definite integral of $1/(a^2cos^2x+b^2sin^2x)$

Question

I am trying to find the definite integral of $\frac{1}{a^2cos^2x+b^2sin^2x}$ over $0$ to $2\pi$.

I used the substitution $tan\ x=t$ but I am getting answer in terms of $tan^{-1}$, and that function over the given limits will be zero. But when I plotted the function to check for area, it is some positive value. I need help with such situations where substitutions and limits give zero answer.

Use https://math.stackexchange.com/questions/2523420/show-that-int-0-pi-f-sin-x-mathrmdx-2-int-0-pi-2f-sin-x-math/2523449#2523449 twice — lab bhattacharjee, Jan 30 '18 at 09:42
thank you, but i am more bothered with doing integration and then applying limits. I should get still the right answer. Where am I going wrong — jnyan, Jan 30 '18 at 09:46
Kindly provide the details of your computation and we'll show you where you went wrong. — uniquesolution, Jan 30 '18 at 09:51
i got it now. The limits must be split because of the nature of the substitution. — jnyan, Jan 30 '18 at 10:00

score 3 · Accepted Answer · 2018-01-30T10:45:10.327

Using your substitution, the indefinite integral seems to be

$$\frac1{ab}\arctan\left(\frac{b\tan x}a\right).$$

But this function has singularities at $x=\dfrac{2k+1}2\pi$, which are in fact jumps by $-\pi$. You can restore the true antiderivative by compensating the discontinuities, using

$$\frac1{ab}\left(\arctan\left(\frac{b\tan x}a\right)+\pi\left\lfloor\frac x\pi+\frac12\right\rfloor\right)$$ and finally the integral from $0$ to $2\pi$ is

$$\frac{2\pi}{ab}.$$

definite integral of $1/(a^2cos^2x+b^2sin^2x)$

1 Answers1