Mixing
Definite Integration ( a little query)
$begingroup$
$$int_0^π frac{xdx}{a^2cos^2x+b^2sin^2x} ,dx$$
Using property
$$int_a^b f(x) ,dx= int_a^b f(a+b-x) ,dx$$
(i can't write it correctly,please check it)
I get, $2I=piint_0^pi frac{dx}{a^2cos^2x+b^2sin^2x} ,dx$
On dividing numerator and denominator of R.H.S by $cos^2x$
I get, $2I=piint_0^pi frac{sec^2xdx}{a^2+b^2tan^2x} ,dx$
Now, solving by substitution method (taking $btan x=t$)
I get
(i have added the image because i was not able to type this correctly)
As the upper limit and lower limit on the function are zero
So, answer should be zero.
But in the solution ( after getting this $2I=piint_0^pi frac{dx}{a^2cos^2x+b^2sin^2x} ,dx$ )they have used the property
$$int_0^2a f(x) ,dx= 2left(int_0^a f(x) ,dxright)$$
Why they didn't ended the solution in the direction in which i did
pardon for my mathjax errors
calculus integration definite-integrals
edited Jan 21 at 12:57
mrtaurho
5,73551540
asked Jan 21 at 12:51
AashishAashish
688
$endgroup$
add a comment |
$begingroup$
$$int_0^π frac{xdx}{a^2cos^2x+b^2sin^2x} ,dx$$
Using property
$$int_a^b f(x) ,dx= int_a^b f(a+b-x) ,dx$$
(i can't write it correctly,please check it)
I get, $2I=piint_0^pi frac{dx}{a^2cos^2x+b^2sin^2x} ,dx$
On dividing numerator and denominator of R.H.S by $cos^2x$
I get, $2I=piint_0^pi frac{sec^2xdx}{a^2+b^2tan^2x} ,dx$
Now, solving by substitution method (taking $btan x=t$)
I get
(i have added the image because i was not able to type this correctly)
As the upper limit and lower limit on the function are zero
So, answer should be zero.
But in the solution ( after getting this $2I=piint_0^pi frac{dx}{a^2cos^2x+b^2sin^2x} ,dx$ )they have used the property
$$int_0^2a f(x) ,dx= 2left(int_0^a f(x) ,dxright)$$
Why they didn't ended the solution in the direction in which i did
pardon for my mathjax errors
calculus integration definite-integrals
edited Jan 21 at 12:57
mrtaurho
5,73551540
asked Jan 21 at 12:51
AashishAashish
688
$endgroup$
5
$begingroup$
HINT: Is the integral continuous over the bounds of the integral? You may have an improper integral.
$endgroup$
– DavidG
Jan 21 at 13:09
2
$begingroup$
One easy way to see why the answer should not be zero is to observe that the integrand is positive and the area under it can't be zero.
$endgroup$
– Shubham Johri
Jan 21 at 13:22
1
$begingroup$
@DavidG please explain what you are saying.. How is that integral isn't continous over the bounds of the integral?
$endgroup$
– Aashish
Jan 21 at 13:29
$begingroup$
Nothing is said about $a,b$.That's a problem if one is zero.
$endgroup$
– FDP
Jan 21 at 13:53
$begingroup$
nothing is said in the original question about a,b
$endgroup$
– Aashish
Jan 21 at 13:56
add a comment |
$begingroup$
$$int_0^π frac{xdx}{a^2cos^2x+b^2sin^2x} ,dx$$
Using property
$$int_a^b f(x) ,dx= int_a^b f(a+b-x) ,dx$$
(i can't write it correctly,please check it)
I get, $2I=piint_0^pi frac{dx}{a^2cos^2x+b^2sin^2x} ,dx$
On dividing numerator and denominator of R.H.S by $cos^2x$
I get, $2I=piint_0^pi frac{sec^2xdx}{a^2+b^2tan^2x} ,dx$
Now, solving by substitution method (taking $btan x=t$)
I get
(i have added the image because i was not able to type this correctly)
As the upper limit and lower limit on the function are zero
So, answer should be zero.
But in the solution ( after getting this $2I=piint_0^pi frac{dx}{a^2cos^2x+b^2sin^2x} ,dx$ )they have used the property
$$int_0^2a f(x) ,dx= 2left(int_0^a f(x) ,dxright)$$
Why they didn't ended the solution in the direction in which i did
pardon for my mathjax errors
calculus integration definite-integrals
edited Jan 21 at 12:57
mrtaurho
5,73551540
asked Jan 21 at 12:51
AashishAashish
688
$endgroup$
$$int_0^π frac{xdx}{a^2cos^2x+b^2sin^2x} ,dx$$
Using property
$$int_a^b f(x) ,dx= int_a^b f(a+b-x) ,dx$$
(i can't write it correctly,please check it)
I get, $2I=piint_0^pi frac{dx}{a^2cos^2x+b^2sin^2x} ,dx$
On dividing numerator and denominator of R.H.S by $cos^2x$
I get, $2I=piint_0^pi frac{sec^2xdx}{a^2+b^2tan^2x} ,dx$
Now, solving by substitution method (taking $btan x=t$)
I get
(i have added the image because i was not able to type this correctly)
As the upper limit and lower limit on the function are zero
So, answer should be zero.
