Q.Evaluate the definite integral $\int_{0}^{\pi/2}\frac{\cos^2 x\,dx}{\cos^2 x+4\sin^2 x}$

The reflection trick $x\to\tfrac{\pi}{2}-x$ does not work here (the denominator is not symmetric), so substitute $t=\tan x$.

With $t=\tan x$: $\cos^2x=\dfrac{1}{1+t^2}$, $\sin^2x=\dfrac{t^2}{1+t^2}$, $dx=\dfrac{dt}{1+t^2}$, and the integrand collapses to $\dfrac{1}{1+4t^2}$. Limits: $0\to\infty$.

$I=\int_0^\infty\frac{dt}{(1+t^2)(1+4t^2)}.$ …

Why not the reflection trick?

Replacing $x\to\tfrac{\pi}{2}-x$ gives $\int_0^{\pi/2}\frac{\sin^2x}{\sin^2x+4\cos^2x}\,dx$. It is tempting to add this to $I$ and cancel the numerators, but the two denominators, $\cos^2x+4\sin^2x$ and $\sin^2x+4\cos^2x$, are different, so the integrands cannot be combined over a common denominator. That route is invalid here; a direct substitution is the honest path.

Substitute $t=\tan x$

Divide numerator and denominator by $\cos^2 x$:

With $t=\tan x$, $dt=\sec^2x\,dx=(1+t^2)\,dx$, so $dx=\dfrac{dt}{1+t^2}$, and as $x$ runs $0\to\tfrac{\pi}{2}$, $t$ runs $0\to\infty$:

Partial fractions …

Method: $t=\tan x$ substitution turning a $\sin/\cos$ ratio into a rational integral

Use this when the reflection ($x\to\tfrac\pi2-x$) trick fails because the two denominators differ — divide by $\cos^2x$ and substitute $t=\tan x$ honestly.

Steps

Step 1: Check the reflection trick — and abandon it if denominators differ.

Adding $I(x)$ and $I(\tfrac\pi2-x)$ only helps if the integrands share a denominator; here $\cos^2x+4\sin^2x$ and its reflection are different, so combining them is invalid.

Step 2: Divide by $\cos^2x$ and substitute. …

Common Mistakes

Mistake 1: Wrongly adding the reflected integral to cancel numerators.

Why it's wrong: the denominators $\cos^2x+4\sin^2x$ and $\sin^2x+4\cos^2x$ differ, so the two integrands cannot be combined; the reflection trick is invalid here. Correct approach: substitute $t=\tan x$ directly.

Mistake 2: Forgetting $dx=\dfrac{dt}{1+t^2}$.

Why it's wrong: with $t=\tan x$, $dt=\sec^2x\,dx=(1+t^2)\,dx$; omitting this drops a whole factor. Correct approach: replace $dx$ properly. …