Question

\[\int\limits_{0}^{\pi \div2} dx/(a ^{2}\cos ^{2}x +b ^{2}\sin ^{2}x)\]

Answer 1

interesting question :)

Answer 2

Are there any conditions set on \(a\) and \(b\)? I would assume both are distinct and non-zero, but there's no way to know for sure.

Here's an attempt using the substitution \(t=\tan\dfrac{x}{2}\).
\[t=\tan\frac{x}{2}\\
dt=\frac{1}{2}\sec^2\frac{x}{2}~dx\]
So you have
\[\cos x=\cos2\left(\frac{x}{2}\right)=\cos^2\frac{x}{2}-\sin^2\frac{x}{2}=\frac{1-t^2}{1+t^2}\\
\sin x=\sin2\left(\frac{x}{2}\right)=2\sin \frac{x}{2}\cos\frac{x}{2}=\frac{2t}{1+t^2}\]
Naturally, you have
\[\cos^2x=\frac{\left(1-t^2\right)^2}{\left(1+t^2\right)^2}\\
\sin^2x=\frac{4t^2}{\left(1+t^2\right)^2}\]
Also, note that
\[dt=\frac{1}{2}\sec^2\frac{x}{2}~dx~~\iff~~2\cos^2\frac{x}{2}~dt=dx~~\iff~~dx=\frac{2}{1+t^2}~dt\]
So, the integral may be equivalent to
\[{\Huge\int_0^1}\frac{\dfrac{2}{1+t^2}}{a^2\dfrac{\left(1-t^2\right)^2}{\left(1+t^2\right)^2}+b^2\dfrac{4t^2}{\left(1+t^2\right)^2}}~dt\]
Let's see if we can simplify that integrand:
\[\begin{align*}\frac{\dfrac{2}{1+t^2}}{a^2\dfrac{\left(1-t^2\right)^2}{\left(1+t^2\right)^2}+b^2\dfrac{4t^2}{\left(1+t^2\right)^2}}&=\frac{\dfrac{2}{1+t^2}}{\dfrac{a^2\left(1-t^2\right)^2+4b^2t^2}{\left(1+t^2\right)^2}}\\\\\\
&=\frac{2}{\dfrac{a^2\left(1-t^2\right)^2+4b^2t^2}{1+t^2}}\\\\\\
&=\frac{2\left(1+t^2\right)}{a^2\left(1-t^2\right)^2+4b^2t^2}\\\\\\
&=\frac{2t^2+2}{a^2t^4+\left(4b^2-2a^2\right)t^2+a^2}
\end{align*}\]
Not much luck there... I would suggest partial fraction decomposition or completing the square, but I doubt either are possible/would help much.

Answer 3

I'm probably just thinking too hard about this...