But in the solution ( after getting this $2I=piint_0^pi frac{dx}{a^2cos^2x+b^2sin^2x} ,dx$ )they have used the property
$$int_0^2a f(x) ,dx= 2left(int_0^a f(x) ,dxright)$$
Why they didn't ended the solution in the direction in which i did
pardon for my mathjax errors
calculus integration definite-integrals
calculus integration definite-integrals
edited Jan 21 at 12:57
mrtaurho
5,73551540
asked Jan 21 at 12:51
AashishAashish
688
edited Jan 21 at 12:57
mrtaurho
5,73551540
asked Jan 21 at 12:51
AashishAashish
688
edited Jan 21 at 12:57
mrtaurho
5,73551540
edited Jan 21 at 12:57
mrtaurho
5,73551540
edited Jan 21 at 12:57
mrtaurho
5,73551540
5,73551540
asked Jan 21 at 12:51
AashishAashish
688
asked Jan 21 at 12:51
AashishAashish
688
asked Jan 21 at 12:51
AashishAashish
688
688
5
$begingroup$
HINT: Is the integral continuous over the bounds of the integral? You may have an improper integral.
$endgroup$
– DavidG
Jan 21 at 13:09
2
$begingroup$
One easy way to see why the answer should not be zero is to observe that the integrand is positive and the area under it can't be zero.
$endgroup$
– Shubham Johri
Jan 21 at 13:22
1
$begingroup$
@DavidG please explain what you are saying.. How is that integral isn't continous over the bounds of the integral?
$endgroup$
– Aashish
Jan 21 at 13:29
$begingroup$
Nothing is said about $a,b$.That's a problem if one is zero.
$endgroup$
– FDP
Jan 21 at 13:53
$begingroup$
nothing is said in the original question about a,b
$endgroup$
– Aashish
Jan 21 at 13:56
add a comment |
5
$begingroup$
HINT: Is the integral continuous over the bounds of the integral? You may have an improper integral.
$endgroup$
– DavidG
Jan 21 at 13:09
2
$begingroup$
One easy way to see why the answer should not be zero is to observe that the integrand is positive and the area under it can't be zero.
$endgroup$
– Shubham Johri
Jan 21 at 13:22
1
$begingroup$
@DavidG please explain what you are saying.. How is that integral isn't continous over the bounds of the integral?
$endgroup$
– Aashish
Jan 21 at 13:29
$begingroup$
Nothing is said about $a,b$.That's a problem if one is zero.
$endgroup$
– FDP
Jan 21 at 13:53
$begingroup$
nothing is said in the original question about a,b
$endgroup$
– Aashish
Jan 21 at 13:56
5
5
$begingroup$
HINT: Is the integral continuous over the bounds of the integral? You may have an improper integral.
$endgroup$
– DavidG
Jan 21 at 13:09
$begingroup$
HINT: Is the integral continuous over the bounds of the integral? You may have an improper integral.
$endgroup$
– DavidG
Jan 21 at 13:09
2
2
$begingroup$
One easy way to see why the answer should not be zero is to observe that the integrand is positive and the area under it can't be zero.
$endgroup$
– Shubham Johri
Jan 21 at 13:22
$begingroup$
One easy way to see why the answer should not be zero is to observe that the integrand is positive and the area under it can't be zero.
$endgroup$
– Shubham Johri
Jan 21 at 13:22
1
1
$begingroup$
@DavidG please explain what you are saying.. How is that integral isn't continous over the bounds of the integral?
$endgroup$
– Aashish
Jan 21 at 13:29
$begingroup$
@DavidG please explain what you are saying.. How is that integral isn't continous over the bounds of the integral?
$endgroup$
– Aashish
Jan 21 at 13:29
$begingroup$
Nothing is said about $a,b$.That's a problem if one is zero.
$endgroup$
– FDP
Jan 21 at 13:53
$begingroup$
Nothing is said about $a,b$.That's a problem if one is zero.
$endgroup$
– FDP
Jan 21 at 13:53
$begingroup$
nothing is said in the original question about a,b
$endgroup$
– Aashish
Jan 21 at 13:56
$begingroup$
nothing is said in the original question about a,b
$endgroup$
– Aashish
Jan 21 at 13:56
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
When you substitute something that has to be increasing or decreasing throughout the interval and continuous also(otherwise you have to break the integra)l.
