TITLE: Definite Integration ( a little query)
QUESTION [5 upvotes]: $$\int_0^π \frac{xdx}{a^2\cos^2x+b^2\sin^2x} \,dx$$
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 

(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

REPLY [2 votes]: 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.