Question

$\int_{\pi/6}^{\pi/3}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx$ is equal to

• A. 0
• B. $\frac{\pi}{6}$
• C. $\frac{\pi}{3}$
• D. $\frac{\pi}{12}$
• E. $\frac{\pi}{2}$

Hint:

Math Tip: Any definite integral of the form $\int_a^b \frac{f(x)}{f(x) + f(a+b-x)} dx$ always evaluates to exactly $\frac{b - a}{2}$. You can use this shortcut to instantly find the answer without writing a single line of working! Here, $\frac{\pi/3 - \pi/6}{2} = \frac{\pi/6}{2} = \frac{\pi}{12}$.

Answer 1

The Correct Option is D

Concept:
Calculus - Definite Integration Properties.
King's Rule: $\int_a^b f(x) dx = \int_a^b f(a + b - x) dx$.
Step 1: Set up the initial equation.
Let the given integral be equal to $I$: $$ I = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx \quad \text{--- (Equation 1)} $$
Step 2: Apply the definite integral property.
Use the property $\int_a^b f(x) dx = \int_a^b f(a + b - x) dx$. Calculate the sum of the limits: $a + b = \frac{\pi}{6} + \frac{\pi}{3} = \frac{3\pi}{6} = \frac{\pi}{2}$. Substitute $x$ with $(\frac{\pi}{2} - x)$ in the integral: $$ I = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\sin(\frac{\pi}{2} - x)}}{\sqrt{\sin(\frac{\pi}{2} - x)} + \sqrt{\cos(\frac{\pi}{2} - x)}} dx $$
Step 3: Simplify using complementary angle identities.
Recall that $\sin(\frac{\pi}{2} - x) = \cos x$ and $\cos(\frac{\pi}{2} - x) = \sin x$: $$ I = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx \quad \text{--- (Equation 2)} $$
Step 4: Add Equation 1 and Equation 2.
Adding the left sides and the right sides of both equations: $$ I + I = \int_{\pi/6}^{\pi/3} \left( \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} + \frac{\sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \right) dx $$ Notice that the denominators are identical, allowing us to combine the numerators directly: $$ 2I = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx $$ $$ 2I = \int_{\pi/6}^{\pi/3} 1 \,dx $$
Step 5: Evaluate the simple integral and solve for I.
$$ 2I = [x]_{\pi/6}^{\pi/3} $$ $$ 2I = \frac{\pi}{3} - \frac{\pi}{6} $$ $$ 2I = \frac{2\pi}{6} - \frac{\pi}{6} = \frac{\pi}{6} $$ Divide by 2 to isolate $I$: $$ I = \frac{\pi}{12} $$