Definite Integration ( a little query)

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

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

• HINT: Is the integral continuous over the bounds of the integral? You may have an improper integral. – user150203 Jan 21 at 13:09
• 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. – Shubham Johri Jan 21 at 13:22
• @DavidG please explain what you are saying.. How is that integral isn't continous over the bounds of the integral? – Aashish Jan 21 at 13:29
• Nothing is said about $a,b$.That's a problem if one is zero. – FDP Jan 21 at 13:53
• nothing is said in the original question about a,b – Aashish Jan 21 at 13:56