# Integral $\int_0^\pi \frac{x\,\operatorname dx}{a^2\cos^2x+b^2\sin^2x}$

Integrate: $$\int_0^\pi \frac{x\,\operatorname dx}{a^2\cos^2x+b^2\sin^2x}$$

## 3 Answers

$$I=\int_0^\pi \frac{x\,\operatorname dx}{a^2\cos^2x+b^2\sin^2x}=\int_0^\pi \frac{(\pi-x)\,\operatorname dx}{a^2\cos^2x+b^2\sin^2x}$$ $$2I=\pi\int_0^\pi \frac{\operatorname dx}{a^2\cos^2x+b^2\sin^2x}=\pi\int_0^\pi \frac{\sec^2x\,\operatorname dx}{a^2+b^2\tan^2x}$$ $$\frac{2I}{\pi}=\int_0^{\pi}\frac{\operatorname d(\tan x)}{a^2+b^2\tan^2x}=\int_0^{\pi/2}\frac{\operatorname d(\tan x)}{a^2+b^2\tan^2x}+\int_{\pi/2}^{\pi}\frac{\operatorname d(\tan x)}{a^2+b^2\tan^2x}$$ $$\frac{2I}{\pi}=\frac1{ab}\arctan\left(\frac{b\tan x}{a}\right)|_0^{\pi/2}+|_{\pi/2}^{\pi}$$ $$I=\ldots\ldots$$

• It's btanx/a. You take the a^2 common so u is btax/a. – deepthought Sep 22 '14 at 8:50

HINT:

Use $$\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$$

Then divide by $\sec^2x$ and set $\tan x=u$

Hint:

We can find the anti-derivative,

\begin{align} \int\frac{\mathrm{d}x}{a^2\cos^2{x}+b^2\sin^2{x}} &=\int\frac{\sec^2{x}\,\mathrm{d}x}{a^2+b^2\tan^2{x}}\\ &=\int\frac{\mathrm{d}u}{a^2+b^2u^2}\\ &=\frac{\arctan{\frac{bu}{a}}}{ab}+\color{grey}{constant}\\ &=\frac{\arctan{\left(\frac{b\tan{x}}{a}\right)}}{ab}+\color{grey}{constant}. \end{align}

Now integrate by parts.

• You missed the x on the numerator of the integrand. It gets tougher with it. – Vinícius Novelli Sep 22 '14 at 7:30
• @user148344 I did not miss the $x$ in the numerator. I removed $x$ from the numerator because the leftover part is easier to integrate. That's why my hint said the next step was to integrate by parts. – David H Sep 22 '14 at 10:35