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

Concept: King Property — For $\int_a^b f(x)\,dx$, substitute $x = a+b-t$; the limits swap and the integral remains unchanged.

Let

Apply the King Property with $a=0$, $b=\pi/2$: replace $x$ by $\frac{\pi}{2} - t$. Then $\sin x = \cos t$, $\cos x = \sin t$, and $dx = -dt$. The limits become $t=\pi/2$ to $t=0$, so

Since the variable is dummy, rename $t$ back to $x$:

Add the two expressions for $I$:

Thus $I = \frac{\pi}{4}$.

The trick here is that the integrand is symmetric in a very particular way — swapping $\sin x$ and $\cos x$ by the substitution $x \to \frac{\pi}{2} - x$ turns the denominator into itself but swaps the numerator. This is the classic setup for the King Property.

Let’s apply it.

1. Define the integral. Let

2. Apply the King Property with $a=0$, $b=\pi/2$. Replace $x$ by $0 + \frac{\pi}{2} - x = \frac{\pi}{2} - x$. Then $dx$ becomes $-dx$, but flipping the limits back gives:

3. Simplify the trig functions. Recall: $\sin(\frac{\pi}{2} - x) = \cos x$ and $\cos(\frac{\pi}{2} - x) = \sin x$. So:

Notice the denominator is the same as before (just the terms swapped), but the numerator is now $\sqrt{\cos x}$ instead of $\sqrt{\sin x}$.

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

and

Adding them:

5. Combine the fractions. The denominators are identical (addition is commutative), so:

6. Evaluate the simple integral.

Hence:

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.

The denominators are identical, so

Step 3: Solve for $I$.

This pattern always gives half the interval length, independent of the specific $f$.

Common Mistakes

Mistake 1: Trying to find an antiderivative of $\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}$ directly.

Why it's wrong: this integrand has no elementary antiderivative; only the reflection trick resolves it. Correct approach: reflect $x\to\tfrac{\pi}{2}-x$ and add.

Mistake 2: Not realising the two denominators are identical after reflection.

Why it's wrong: $\sqrt{\sin x}+\sqrt{\cos x}$ and $\sqrt{\cos x}+\sqrt{\sin x}$ are the same, so the fractions add to $1$; treating them as different blocks the cancellation. Correct approach: add to get $2I=\int_0^{\pi/2}1\,dx=\tfrac{\pi}{2}$, so $I=\tfrac{\pi}{4}$.