Question

Evaluate :

\[ \int_{0}^{\pi/2} \frac{\sin^{2} x}{1+\sin x \cos x}\, dx \]

Hint:

Use the king property to add the integral to its mirror image so the numerator becomes 1, then substitute t = tan x.

Answer 1

Step 1: Name the integral and use the king property:
Let \( I = \int_{0}^{\pi/2} \frac{\sin^{2} x}{1+\sin x \cos x}\, dx \).
The rule \( \int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx \) lets us replace \( x \) by \( \frac{\pi}{2}-x \).
Then \( \sin x \) becomes \( \cos x \) and \( \cos x \) becomes \( \sin x \).
\[ I = \int_{0}^{\pi/2} \frac{\cos^{2} x}{1+\sin x \cos x}\, dx \]

Step 2: Add the two forms of I:
Adding the two expressions for \( I \) gives \( \sin^2 x + \cos^2 x = 1 \) in the numerator.
\[ 2I = \int_{0}^{\pi/2} \frac{dx}{1+\sin x \cos x} \]

Step 3: Divide the top and bottom by cos^2 x:
Divide numerator and denominator by \( \cos^2 x \).
The numerator becomes \( \sec^2 x \).
The denominator becomes \( \sec^2 x + \tan x = 1 + \tan^2 x + \tan x \).
\[ 2I = \int_{0}^{\pi/2} \frac{\sec^{2} x}{\tan^{2} x + \tan x + 1}\, dx \]

Step 4: Substitute t = tan x:
Put \( t = \tan x \), so \( dt = \sec^2 x\, dx \).
When \( x = 0 \), \( t = 0 \). When \( x \to \frac{\pi}{2} \), \( t \to \infty \).
\[ 2I = \int_{0}^{\infty} \frac{dt}{t^{2}+t+1} = \int_{0}^{\infty} \frac{dt}{\left(t+\frac{1}{2}\right)^{2}+\frac{3}{4}} \]

Step 5: Integrate and put in the limits:
Use \( \int \frac{du}{u^2+a^2} = \frac{1}{a}\tan^{-1}\frac{u}{a} \) with \( a = \frac{\sqrt{3}}{2} \).
\[ 2I = \frac{2}{\sqrt{3}} \left[ \tan^{-1}\frac{2t+1}{\sqrt{3}} \right]_{0}^{\infty} = \frac{2}{\sqrt{3}} \left( \frac{\pi}{2} - \frac{\pi}{6} \right) = \frac{2}{\sqrt{3}} \cdot \frac{\pi}{3} = \frac{2\pi}{3\sqrt{3}} \]

Final Answer:
Dividing by 2 gives the value of the integral (a numerical check gives 0.6046).
\[ \boxed{I = \frac{\pi}{3\sqrt{3}} = \frac{\pi\sqrt{3}}{9}} \]