Q.Evaluate $\int_0^{\pi/2} \dfrac{\sin^4 x}{\sin^4 x + \cos^4 x}\, dx$

The key idea is the symmetry property of definite integrals: $\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx$.

Step 1: Let $I = \int_0^{\pi/2} \frac{\sin^4 x}{\sin^4 x + \cos^4 x}\, dx$.

Step 2: Replace $x$ with $\frac{\pi}{2} - x$. Since $\sin(\frac{\pi}{2} - x) = \cos x$ and $\cos(\frac{\pi}{2} - x) = \sin x$, we get:

$I = \int_0^{\pi/2} \frac{\cos^4 x}{\cos^4 x + \sin^4 x}\, dx.$ …

The key insight here is a symmetry trick that works beautifully for integrals over $[0, \pi/2]$ when the integrand involves $\sin$ and $\cos$ in a balanced way. Instead of grinding through trigonometric identities, we can exploit the fact that $\sin x$ and $\cos x$ swap roles when we replace $x$ by $\pi/2 - x$.

Let’s see why this works.

1. Define the integral and apply the substitution $x \to \frac{\pi}{2} - x$. Let

Now make the substitution $t = \frac{\pi}{2} - x$. Then $dx = -dt$, and when $x = 0$, $t = \pi/2$; when $x = \pi/2$, $t = 0$. So

Since $\sin(\pi/2 - t) = \cos t$ and $\cos(\pi/2 - t) = \sin t$, the denominator is symmetric. Renaming $t$ back to $x$, we get

2. Add the two forms of $I$. We now have two expressions for the same $I$:

Adding them:

3. Evaluate the simple integral. …

Method: The "King" Property $\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx$ for $\sin$/$\cos$ Swaps

Use this for integrals over $[0,\tfrac\pi2]$ where replacing $x$ by $\tfrac\pi2-x$ swaps $\sin$ and $\cos$: adding the original and reflected integrals collapses the denominator.

Steps

Step 1: Form the reflected integral.

Let $I=\int_0^{\pi/2}\frac{\sin^4x}{\sin^4x+\cos^4x}\,dx$. Substituting $x\to\frac\pi2-x$ swaps $\sin\leftrightarrow\cos$, giving $I=\int_0^{\pi/2}\frac{\cos^4x}{\cos^4x+\sin^4x}\,dx$. …

Common Mistakes

Mistake 1: Expanding $\sin^4x+\cos^4x$ and grinding through identities.

Why it's wrong: it is far longer and error-prone when the symmetry trick gives the answer in two lines. Correct approach: use $x\to\frac\pi2-x$.

Mistake 2: Not noticing the denominators match after reflection. …