Question

Evaluate \[ \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{x\sin x}{1+\cos2x}\,dx \]

• A. \(\frac{\pi}{\sqrt2}\)
• B. \(-\frac{\pi}{\sqrt2}\)
• C. \(\sqrt2\pi\)
• D. \(-\sqrt2\pi\)

Hint:

For definite integrals involving \(x\) and symmetric limits, always test the transformation \(x\to a+b-x\).

Answer 1

The Correct Option is B

Concept: For definite integrals over symmetric intervals, property \[ \int_a^b f(x)\,dx=\int_a^b f(a+b-x)\,dx \] can simplify the expression significantly. Also use identity \[ 1+\cos2x=2\cos^2x \]

Step 1: Simplify denominator.
\[ I= \int_{\pi/4}^{3\pi/4} \frac{x\sin x}{2\cos^2x}\,dx \] \[ = \frac12 \int_{\pi/4}^{3\pi/4} x\tan x\sec x\,dx \]

Step 2: Apply property of definite integrals.
Using substitution \[ x=\pi-u \] Then \[ I= \frac12 \int_{\pi/4}^{3\pi/4} (\pi-x)(-\tan x)\sec x\,dx \] Adding both forms, \[ 2I = -\frac{\pi}{2} \int_{\pi/4}^{3\pi/4} \tan x\sec x\,dx \]

Step 3: Integrate.
Since \[ \int\tan x\sec xdx=\sec x \] Thus \[ 2I = -\frac{\pi}{2} [\sec x]_{\pi/4}^{3\pi/4} \] \[ = -\frac{\pi}{2} (-\sqrt2-\sqrt2) \] \[ = \pi\sqrt2 \] Hence \[ I= -\frac{\pi}{\sqrt2} \] Therefore \[ \boxed{-\frac{\pi}{\sqrt2}} \]