Question

\( \int_{\pi/4}^{3\pi/4} \frac{x}{1+\sin x} \, dx \) is equal to

• A. \( \pi(\sqrt{2}-1) \)
• B. \( \pi(\sqrt{2}+1) \)
• C. \( 2\pi(\sqrt{2}-1) \)
• D. \( 2\pi(\sqrt{2}+1) \)
• E. \( \frac{\pi}{\sqrt{2}+1} \)

Hint:

For integrals involving \(x\) in numerator, always try \(x \to a+b-x\) substitution to exploit symmetry.

Answer 1

The Correct Option is B

Concept: This is a definite integral where symmetry and substitution are extremely useful. A standard trick is to use the identity: \[ I = \int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx \] to simplify expressions involving \(x\).

Step 1: Let: \[ I = \int_{\pi/4}^{3\pi/4} \frac{x}{1+\sin x} dx \]

Step 2: Apply substitution: \[ x \to \pi - x \] Limits remain same: \[ I = \int_{\pi/4}^{3\pi/4} \frac{\pi - x}{1+\sin(\pi - x)} dx \]

Step 3: Use identity: \[ \sin(\pi - x) = \sin x \] So: \[ I = \int_{\pi/4}^{3\pi/4} \frac{\pi - x}{1+\sin x} dx \]

Step 4: Add both expressions: \[ 2I = \int_{\pi/4}^{3\pi/4} \frac{x + (\pi - x)}{1+\sin x} dx \] \[ 2I = \int_{\pi/4}^{3\pi/4} \frac{\pi}{1+\sin x} dx \]

Step 5: Take constant outside: \[ 2I = \pi \int_{\pi/4}^{3\pi/4} \frac{1}{1+\sin x} dx \]

Step 6: Simplify integrand: \[ \frac{1}{1+\sin x} = \frac{1-\sin x}{1-\sin^2 x} = \frac{1-\sin x}{\cos^2 x} \]

Step 7: Split: \[ = \frac{1}{\cos^2 x} - \frac{\sin x}{\cos^2 x} = \sec^2 x - \sec x \tan x \]

Step 8: Integrate: \[ \int \sec^2 x dx = \tan x \] \[ \int \sec x \tan x dx = \sec x \] So: \[ \int \frac{1}{1+\sin x} dx = \tan x - \sec x \]

Step 9: Apply limits: At \(x=3\pi/4\): \[ \tan = -1,\quad \sec = -\sqrt{2} \Rightarrow -1 - (-\sqrt{2}) = -1 + \sqrt{2} \] At \(x=\pi/4\): \[ \tan = 1,\quad \sec = \sqrt{2} \Rightarrow 1 - \sqrt{2} \] Difference: \[ (-1+\sqrt{2}) - (1-\sqrt{2}) = -2 + 2\sqrt{2} = 2(\sqrt{2}-1) \]

Step 10: Substitute: \[ 2I = \pi \cdot 2(\sqrt{2}-1) \]

Step 11: Final: \[ I = \pi(\sqrt{2}+1) \]