Problem Statement
સાબિત કરો (Prove that): $$\int_{0}^{\pi/2} \frac{\sin^2 x}{\sin x + \cos x} \, dx = \frac{1}{\sqrt{2}} \log(\sqrt{2} + 1)$$
Verified Solution & Marking Scheme
Apply definite integral property ∫ f(x)dx = ∫ f(a-x)dx
Let $$I = \int_{0}^{\pi/2} \frac{\sin^2 x}{\sin x + \cos x} \, dx \quad \text{--- (1)}$$
Using $\int_0^a f(x)dx = \int_0^a f(a - x)dx$:
$$I = \int_{0}^{\pi/2} \frac{\sin^2(\frac{\pi}{2} - x)}{\sin(\frac{\pi}{2} - x) + \cos(\frac{\pi}{2} - x)} \, dx = \int_{0}^{\pi/2} \frac{\cos^2 x}{\cos x + \sin x} \, dx \quad \text{--- (2)}$$
Add (1) and (2)
$$2I = \int_{0}^{\pi/2} \frac{\sin^2 x + \cos^2 x}{\sin x + \cos x} \, dx = \int_{0}^{\pi/2} \frac{1}{\sin x + \cos x} \, dx$$
Convert denominator into single trigonometric term
$$\sin x + \cos x = \sqrt{2}\left(\frac{1}{\sqrt{2}}\cos x + \frac{1}{\sqrt{2}}\sin x\right) = \sqrt{2}\cos\left(x - \frac{\pi}{4}\right)$$
$$2I = \frac{1}{\sqrt{2}} \int_{0}^{\pi/2} \sec\left(x - \frac{\pi}{4}\right) dx$$
$$2I = \frac{1}{\sqrt{2}} \left[ \ln\left| \sec\left(x - \frac{\pi}{4}\right) + \tan\left(x - \frac{\pi}{4}\right) \right| \right]_{0}^{\pi/2}$$
At $x = \pi/2$: $\sec(\pi/4) + \tan(\pi/4) = \sqrt{2} + 1$.
At $x = 0$: $\sec(-\pi/4) + \tan(-\pi/4) = \sqrt{2} - 1$.
$$2I = \frac{1}{\sqrt{2}} \left[ \ln(\sqrt{2} + 1) - \ln(\sqrt{2} - 1) \right] = \frac{1}{\sqrt{2}} \ln\left(\frac{\sqrt{2} + 1}{\sqrt{2} - 1}\right)$$
Since $\frac{\sqrt{2} + 1}{\sqrt{2} - 1} = (\sqrt{2} + 1)^2$:
$$2I = \frac{1}{\sqrt{2}} \ln(\sqrt{2} + 1)^2 = \frac{2}{\sqrt{2}} \ln(\sqrt{2} + 1) \implies I = \frac{1}{\sqrt{2}}\ln(\sqrt{2} + 1)$$