What is the solution for $int_{0}^{pi/2}frac{cos^2x}{cos^2x+4sin^2x},dx$












3














I used the rule :$$int_{0}^{a}f(x) = int_{0}^{a}f(a-x),dx$$
And got :$$int_{0}^{pi/2}frac{sin^2x}{sin^2x+4cos^2x},dx$$
Then, I added both the question and the above integrand, and I got :$$int_{0}^{pi/2}frac{dx}{5}$$
Solving the above, I finally got the answer as $$frac{pi}{20}$$
But, according to my textbook, the answer is $$frac{pi}{6}$$
Where did I go wrong?






integration definite-integrals trigonometric-integrals







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  • 1




    It's not true that $frac{a}{b+c} + frac{d}{e+f} = frac{a+d}{b+c+e+f}$.
    – mathworker21
    2 days ago






  • 1




    Both denominators are different. You can't attach them together to get rid of denominator.
    – Nosrati
    2 days ago








  • 1




    $$cos^2x+4sin^2xnesin^2x+4cos^2x$$
    – Did
    2 days ago
















3














I used the rule :$$int_{0}^{a}f(x) = int_{0}^{a}f(a-x),dx$$
And got :$$int_{0}^{pi/2}frac{sin^2x}{sin^2x+4cos^2x},dx$$
Then, I added both the question and the above integrand, and I got :$$int_{0}^{pi/2}frac{dx}{5}$$
Solving the above, I finally got the answer as $$frac{pi}{20}$$
But, according to my textbook, the answer is $$frac{pi}{6}$$
Where did I go wrong?






integration definite-integrals trigonometric-integrals







share|cite|improve this question









New contributor




WildPeace007 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 1




    It's not true that $frac{a}{b+c} + frac{d}{e+f} = frac{a+d}{b+c+e+f}$.
    – mathworker21
    2 days ago






  • 1




    Both denominators are different. You can't attach them together to get rid of denominator.
    – Nosrati
    2 days ago








  • 1




    $$cos^2x+4sin^2xnesin^2x+4cos^2x$$
    – Did
    2 days ago














3












3








3







I used the rule :$$int_{0}^{a}f(x) = int_{0}^{a}f(a-x),dx$$
And got :$$int_{0}^{pi/2}frac{sin^2x}{sin^2x+4cos^2x},dx$$
Then, I added both the question and the above integrand, and I got :$$int_{0}^{pi/2}frac{dx}{5}$$
Solving the above, I finally got the answer as $$frac{pi}{20}$$
But, according to my textbook, the answer is $$frac{pi}{6}$$
Where did I go wrong?






integration definite-integrals trigonometric-integrals







share|cite|improve this question









New contributor




WildPeace007 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











I used the rule :$$int_{0}^{a}f(x) = int_{0}^{a}f(a-x),dx$$
And got :$$int_{0}^{pi/2}frac{sin^2x}{sin^2x+4cos^2x},dx$$
Then, I added both the question and the above integrand, and I got :$$int_{0}^{pi/2}frac{dx}{5}$$
Solving the above, I finally got the answer as $$frac{pi}{20}$$
But, according to my textbook, the answer is $$frac{pi}{6}$$
Where did I go wrong?






integration definite-integrals trigonometric-integrals




integration definite-integrals trigonometric-integrals






share|cite|improve this question









New contributor




WildPeace007 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




WildPeace007 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 2 days ago









Chinnapparaj R

5,3131827




5,3131827






New contributor




WildPeace007 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 2 days ago









WildPeace007WildPeace007

183




183




New contributor




WildPeace007 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





WildPeace007 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






WildPeace007 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 1




    It's not true that $frac{a}{b+c} + frac{d}{e+f} = frac{a+d}{b+c+e+f}$.
    – mathworker21
    2 days ago






  • 1




    Both denominators are different. You can't attach them together to get rid of denominator.
    – Nosrati
    2 days ago








  • 1




    $$cos^2x+4sin^2xnesin^2x+4cos^2x$$
    – Did
    2 days ago














  • 1




    It's not true that $frac{a}{b+c} + frac{d}{e+f} = frac{a+d}{b+c+e+f}$.
    – mathworker21
    2 days ago






