Q.By using the properties of definite integrals, evaluate the integral $\int_{0}^{\pi/2}\frac{\cos^5 x}{\sin^5 x+\cos^5 x}\,dx$

Concept: Definite Integral Symmetry — when the integrand has the form $\frac{f(x)}{f(x)+f(a-x)}$, the integral over $[0,a]$ equals $\frac{a}{2}$.

Let $I = \int_{0}^{\pi/2}\frac{\cos^5 x}{\sin^5 x+\cos^5 x}\,dx$.

Step 1: Use the property $\int_{0}^{a} f(x)\,dx = \int_{0}^{a} f(a-x)\,dx$.

Replace $x$ by $\frac{\pi}{2} - x$:

$I = \int_{0}^{\pi/2}\frac{\cos^5(\frac{\pi}{2}-x)}{\sin^5(\frac{\pi}{2}-x)+\cos^5(\frac{\pi}{2}-x)}\,dx = \int_{0}^{\pi/2}\frac{\sin^5 x}{\cos^5 x+\sin^5 x}\,dx.$ …

The key insight here is that the integrand has a special structure when you replace $x$ by $\frac{\pi}{2} - x$. This is a classic trick for integrals over $[0, \pi/2]$ where the numerator and denominator are symmetric powers of sine and cosine.

Why this works:

For any function $f(x)$ that is continuous on $[0, a]$, we have the property

Here $a = \pi/2$, and the integrand is $\frac{\cos^5 x}{\sin^5 x + \cos^5 x}$. When we replace $x$ by $\frac{\pi}{2} - x$, $\cos x$ becomes $\sin x$ and $\sin x$ becomes $\cos x$, so the integrand flips to $\frac{\sin^5 x}{\cos^5 x + \sin^5 x}$. That is exactly the "complement" of the original fraction — their sum is simply $1$. This lets us add the two forms and solve for the integral.

Let’s work through it step by step.

1. Define the integral Let

2. Apply the substitution $x \to \frac{\pi}{2} - x$ Using the property $\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx$, set $t = \frac{\pi}{2} - x$. Then $dx = -dt$, and when $x=0$, $t=\pi/2$; when $x=\pi/2$, $t=0$. The integral becomes

Reversing the limits removes the minus sign:

3. Simplify the trigonometric expressions Recall: $\cos(\frac{\pi}{2} - t) = \sin t$ and $\sin(\frac{\pi}{2} - t) = \cos t$. So the integrand becomes

Since $t$ is a dummy variable, we can rename it back to $x$:

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

Adding them: …

Method: The $\dfrac{f}{f+g}$ complementary-integral trick

For $\displaystyle\int_0^a \frac{f(x)}{f(x)+f(a-x)}\,dx$ (with $f,g$ swapping under $x\to a-x$), reflection produces the complementary fraction; the two add to $1$.

Steps

Step 1: Set $I$ and reflect with $x\to a-x$.

where the reflection swaps $f\leftrightarrow g$ (e.g. $\sqrt{\sin}\leftrightarrow\sqrt{\cos}$).

Step 2: Add the two forms. …

Common Mistakes

Mistake 1: Attempting a $t=\tan x$ or half-angle substitution to integrate directly.

Why it's wrong: $\frac{\cos^5x}{\sin^5x+\cos^5x}$ leads to an intractable rational function; the reflection property is the intended, far shorter route. Correct approach: use $\int_0^a f(x)=\int_0^a f(a-x)$.

Mistake 2: Not simplifying the added fractions to $1$. …