:28
$begingroup$
Because we have to integrate with respect to y back to get the original right? And then you have the constant of integration
$endgroup$
– C. Cristi
Jan 26 at 8:28
$begingroup$
@C.Cristi I have updated the answer.
$endgroup$
– Zacky
Jan 26 at 9:58
$begingroup$
@C.Cristi I have updated the answer.
$endgroup$
– Zacky
Jan 26 at 9:58
add a comment |