  • 1




    Both denominators are different. You can't attach them together to get rid of denominator.
    – Nosrati
    2 days ago








  • 1




    $$cos^2x+4sin^2xnesin^2x+4cos^2x$$
    – Did
    2 days ago








1




1




It's not true that $frac{a}{b+c} + frac{d}{e+f} = frac{a+d}{b+c+e+f}$.
– mathworker21
2 days ago




It's not true that $frac{a}{b+c} + frac{d}{e+f} = frac{a+d}{b+c+e+f}$.
– mathworker21
2 days ago




1




1




Both denominators are different. You can't attach them together to get rid of denominator.
– Nosrati
2 days ago






Both denominators are different. You can't attach them together to get rid of denominator.
– Nosrati
2 days ago






1




1




$$cos^2x+4sin^2xnesin^2x+4cos^2x$$
– Did
2 days ago




$$cos^2x+4sin^2xnesin^2x+4cos^2x$$
– Did
2 days ago










1 Answer
1






active

oldest

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5














So what you have done is apparently wrong. The equality you used is true, the problemes lies elsewhere.



In fact when you add the two terms you found :



$$A =frac{cos^2x}{cos^2x+4sin^2x}$$
$$B =frac{sin^2x}{sin^2x+4cos^2x}$$



$$implies A + B = frac {3 cos(x)^4 + 3 sin(x)^4+1}{(3 sin(x)^2 +1) cdot (3 cos(x)^2 + 1)} $$



Here is an attempt to solve the exercice:



$$I = int_{0}^{pi/2}frac{cos^2x}{cos^2x+4sin^2x}dx = int_{0}^{pi/2}frac{1}{1+4tan^2x}dx $$
we can do this because the interval of integration permits it.
Then we change the variable of integration ; the substitution is the following :



$$ u = tan x$$
we get :



$$I =int_{0}^{infty}frac{1}{(1 +4u^2)(1 + u^2)}du = int_{0}^{infty} frac {4}{3(4u^2 + 1)} - frac {1}{3(u^2+1)} du $$



from there we can easily conclude.



In fact you just have to know this formula :



$$int frac 1 {x^2 + a^2} = frac 1 {|a|} arctan left(frac x a right) + C $$
Where $C$ is the constant of integration.



And thus we get :



$$int_{0}^{infty} frac {4}{3(4u^2 + 1)} du = frac 2 3 arctan(2u)biggrvert_0^{infty} = pi / 3$$
$$int_{0}^{infty} frac {1}{3(u^2+1)} du = frac 1 3 arctan u biggrvert_0^{infty} = pi/6 $$






share|cite|improve this answer



















  • 1




    In fact,$$int frac 1 {x^2 + a^2} = frac 1{|a|} arctan frac x a $$
    – lab bhattacharjee
    2 days ago












  • yeah, you're right, my bad
    – Marine Galantin
    2 days ago






  • 1




    In case anyone's wondering why we don't use the usual substitution $t=tan x/2$, the fact that $sin x,,cos x$ only appear squared in the integrand halves its period.
    – J.G.
    2 days ago











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1 Answer
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1 Answer
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active

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active

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5














So what you have done is apparently wrong. The equality you used is true, the problemes lies elsewhere.



In fact when you add the two terms you found :



$$A =frac{cos^2x}{cos^2x+4sin^2x}$$
$$B =frac{sin^2x}{sin^2x+4cos^2x}$$



$$implies A + B = frac {3 cos(x)^4 + 3 sin(x)^4+1}{(3 sin(x)^2 +1) cdot (3 cos(x)^2 + 1)} $$



Here is an attempt to solve the exercice:



$$I = int_{0}^{pi/2}frac{cos^2x}{cos^2x+4sin^2x}dx = int_{0}^{pi/2}frac{1}{1+4tan^2x}dx $$
we can do this because the interval of integration permits it.
Then we change the variable of integration ; the substitution is the following :



$$ u = tan x$$
we get :



$$I =int_{0}^{infty}frac{1}{(1 +4u^2)(1 + u^2)}du = int_{0}^{infty} frac {4}{3(4u^2 + 1)} - frac {1}{3(u^2+1)} du $$



from there we can easily conclude.



