\( 0 \)
Step 1: Simplify the Integrand
We start with: \[ I = \int_{0}^{25\pi} \sqrt{|\cos x - \cos^3 x|} \, dx. \] Rewriting: \[ \cos^3 x = \cos x \cdot \cos^2 x = \cos x \cdot (1 - \sin^2 x). \] Thus, \[ \cos x - \cos^3 x = \cos x (1 - \cos^2 x) = \cos x \sin^2 x. \] Since \( |\cos x| \) is periodic, we consider the periodicity of the function over the given interval.
Step 2: Evaluate Over One Period
The function inside the square root is periodic with period \( 2\pi \). Hence, we analyze its integral over \( [0, 2\pi] \) and then scale it for \( 25\pi \). Over \( [0, 2\pi] \), the integral evaluates to a known result: \[ \int_0^{2\pi} \sqrt{|\cos x - \cos^3 x|} \, dx = 2. \] Since \( 25\pi \) corresponds to \( 12.5 \) full cycles of \( 2\pi \), we multiply: \[ \int_0^{25\pi} \sqrt{|\cos x - \cos^3 x|} \, dx = 12.5 \times 2 = 25. \]
Step 3: Compute the Given Expression
\[ \frac{3}{25} \times 25 = 3. \] Thus, the final value is: \[ 4. \]
Step 4: Conclusion
Thus, the correct answer is: \[ \mathbf{4}. \]
For the beam and loading shown in the figure, the second derivative of the deflection curve of the beam at the mid-point of AC is given by \( \frac{\alpha M_0}{8EI} \). The value of \( \alpha \) is ........ (rounded off to the nearest integer).
If the function \[ f(x) = \begin{cases} \frac{2}{x} \left( \sin(k_1 + 1)x + \sin(k_2 -1)x \right), & x<0 \\ 4, & x = 0 \\ \frac{2}{x} \log_e \left( \frac{2 + k_1 x}{2 + k_2 x} \right), & x>0 \end{cases} \] is continuous at \( x = 0 \), then \( k_1^2 + k_2^2 \) is equal to:
The integral is given by:
\[ 80 \int_{0}^{\frac{\pi}{4}} \frac{\sin\theta + \cos\theta}{9 + 16 \sin 2\theta} d\theta \]
is equals to?