Question

Evaluate \[ \int_{0}^{\pi} \left( \cos^2\left(\frac{3\pi}{8}-\frac{x}{4}\right) - \cos^2\left(\frac{11\pi}{8}+\frac{x}{4}\right) \right)\,dx \] is equal to:

• A. \(\dfrac{1}{\sqrt{2}}\)
• B. \(2\sqrt{2}\)
• C. \(\sqrt{2}\)
• D. \(2\)

Hint:

Use \[ \cos^2A-\cos^2B=-\sin(A+B)\sin(A-B) \] to simplify trigonometric definite integrals quickly.

Answer 1

The Correct Option is C

Step 1: Use the identity for difference of squares of cosine.
Let \[ A=\frac{3\pi}{8}-\frac{x}{4} \] and \[ B=\frac{11\pi}{8}+\frac{x}{4} \] We know that \[ \cos^2A-\cos^2B = -\sin(A+B)\sin(A-B) \]

Step 2: Find \(A+B\) and \(A-B\).
\[ A+B= \frac{3\pi}{8}-\frac{x}{4}+\frac{11\pi}{8}+\frac{x}{4} \] \[ A+B=\frac{14\pi}{8} \] \[ A+B=\frac{7\pi}{4} \] Also, \[ A-B= \frac{3\pi}{8}-\frac{x}{4}-\frac{11\pi}{8}-\frac{x}{4} \] \[ A-B=-\pi-\frac{x}{2} \]

Step 3: Simplify the integrand.
Now, \[ -\sin(A+B)\sin(A-B) = -\sin\left(\frac{7\pi}{4}\right)\sin\left(-\pi-\frac{x}{2}\right) \] Since \[ \sin\left(\frac{7\pi}{4}\right)=-\frac{1}{\sqrt{2}}, \] and \[ \sin\left(-\pi-\frac{x}{2}\right)=\sin\frac{x}{2}, \] we get \[ -\sin(A+B)\sin(A-B) = -\left(-\frac{1}{\sqrt{2}}\right)\sin\frac{x}{2} \] \[ = \frac{1}{\sqrt{2}}\sin\frac{x}{2} \]

Step 4: Evaluate the integral.
Therefore, \[ I= \int_{0}^{\pi}\frac{1}{\sqrt{2}}\sin\frac{x}{2}\,dx \] \[ I= \frac{1}{\sqrt{2}}\int_{0}^{\pi}\sin\frac{x}{2}\,dx \] Now, \[ \int \sin\frac{x}{2}\,dx=-2\cos\frac{x}{2} \] So, \[ I= \frac{1}{\sqrt{2}} \left[-2\cos\frac{x}{2}\right]_{0}^{\pi} \] \[ = \frac{1}{\sqrt{2}} \left[-2\cos\frac{\pi}{2}+2\cos0\right] \] \[ = \frac{1}{\sqrt{2}}(0+2) \] \[ = \frac{2}{\sqrt{2}} \] \[ = \sqrt{2} \]

Step 5: Final conclusion.
Therefore, \[ \boxed{\sqrt{2}} \]