In fact you just have to know this formula :



$$int frac 1 {x^2 + a^2} = frac 1 {|a|} arctan left(frac x a right) + C $$
Where $C$ is the constant of integration.



And thus we get :



$$int_{0}^{infty} frac {4}{3(4u^2 + 1)} du = frac 2 3 arctan(2u)biggrvert_0^{infty} = pi / 3$$
$$int_{0}^{infty} frac {1}{3(u^2+1)} du = frac 1 3 arctan u biggrvert_0^{infty} = pi/6 $$






share|cite|improve this answer



















  • 1




    In fact,$$int frac 1 {x^2 + a^2} = frac 1{|a|} arctan frac x a $$
    – lab bhattacharjee
    2 days ago












  • yeah, you're right, my bad
    – Marine Galantin
    2 days ago






  • 1




    In case anyone's wondering why we don't use the usual substitution $t=tan x/2$, the fact that $sin x,,cos x$ only appear squared in the integrand halves its period.
    – J.G.
    2 days ago
















5














So what you have done is apparently wrong. The equality you used is true, the problemes lies elsewhere.



In fact when you add the two terms you found :



$$A =frac{cos^2x}{cos^2x+4sin^2x}$$
$$B =frac{sin^2x}{sin^2x+4cos^2x}$$



$$implies A + B = frac {3 cos(x)^4 + 3 sin(x)^4+1}{(3 sin(x)^2 +1) cdot (3 cos(x)^2 + 1)} $$



Here is an attempt to solve the exercice:



$$I = int_{0}^{pi/2}frac{cos^2x}{cos^2x+4sin^2x}dx = int_{0}^{pi/2}frac{1}{1+4tan^2x}dx $$
we can do this because the interval of integration permits it.
Then we change the variable of integration ; the substitution is the following :



$$ u = tan x$$
we get :



$$I =int_{0}^{infty}frac{1}{(1 +4u^2)(1 + u^2)}du = int_{0}^{infty} frac {4}{3(4u^2 + 1)} - frac {1}{3(u^2+1)} du $$



from there we can easily conclude.



In fact you just have to know this formula :



$$int frac 1 {x^2 + a^2} = frac 1 {|a|} arctan left(frac x a right) + C $$
Where $C$ is the constant of integration.



And thus we get :



$$int_{0}^{infty} frac {4}{3(4u^2 + 1)} du = frac 2 3 arctan(2u)biggrvert_0^{infty} = pi / 3$$
$$int_{0}^{infty} frac {1}{3(u^2+1)} du = frac 1 3 arctan u biggrvert_0^{infty} = pi/6 $$






share|cite|improve this answer



















  • 1




    In fact,$$int frac 1 {x^2 + a^2} = frac 1{|a|} arctan frac x a $$
    – lab bhattacharjee
    2 days ago












  • yeah, you're right, my bad
    – Marine Galantin
    2 days ago






  • 1




    In case anyone's wondering why we don't use the usual substitution $t=tan x/2$, the fact that $sin x,,cos x$ only appear squared in the integrand halves its period.
    – J.G.
    2 days ago














5












5








5






So what you have done is apparently wrong. The equality you used is true, the problemes lies elsewhere.



In fact when you add the two terms you found :



$$A =frac{cos^2x}{cos^2x+4sin^2x}$$
$$B =frac{sin^2x}{sin^2x+4cos^2x}$$



$$implies A + B = frac {3 cos(x)^4 + 3 sin(x)^4+1}{(3 sin(x)^2 +1) cdot (3 cos(x)^2 + 1)} $$



Here is an attempt to solve the exercice:



$$I = int_{0}^{pi/2}frac{cos^2x}{cos^2x+4sin^2x}dx = int_{0}^{pi/2}frac{1}{1+4tan^2x}dx $$
we can do this because the interval of integration permits it.
Then we change the variable of integration ; the substitution is the following :



$$ u = tan x$$
we get :



$$I =int_{0}^{infty}frac{1}{(1 +4u^2)(1 + u^2)}du = int_{0}^{infty} frac {4}{3(4u^2 + 1)} - frac {1}{3(u^2+1)} du $$



from there we can easily conclude.



