Q.Evaluate $\displaystyle\int_{0}^{\frac{\pi}{2}}\dfrac{dx}{1+5\cos^2x}$

Converts the integrand to a function of $\tan x$ by dividing through by $\cos^2x$, then substitutes $t=\tan x$ to reduce it to a standard $\displaystyle\int\dfrac{dt}{t^2+a^2}$ form.

1. Given $I=\displaystyle\int_0^{\pi/2}\dfrac{dx}{1+5\cos^2x}$.
2. Divide the numerator and denominator of the integrand by $\cos^2x$ (valid since $\cos x\ne0$ on $[0,\pi/2)$): $\dfrac{1}{1+5\cos^2x}=\dfrac{\sec^2x}{\sec^2x+5}$.
3. Use $\sec^2x=1+\tan^2x$: $\sec^2x+5=1+\tan^2x+5=\tan^2x+6$.
4. So $I=\displaystyle\int_0^{\pi/2}\dfrac{\sec^2x}{\tan^2x+6}\,dx$.
5. Substitute $t=\tan x$, so $dt=\sec^2x\,dx$. Change limits: $x=0\Rightarrow t=\tan0=0$; $x\to\dfrac{\pi}{2}^-\Rightarrow t\to\infty$.
6. So $I=\displaystyle\int_0^{\infty}\dfrac{dt}{t^2+6}$. …