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$$\int_0^π \frac{xdx}{a^2\cos^2x+b^2\sin^2x} \,dx$$

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, $2I=\pi\int_0^\pi \frac{dx}{a^2\cos^2x+b^2\sin^2x} \,dx$

On dividing numerator and denominator of R.H.S by $\cos^2x$ I get, $2I=\pi\int_0^\pi \frac{\sec^2xdx}{a^2+b^2\tan^2x} \,dx$

Now, solving by substitution method (taking $b\tan x=t$)

I get enter image description here

(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 $2I=\pi\int_0^\pi \frac{dx}{a^2\cos^2x+b^2\sin^2x} \,dx$ )they have used the property

$$\int_0^2a f(x) \,dx= 2\left(\int_0^a f(x) \,dx\right)$$

Why they didn't ended the solution in the direction in which i did

pardon for my mathjax errors

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    $\begingroup$ HINT: Is the integral continuous over the bounds of the integral? You may have an improper integral. $\endgroup$ – user150203 Jan 21 at 13:09
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    $\begingroup$ 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
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    $\begingroup$ @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
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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.

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    $\begingroup$ I got some idea $\endgroup$ – Aashish Jan 21 at 13:26
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    $\begingroup$ 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
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    $\begingroup$ Here tutorial.math.lamar.edu/Classes/CalcI/… $\endgroup$ – Lalla95 Jan 21 at 14:12

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