Here you have taken tan(x) which changes on π/2.
answered Jan 21 at 13:09
Lalla95Lalla95
564
$endgroup$
1
$begingroup$
I got some idea
$endgroup$
– Aashish
Jan 21 at 13:26
1
$begingroup$
I wanna know more about this because things are still not clear...please give me some source reference
$endgroup$
– Aashish
Jan 21 at 13:27
1
$begingroup$
Here tutorial.math.lamar.edu/Classes/CalcI/…
$endgroup$
– Lalla95
Jan 21 at 14:12
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
When you substitute something that has to be increasing or decreasing throughout the interval and continuous also(otherwise you have to break the integra)l.
Here you have taken tan(x) which changes on π/2.
answered Jan 21 at 13:09
Lalla95Lalla95
564
$endgroup$
1
$begingroup$
I got some idea
$endgroup$
– Aashish
Jan 21 at 13:26
1
$begingroup$
I wanna know more about this because things are still not clear...please give me some source reference
$endgroup$
– Aashish
Jan 21 at 13:27
1
$begingroup$
Here tutorial.math.lamar.edu/Classes/CalcI/…
$endgroup$
– Lalla95
Jan 21 at 14:12
add a comment |
$begingroup$
When you substitute something that has to be increasing or decreasing throughout the interval and continuous also(otherwise you have to break the integra)l.
Here you have taken tan(x) which changes on π/2.
answered Jan 21 at 13:09
Lalla95Lalla95
564
$endgroup$
1
$begingroup$
I got some idea
$endgroup$
– Aashish
Jan 21 at 13:26
1
$begingroup$
I wanna know more about this because things are still not clear...please give me some source reference
$endgroup$
– Aashish
Jan 21 at 13:27
1
$begingroup$
Here tutorial.math.lamar.edu/Classes/CalcI/…
$endgroup$
– Lalla95
Jan 21 at 14:12
add a comment |
$begingroup$
When you substitute something that has to be increasing or decreasing throughout the interval and continuous also(otherwise you have to break the integra)l.
Here you have taken tan(x) which changes on π/2.
answered Jan 21 at 13:09
Lalla95Lalla95
564
$endgroup$
When you substitute something that has to be increasing or decreasing throughout the interval and continuous also(otherwise you have to break the integra)l.
Here you have taken tan(x) which changes on π/2.
answered Jan 21 at 13:09
Lalla95Lalla95
564
answered Jan 21 at 13:09
Lalla95Lalla95
564
answered Jan 21 at 13:09
Lalla95Lalla95
564
answered Jan 21 at 13:09
Lalla95Lalla95
564
564
1
$begingroup$
I got some idea
$endgroup$
– Aashish
Jan 21 at 13:26
1
$begingroup$
I wanna know more about this because things are still not clear...please give me some source reference
$endgroup$
– Aashish
Jan 21 at 13:27
1
$begingroup$
Here tutorial.math.lamar.edu/Classes/CalcI/…
$endgroup$
– Lalla95
Jan 21 at 14:12
add a comment |
1
$begingroup$
I got some idea
$endgroup$
– Aashish
Jan 21 at 13:26
1
$begingroup$
I wanna know more about this because things are still not clear...please give me some source reference
$endgroup$
– Aashish
Jan 21 at 13:27
1
$begingroup$
Here tutorial.math.lamar.edu/Classes/CalcI/…
$endgroup$
– Lalla95
Jan 21 at 14:12
1
1
$begingroup$
I got some idea
$endgroup$
– Aashish
Jan 21 at 13:26
$begingroup$
I got some idea
$endgroup$
– Aashish
Jan 21 at 13:26
1
1
$begingroup$
I wanna know more about this because things are still not clear...please give me some source reference
$endgroup$
– Aashish
Jan 21 at 13:27
$begingroup$
I wanna know more about this because things are still not clear...please give me some source reference
$endgroup$
– Aashish
Jan 21 at 13:27
1
1
$begingroup$
Here tutorial.math.lamar.edu/Classes/CalcI/…
$endgroup$
– Lalla95
Jan 21 at 14:12
$begingroup$
Here tutorial.math.lamar.edu/Classes/CalcI/…
$endgroup$
– Lalla95
Jan 21 at 14:12
add a comment |
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5
$begingroup$
HINT: Is the integral continuous over the bounds of the integral? You may have an improper integral.
$endgroup$
– DavidG
Jan 21 at 13:09
2
$begingroup$
One easy way to see why the answer should not be zero is to observe that the integrand is positive and the area under it can't be zero.
$endgroup$
– Shubham Johri
Jan 21 at 13:22
1
$begingroup$
@DavidG please explain what you are saying.. How is that integral isn't continous over the bounds of the integral?
$endgroup$
– Aashish
Jan 21 at 13:29
$begingroup$
Nothing is said about $a,b$.That's a problem if one is zero.
$endgroup$
– FDP
Jan 21 at 13:53
$begingroup$
nothing is said in the original question about a,b
$endgroup$
– Aashish
Jan 21 at 13:56