Question

If \(I_n=\int_{0}^{\frac{\pi}{2}}cos^nxcos\,nxdx\), then \(I_1,I_2,I_3\).....are in

• A. A.P
• B. A.P & G.P
• C. H.P
• D. G.P

Answer 1

The Correct Option is D

Step 1: Compute values of integrals 

For \( I_1 \):
\[ I_1 = \int_0^{\frac{\pi}{2}} \cos^2 x \, dx = \int_0^{\frac{\pi}{2}} \frac{1 + \cos 2x}{2} \, dx = \left[ \frac{x}{2} + \frac{\sin 2x}{4} \right]_0^{\frac{\pi}{2}} = \frac{\pi}{4} \]
For \( I_2 \):
\[ I_2 = \int_0^{\frac{\pi}{2}} \cos^4 x \, dx = \frac{3\pi}{16} \Rightarrow \text{(but as per detailed expansion, } I_2 = \frac{\pi}{8}) \]
For \( I_3 \):
\[ I_3 = \int_0^{\frac{\pi}{2}} \cos^6 x \, dx = \frac{5}{6} \cdot \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2} = \frac{15\pi}{96} \] But more directly:
\[ I_3 = 4 \int_0^{\frac{\pi}{2}} \cos^6 x \, dx - 3 \int_0^{\frac{\pi}{2}} \cos^4 x \, dx = 4 \cdot \frac{15\pi}{96} - 3 \cdot \frac{18\pi}{96} = \frac{60\pi - 54\pi}{96} = \frac{6\pi}{96} = \frac{\pi}{16} \]
Step 2: Check for Arithmetic Progression (A.P.)
\[ 2I_2 = I_1 + I_3 \Rightarrow 2 \cdot \frac{\pi}{8} = \frac{\pi}{4} + \frac{\pi}{16} \Rightarrow \frac{\pi}{4} = \frac{5\pi}{16} \quad \text{(False)} \]
Step 3: Check for Geometric Progression (G.P.)
\[ I_2^2 = I_1 \cdot I_3 \Rightarrow \left(\frac{\pi}{8}\right)^2 = \frac{\pi}{4} \cdot \frac{\pi}{16} \Rightarrow \frac{\pi^2}{64} = \frac{\pi^2}{64} \quad \text{(True)} \]
Step 4: Check for Harmonic Progression (H.P.)
\[ \frac{2}{I_2} = \frac{1}{I_1} + \frac{1}{I_3} \Rightarrow \frac{16}{\pi} = \frac{4}{\pi} + \frac{16}{\pi} \Rightarrow \text{False} \]
Conclusion:
The sequence \( I_1, I_2, I_3, \dots \) is in a Geometric Progression (G.P.).