# 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, 2019 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. Jan 21, 2019 at 13:22
• @DavidG please explain what you are saying.. How is that integral isn't continous over the bounds of the integral? Jan 21, 2019 at 13:29
• Nothing is said about $a,b$.That's a problem if one is zero.
– FDP
Jan 21, 2019 at 13:53
• nothing is said in the original question about a,b Jan 21, 2019 at 13:56