In fact you just have to know this formula :



$$int frac 1 {x^2 + a^2} = frac 1 {|a|} arctan left(frac x a right) + C $$
Where $C$ is the constant of integration.



And thus we get :



$$int_{0}^{infty} frac {4}{3(4u^2 + 1)} du = frac 2 3 arctan(2u)biggrvert_0^{infty} = pi / 3$$
$$int_{0}^{infty} frac {1}{3(u^2+1)} du = frac 1 3 arctan u biggrvert_0^{infty} = pi/6 $$






share|cite|improve this answer














So what you have done is apparently wrong. The equality you used is true, the problemes lies elsewhere.



In fact when you add the two terms you found :



$$A =frac{cos^2x}{cos^2x+4sin^2x}$$
$$B =frac{sin^2x}{sin^2x+4cos^2x}$$



$$implies A + B = frac {3 cos(x)^4 + 3 sin(x)^4+1}{(3 sin(x)^2 +1) cdot (3 cos(x)^2 + 1)} $$



Here is an attempt to solve the exercice:



$$I = int_{0}^{pi/2}frac{cos^2x}{cos^2x+4sin^2x}dx = int_{0}^{pi/2}frac{1}{1+4tan^2x}dx $$
we can do this because the interval of integration permits it.
Then we change the variable of integration ; the substitution is the following :



$$ u = tan x$$
we get :



$$I =int_{0}^{infty}frac{1}{(1 +4u^2)(1 + u^2)}du = int_{0}^{infty} frac {4}{3(4u^2 + 1)} - frac {1}{3(u^2+1)} du $$



from there we can easily conclude.



In fact you just have to know this formula :



$$int frac 1 {x^2 + a^2} = frac 1 {|a|} arctan left(frac x a right) + C $$
Where $C$ is the constant of integration.



And thus we get :



$$int_{0}^{infty} frac {4}{3(4u^2 + 1)} du = frac 2 3 arctan(2u)biggrvert_0^{infty} = pi / 3$$
$$int_{0}^{infty} frac {1}{3(u^2+1)} du = frac 1 3 arctan u biggrvert_0^{infty} = pi/6 $$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 3 hours ago









DavidG

1,836620




1,836620










answered 2 days ago









Marine GalantinMarine Galantin

795215




795215








  • 1




    In fact,$$int frac 1 {x^2 + a^2} = frac 1{|a|} arctan frac x a $$
    – lab bhattacharjee
    2 days ago












  • yeah, you're right, my bad
    – Marine Galantin
    2 days ago






  • 1




    In case anyone's wondering why we don't use the usual substitution $t=tan x/2$, the fact that $sin x,,cos x$ only appear squared in the integrand halves its period.
    – J.G.
    2 days ago














  • 1




    In fact,$$int frac 1 {x^2 + a^2} = frac 1{|a|} arctan frac x a $$
    – lab bhattacharjee
    2 days ago












  • yeah, you're right, my bad
    – Marine Galantin
    2 days ago






  • 1




    In case anyone's wondering why we don't use the usual substitution $t=tan x/2$, the fact that $sin x,,cos x$ only appear squared in the integrand halves its period.
    – J.G.
    2 days ago








1




1




In fact,$$int frac 1 {x^2 + a^2} = frac 1{|a|} arctan frac x a $$
– lab bhattacharjee
2 days ago






In fact,$$int frac 1 {x^2 + a^2} = frac 1{|a|} arctan frac x a $$
– lab bhattacharjee
2 days ago














yeah, you're right, my bad
– Marine Galantin
2 days ago




yeah, you're right, my bad
– Marine Galantin
2 days ago




1




1




In case anyone's wondering why we don't use the usual substitution $t=tan x/2$, the fact that $sin x,,cos x$ only appear squared in the integrand halves its period.
– J.G.
2 days ago




In case anyone's wondering why we don't use the usual substitution $t=tan x/2$, the fact that $sin x,,cos x$ only appear squared in the integrand halves its period.
– J.G.
2 days ago










WildPeace007 is a new contributor. Be nice, and check out our Code of Conduct.










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